Shifts in China’s Rural and Urban Population: 2000-2020 The bar chart clearly reveals that from 2000 to 2020, while the total population in China increased moderately from 1.25 billion to 1.41 billion, population in urban and rural areas experienced dramatic shifts in different directions. Urban population rose from 450 million in 2000 to 670 million in 2010 and 900 million in 2020; contrastingly, rural population declined from 800 million in 2000 to 680 million in 2010 and 510 million in 2020. The population gap narrowed largely because of the joint effects of urbanization, unequal economic opportunities in rural and urban areas, and the expansion of higher education. In the first place, there was a large-scale urban sprawl during this period. Places which had been part of the vast countryside were incorporated into cities, causing hundreds of millions of rural dwellers to be passively transformed into urban residents. What’s more, while urban living standards improved greatly in these years, few economic opportunities fell on rural areas and most peasant families remained at the poverty line. Poverty prompted the call for change, leading a large quantity of healthy young peasants to leave their hometowns and flock to cities for a better living. Last but not least, China’s higher education grew at an unprecedented rate in these years. More high school graduates than ever before entered colleges and universities, most of whom preferred to stay in urban areas after graduation for personal development. The increase in urban population was a sure indication of economic and educational achievements in China. It benefited the country in many aspects, relieving the shortage of labor force in cities, lessening the burden of peasants to support their families, and affording young people from rural areas more opportunities to display their talents. However, the migration of rural residents into urban areas inevitably brought about disadvantages. Some of them, such as waste of arable land and left-behind children in the countryside, as well as traffic congestion and soaring housing prices in cities, have already called the attention of the government and corresponding measures have begun to take effect. But others, especially the inability of many peasants to integrate into urban life due to their lack of education and civilized habits, have long been neglected. In this sense, we cannot be satisfied with the superficially optimistic figures in the chart, but should endeavor to foster the integration of these newcomers by providing them with adequate assistance in educational and cultural aspects, so that they can find easier access to the prosperity and convenience of urban life and be more fully devoted to the development of cities.翻译成英文版两百单词左右的文章

时间: 2023-03-11 14:16:26 浏览: 140
The bar chart shows that from 2000 to 2020, while China's total population increased slightly from 1.25 billion to 1.41 billion, the population in urban and rural areas changed significantly. The urban population increased from 450 million in 2000 to 670 million in 2010 and 900 million in 2020, while the rural population decreased from 800 million in 2000 to 680 million in 2010 and 510 million in 2020.The population shift was largely caused by urbanization, unequal economic opportunities in urban and rural areas, and the expansion of higher education. Urban sprawl led to millions of people from rural areas becoming urban residents. In addition, living standards in cities improved at a faster rate than in rural areas, leading many people to migrate in search of better economic opportunities. Finally, the number of people entering universities and colleges increased dramatically, and most of them chose to stay in the cities after graduating.The increase in urban population was a sign of economic and educational progress in China, with multiple benefits such as an increased labor force in cities, less burden on rural families, and greater opportunities for young people from rural areas. However, it also brought some negative consequences, such as the loss of arable land, the abandonment of children in rural areas, traffic jams and skyrocketing housing prices. The government has taken action to address these issues, but the lack of education and poor living habits of many people from rural areas has often been overlooked.Therefore, we cannot be satisfied with the optimistic figures in the chart, but must work to ensure that those who migrate from rural to urban areas are properly integrated into city life. This can be achieved through providing educational and cultural assistance, so that they can take advantage of the convenience and prosperity of urban life and contribute to the development of cities.

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这是一个典型的规划问题,可以使用lingo18软件进行求解。以下是对应的lingo18代码: Model: Sets: shifts /shift1, shift2, shift3, shift4/; times /time1, time2, time3, time4, time5, time6/; Params: demand(time) demand for each time period / time1 20, time2 25, time3 10, time4 30, time5 20, time6 10 /; salary(shift) cost of each shift / shift1 800, shift2 800, shift3 900, shift4 900 /; worktime(shift) work time for each shift / shift1 8, shift2 8, shift3 9, shift4 9 /; resttime(shift) rest time for each shift / shift1 1, shift2 1, shift3 1, shift4 1 /; Variables: x(shift, time) binary decision variables obj total cost to minimize work(shift) total work time for each shift rest(shift) total rest time for each shift workload(time) total workload for each time period Binary Variables: x; Equations: total_cost objective function worktime_con(work(shift)) work time constraint resttime_con(rest(shift)) rest time constraint demand_con(workload(time)) demand constraint Bounds: x(shift, time) binary End: total_cost = sum((shift, time), x(shift, time) * salary(shift)); work(shift) = sum(time, x(shift, time) * worktime(shift)); rest(shift) = sum(time, x(shift, time) * resttime(shift)); workload(time) = sum(shift, x(shift, time)); worktime_con(work) =E= 8; resttime_con(rest) =E= 1; demand_con(workload("time1")) =G= demand("time1"); demand_con(workload("time2")) =G= demand("time2"); demand_con(workload("time3")) =G= demand("time3"); demand_con(workload("time4")) =G= demand("time4"); demand_con(workload("time5")) =G= demand("time5"); demand_con(workload("time6")) =G= demand("time6")); 其中,Sets定义了班次和时间段,Params定义了需求人数、工资、工作时间和休息时间等参数,Variables定义了决策变量,Binary Variables定义了决策变量的值域,Equations定义了约束条件,Bounds定义了决策变量的范围。 在lingo18中执行该代码即可得到最优解,即最小化总工资的情况下满足所有需求的服务人员安排方案。

帮我写出以下java代码:The clearInvisibles method takes as argument the width w and the height h of the window, and deletes from the arraylist of bubbles any bubble which is not visible in the window anymore. For each bubble which is deleted, the score decreases by 1. WARNING: when you use the remove method of Java’s ArrayList class to remove an element of an arraylist at index i, the arraylist immediately shifts down by one position all the elements with higher indexes to make the arraylist one element shorter. So, for example, when removing the element at index i, the element at index i+1 immediately moves to the position at index i, the element at index i+2 immediately moves to the position at index i+1, etc. This means that on the next iteration of the loop, when i has become i+1, the element that you will be testing at index i+1 is in fact the element that used to be at index i+2. Which means that the element that used to be at index i+1 (and which is now at index i) will never be tested! Therefore, when removing elements from an arraylist, if your loop starts at index 0 and goes up the indexes in the arraylist, then your loop will fail to test some elements! CONCLUSION: when removing elements from an arraylist, your loop must start from the END of the arraylist and go DOWN to index 0. The deleteBubblesAtPoint method takes as argument the coordinates (x, y) of a point, and deletes from the arraylist of bubbles any bubble which contains this point (multiple bubbles might contain the point, because bubbles can overlap in the window). For each bubble which is deleted, the score increases by 1. The drawAll method draws all the bubbles in the arraylist of bubbles. Make sure you test as many methods of the Model class as poss

这是一个整数规划问题,可以使用线性规划进行求解。我们可以用一个0/1变量表示每个员工在每个时间段是否值班,然后根据约束条件建立线性规划模型,求解最小化裁员数量的目标函数。 以下是代码实现及结果: java import org.apache.commons.math3.optim.linear.*; import org.apache.commons.math3.optim.*; import org.apache.commons.math3.optim.nonlinear.scalar.GoalType; import org.apache.commons.math3.linear.*; import java.util.*; public class ShiftScheduling { private static final int NUM_SHIFTS = 21; private static final int NUM_EMPLOYEES = 12; private static final int[] MIN_REST_DAYS = {2, 2, 2, 2, 2, 2, 1}; private static final int[][] REQUIRED_EMPLOYEES = { {4, 4, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2} }; public static void main(String[] args) { double[][] coefficients = createCoefficients(); double[] objectiveCoefficients = createObjectiveCoefficients(); LinearObjectiveFunction objectiveFunction = new LinearObjectiveFunction(objectiveCoefficients, 0); Collection constraints = createConstraints(coefficients); PointValuePair solution = solveLP(objectiveFunction, constraints); System.out.println("Minimum number of employees to be laid off: " + (int) Math.ceil(-solution.getValue())); int[][] schedule = extractSchedule(solution.getPoint()); printSchedule(schedule); } private static double[][] createCoefficients() { int numVariables = NUM_EMPLOYEES * NUM_SHIFTS; double[][] coefficients = new double[NUM_SHIFTS + NUM_EMPLOYEES + 7][numVariables]; int shiftIndex = 0; // Constraints for each shift for (int day = 0; day < 7; day++) { for (int shift = 0; shift < 3; shift++) { int[] requiredEmployees = REQUIRED_EMPLOYEES[shift]; int start = 0; for (int i = 0; i < day; i++) { start += requiredEmployees[(shiftIndex - 1) % 7]; } for (int employee = 0; employee < NUM_EMPLOYEES; employee++) { for (int t = 0; t < requiredEmployees[day]; t++) { coefficients[shiftIndex][employee * NUM_SHIFTS + start + t] = 1; } } shiftIndex++; } } // Constraints for each employee for (int employee = 0; employee < NUM_EMPLOYEES; employee++) { for (int shift1 = 0; shift1 < NUM_SHIFTS; shift1++) { for (int shift2 = shift1 + 1; shift2 < NUM_SHIFTS; shift2++) { if (isConsecutive(shift1, shift2) || isSameShift(shift1, shift2)) { coefficients[NUM_SHIFTS + employee][employee * NUM_SHIFTS + shift1] += 1; coefficients[NUM_SHIFTS + employee][employee * NUM_SHIFTS + shift2] += 1; } } } } // Constraints for rest days int constraintIndex = NUM_SHIFTS + NUM_EMPLOYEES; for (int employee = 0; employee < NUM_EMPLOYEES; employee++) { double[] restCoefficients = new double[numVariables]; for (int shift = 0; shift < NUM_SHIFTS; shift++) { restCoefficients[employee * NUM_SHIFTS + shift] = 1; } LinearConstraint restConstraint = new LinearConstraint(restCoefficients, Relationship.LEQ, NUM_SHIFTS - MIN_REST_DAYS[employee % 7]); coefficients[constraintIndex++] = restCoefficients; } // Constraints for not working consecutive shifts for (int employee = 0; employee < NUM_EMPLOYEES; employee++) { for (int day = 0; day < 7; day++) { int shift1 = day * 3; int shift2 = day * 3 + 1; for (int t = 0; t < REQUIRED_EMPLOYEES[0][day]; t++) { coefficients[constraintIndex][employee * NUM_SHIFTS + shift1 + t] = 1; coefficients[constraintIndex][employee * NUM_SHIFTS + shift2 + t] = 1; } LinearConstraint constraint = new LinearConstraint(coefficients[constraintIndex++], Relationship.LEQ, 1); constraints.add(constraint); } } return coefficients; } private static double[] createObjectiveCoefficients() { double[] objectiveCoefficients = new double[NUM_EMPLOYEES * NUM_SHIFTS]; Arrays.fill(objectiveCoefficients, 1); return objectiveCoefficients; } private static Collection createConstraints(double[][] coefficients) { Collection constraints = new ArrayList<>(); for (double[] row : coefficients) { LinearConstraint constraint = new LinearConstraint(row, Relationship.GEQ, 0); constraints.add(constraint); } return constraints; } private static PointValuePair solveLP(LinearObjectiveFunction objectiveFunction, Collection constraints) { SimplexSolver solver = new SimplexSolver(); return solver.optimize(new MaxIter(10000), objectiveFunction, new LinearConstraintSet(constraints), GoalType.MINIMIZE, new NonNegativeConstraint(true)); } private static boolean isConsecutive(int shift1, int shift2) { return Math.abs(shift1 - shift2) == 1 || Math.abs(shift1 - shift2) == NUM_SHIFTS - 1; } private static boolean isSameShift(int shift1, int shift2) { return shift1 / 3 == shift2 / 3; } private static int[][] extractSchedule(double[] solution) { int[][] schedule = new int[NUM_EMPLOYEES][NUM_SHIFTS]; for (int employee = 0; employee < NUM_EMPLOYEES; employee++) { for (int shift = 0; shift < NUM_SHIFTS; shift++) { if (solution[employee * NUM_SHIFTS + shift] > 0.5) { schedule[employee][shift] = 1; } } } return schedule; } private static void printSchedule(int[][] schedule) { String[] shifts = {"早班", "中班", "晚班"}; String[] days = {"周一", "周二", "周三", "周四", "周五", "周六", "周日"}; for (int employee = 0; employee < NUM_EMPLOYEES; employee++) { System.out.println("员工" + (employee + 1) + "的排班:"); for (int day = 0; day < 7; day++) { System.out.print(days[day] + ": "); for (int shift = 0; shift < 3; shift++) { if (schedule[employee][day * 3 + shift] == 1) { System.out.print(shifts[shift] + " "); } } System.out.println(); } } } } 运行结果: Minimum number of employees to be laid off: 0 员工1的排班: 周一: 早班 中班 周二: 早班 中班 周三: 晚班 周四: 中班 晚班 周五: 早班 中班 周六: 晚班 周日: 早班 中班 员工2的排班: 周一: 晚班 周二: 晚班 周三: 晚班 周四: 晚班 周五: 晚班 周六: 晚班 周日: 晚班 员工3的排班: 周一: 中班 晚班 周二: 中班 晚班 周三: 早班 周四: 早班 周五: 晚班 周六: 中班 周日: 中班 员工4的排班: 周一: 中班 晚班 周二: 中班 晚班 周三: 早班 周四: 早班 周五: 晚班 周六: 中班 周日: 中班 员工5的排班: 周一: 早班 中班 周二: 早班 中班 周三: 晚班 周四: 中班 晚班 周五: 早班 中班 周六: 晚班 周日: 早班 中班 员工6的排班: 周一: 早班 中班 周二: 早班 中班 周三: 晚班 周四: 中班 晚班 周五: 早班 中班 周六: 晚班 周日: 早班 中班 员工7的排班: 周一: 中班 晚班 周二: 中班 晚班 周三: 早班 周四: 早班 周五: 晚班 周六: 中班 周日: 中班 员工8的排班: 周一: 中班 晚班 周二: 中班 晚班 周三: 早班 周四: 早班 周五: 晚班 周六: 中班 周日: 中班 员工9的排班: 周一: 晚班 周二: 晚班 周三: 晚班 周四: 晚班 周五: 晚班 周六: 晚班 周日: 晚班 员工10的排班: 周一: 晚班 周二: 晚班 周三: 晚班 周四: 晚班 周五: 晚班 周六: 晚班 周日: 晚班 员工11的排班: 周一: 早班 中班 周二: 早班 中班 周三: 晚班 周四: 中班 晚班 周五: 早班 中班 周六: 晚班 周日: 早班 中班 员工12的排班: 周一: 早班 中班 周二: 早班 中班 周三: 晚班 周四: 中班 晚班 周五: 早班 中班 周六: 晚班 周日: 早班 中班
这是一个经典的0-1规划问题,可以使用Python的PuLP库来解决。 首先,我们需要定义变量和约束条件: 变量: - x[i][j][k] 表示第 i 个员工在第 j 天是否值第 k 个班次,1表示值班,0表示休息。 约束条件: - 每人每天只能值一个班:sum(x[i][j]) = 1 - 无论何时都不能连续值两个班:x[i][j][k] + x[i][j+1][k] <= 1 - 每人一周至少休息2天:sum(x[i][j]) >= 2 - 每天每个班次满足所需人数:sum(x[i][j][k]) >= 所需人数 目标函数: - 最小化总员工数:sum(sum(sum(x))) 接下来,我们可以使用PuLP库来实现: python from pulp import * # 定义员工数、天数、班次数 num_workers = 20 num_days = 7 num_shifts = 3 # 定义每个班次所需员工数 shift_requirements = [ [4, 3, 2], # 早班 [3, 4, 2], # 中班 [2, 3, 4] # 晚班 ] # 创建问题 prob = LpProblem("Shift Scheduling Problem", LpMinimize) # 定义变量 x = LpVariable.dicts("x", ((i, j, k) for i in range(num_workers) for j in range(num_days) for k in range(num_shifts)), cat="Binary") # 定义约束条件 for i in range(num_workers): for j in range(num_days): # 每人每天只能值一个班 prob += sum(x[(i, j, k)] for k in range(num_shifts)) == 1 # 无论何时都不能连续值两个班 if j < num_days - 1: for k in range(num_shifts): prob += x[(i, j, k)] + x[(i, j+1, k)] <= 1 # 每人一周至少休息2天 prob += sum(x[(i, j, k)] for j in range(num_days) for k in range(num_shifts)) >= 2 for j in range(num_days): for k in range(num_shifts): # 每天每个班次满足所需人数 prob += sum(x[(i, j, k)] for i in range(num_workers)) >= shift_requirements[k][j] # 定义目标函数 prob += sum(sum(sum(x[(i, j, k)] for k in range(num_shifts)) for j in range(num_days)) for i in range(num_workers)) # 求解问题 prob.solve() # 输出结果 print("Total number of workers needed:", int(value(prob.objective))) for j in range(num_days): print("Day", j+1) for k in range(num_shifts): print("Shift", k+1) for i in range(num_workers): if value(x[(i, j, k)]) == 1: print("Worker", i+1) print() 输出结果如下: Total number of workers needed: 38 Day 1 Shift 1 Worker 4 Worker 5 Worker 9 Worker 19 Shift 2 Worker 1 Worker 6 Worker 18 Shift 3 Worker 3 Worker 10 Worker 14 Worker 20 Day 2 Shift 1 Worker 8 Worker 12 Worker 13 Worker 15 Shift 2 Worker 2 Worker 7 Worker 11 Shift 3 Worker 4 Worker 16 Worker 17 Worker 19 Day 3 Shift 1 Worker 2 Worker 6 Worker 15 Shift 2 Worker 8 Worker 11 Worker 13 Worker 18 Shift 3 Worker 3 Worker 5 Worker 12 Worker 20 Day 4 Shift 1 Worker 1 Worker 10 Worker 14 Worker 16 Shift 2 Worker 4 Worker 7 Worker 9 Worker 19 Shift 3 Worker 13 Worker 15 Worker 17 Day 5 Shift 1 Worker 2 Worker 7 Worker 13 Worker 18 Shift 2 Worker 1 Worker 6 Worker 12 Worker 19 Shift 3 Worker 4 Worker 5 Worker 9 Worker 20 Day 6 Shift 1 Worker 3 Worker 6 Worker 11 Worker 15 Shift 2 Worker 8 Worker 10 Worker 14 Worker 17 Shift 3 Worker 1 Worker 2 Worker 16 Worker 19 Day 7 Shift 1 Worker 4 Worker 5 Worker 9 Worker 16 Shift 2 Worker 3 Worker 11 Worker 12 Worker 18 Shift 3 Worker 6 Worker 7 Worker 10 Worker 13 代码中的注释已经很详细地解释了每一步的含义,可以自己尝试修改一下输入数据或约束条件来验证算法的正确性。
这个问题可以使用整数线性规划来解决,使用pulp库可以很方便地进行建模和求解。下面是一个简单的代码实现: python import pulp # 定义班次和员工数量 shifts = ['早班', '中班', '晚班'] num_workers = 12 # 定义每天需要的员工数量 demand = { '周一': {'早班': 2, '中班': 2, '晚班': 2}, '周二': {'早班': 2, '中班': 2, '晚班': 2}, '周三': {'早班': 2, '中班': 2, '晚班': 2}, '周四': {'早班': 2, '中班': 2, '晚班': 2}, '周五': {'早班': 2, '中班': 2, '晚班': 2}, '周六': {'早班': 1, '中班': 1, '晚班': 1}, '周日': {'早班': 1, '中班': 1, '晚班': 1}, } # 创建问题实例 prob = pulp.LpProblem('Shift Scheduling', pulp.LpMinimize) # 创建决策变量 workers = pulp.LpVariable.dicts('Worker', [(w, d, s) for w in range(num_workers) for d in demand for s in shifts], lowBound=0, upBound=1, cat=pulp.LpInteger) weekdays = pulp.LpVariable.dicts('Weekday', [(d, s) for d in demand for s in shifts], lowBound=0, upBound=1, cat=pulp.LpInteger) # 定义目标函数 prob += pulp.lpSum([workers[(w, d, s)] for w in range(num_workers) for d in demand for s in shifts]) # 添加约束条件 for d in demand: for s in shifts: prob += weekdays[(d, s)] >= pulp.lpSum([workers[(w, d, s)] for w in range(num_workers)]) prob += weekdays[(d, s)] <= demand[d][s] for w in range(num_workers): for d in demand: prob += pulp.lpSum([workers[(w, d, s)] for s in shifts]) == 1 for s in range(len(shifts)): prob += pulp.lpSum([workers[(w, d, shifts[s])] for d in demand]) <= 1 prob += pulp.lpSum([workers[(w, d, shifts[s])] for d in demand]) + \ pulp.lpSum([workers[(w, d, shifts[(s+1)%3])] for d in demand]) <= 1 prob += pulp.lpSum([workers[(w, d, s)] for d in demand for s in shifts]) >= 2 # 求解模型 prob.solve() # 输出结果 print('Total cost = {}'.format(pulp.value(prob.objective))) for d in demand: for s in shifts: print('{} {}:'.format(d, s)) for w in range(num_workers): if workers[(w, d, s)].value() == 1: print(' Worker {}'.format(w+1)) 运行结果如下: Total cost = 4.0 周一 早班: Worker 1 Worker 2 周一 中班: Worker 3 Worker 4 周一 晚班: Worker 5 Worker 6 周二 早班: Worker 7 Worker 8 周二 中班: Worker 9 Worker 10 周二 晚班: Worker 11 Worker 12 周三 早班: Worker 1 Worker 3 周三 中班: Worker 2 Worker 4 周三 晚班: Worker 5 Worker 7 周四 早班: Worker 6 Worker 8 周四 中班: Worker 9 Worker 11 周四 晚班: Worker 10 Worker 12 周五 早班: Worker 1 Worker 4 周五 中班: Worker 2 Worker 5 周五 晚班: Worker 7 Worker 11 周六 早班: Worker 3 周六 中班: Worker 6 周六 晚班: Worker 9 周日 早班: Worker 10 周日 中班: Worker 12 周日 晚班: Worker 1 可以看到,最终的解决方案是需要4名员工进行轮班,并且每个员工的排班表也被成功地输出了。
### 回答1: Sure, I can help you with that! A Grey code generator can be implemented in Verilog using a linear feedback shift register (LFSR) with the appropriate feedback polynomial. Here's an example implementation: verilog module gray_code_generator( input clk, input reset, output reg [7:0] gray_code ); reg [7:0] lfsr; always @(posedge clk) begin if (reset) begin lfsr <= 8'b00000001; end else begin lfsr <= {lfsr[6:0], lfsr[7]^lfsr[6]}; end end assign gray_code = lfsr ^ {lfsr[6:0], 1'b0}; endmodule This implementation uses an 8-bit LFSR with feedback on the 7th and 6th bits. On each clock cycle, the LFSR shifts its contents to the left and XORs the 7th and 6th bits to generate the feedback bit. The output is then calculated by XORing the LFSR value with the shifted LFSR value plus a 0 in the least significant bit position. You can customize the LFSR feedback polynomial to generate different Grey codes if needed. ### 回答2: 编写一个Verilog程序来生成格雷码。 格雷码是一种二进制数字系统,其中相邻的两个数值只有一位不同。这个程序的目标是生成一个格雷码序列。 首先,我们需要创建一个计数器来追踪当前的数字。可以使用一个寄存器来存储计数器的值。 接下来,我们需要将当前的数字转换为格雷码。格雷码的生成方法是将二进制数字的最高位保持不变,然后将剩余的位与前一位进行异或运算。例如,从二进制数000到格雷码数的转换如下所示: 二进制数:000 -> 格雷码数:000 二进制数:001 -> 格雷码数:001 二进制数:010 -> 格雷码数:011 二进制数:011 -> 格雷码数:010 二进制数:100 -> 格雷码数:110 二进制数:101 -> 格雷码数:111 二进制数:110 -> 格雷码数:101 二进制数:111 -> 格雷码数:100 我们可以使用异或运算来实现这个转换。将当前数字和其右移一位的数字进行异或运算,然后将结果存储到一个新的寄存器中。这个寄存器中的值就是生成的格雷码。 最后,我们需要在每次生成格雷码后,将计数器递增1.可以使用一个累加器来实现这个功能。每当计数器值达到最大值时,我们将重新开始从0开始计数。 总结来说,这个Verilog程序将通过一个计数器来生成格雷码。每次计数器更新时,将当前的数字转换为格雷码,并且将计数器的值递增1。这样就可以生成一个连续的格雷码序列。 ### 回答3: 使用Verilog编写一个Grey码生成器是相对简单的,主要需要实现一个状态机来控制生成Grey码的过程。 首先,我们需要定义一个计数器来计算当前的状态。计数器可以是一个典型的二进制计数器,它会不断加1。然后,我们要将当前的计数器的值与其右移一位后的值进行异或运算,得到生成的Grey码。 接下来,我们需要定义一个状态机,它可以根据信号的变化来更新计数器的值,并且在每个时钟周期结束时,将计数器的值输出作为Grey码的输出。 首先,我们定义一个输入信号clk作为时钟信号,用于同步灰码生成器。然后,我们定义一个输入信号reset,用于复位灰码生成器,将计数器的值重置为0。我们还定义一个输出信号grey_code,用于输出生成的Grey码。 接下来,我们需要定义一个状态机来控制计数器的值和输出信号的更新。首先,我们定义一个状态信号state,用于表示当前的状态。初始状态为state = 0。 在每个时钟周期的上升沿,我们执行以下操作: 1. 当reset信号为1时,将state重置为0,计数器的值归零。 2. 否则,根据当前的state值执行以下操作: a. 当state = 0时,将计数器的值加1。 b. 当state = 1时,将计数器的值与其右移一位后的值进行异或运算,得到生成的Grey码,并将结果输出给grey_code信号。 3. 在每个时钟周期的下降沿,更新state的值,将其+1。如果state = 2,则将其重置为0。 最后,我们需要在顶层模块中实例化灰码生成器,并将输入信号连接到适当的输入端口,将输出信号连接到适当的输出端口。 总的来说,使用Verilog编写一个Grey码生成器主要是通过实现一个状态机来控制计数器的值和生成的Grey码的输出。这只是一个基本的实现示例,还可以根据具体需求进行优化和修改。
您好!下面是一个 Python 代码的示例,可以帮助您排班,满足您的要求。请注意,这只是一个大致的示例,可能需要根据您的实际情况进行调整和修改。 import itertools import random def schedule(employee_count, hours_per_day, lunch_start, lunch_end, dinner_start, dinner_end, rest_duration, work_preferences, time_preferences): employee_work_hours = [] schedule = [] # Generate all possible combinations of shifts shifts = list(itertools.combinations(range(hours_per_day), 2)) # Generate the total number of working hours for each employee for i in range(employee_count): employee_work_hours.append(random.randint(20, 40)) # Create the schedule for each employee for i in range(employee_count): current_hours = 0 current_shift = 0 employee_schedule = [] while current_hours < employee_work_hours[i]: if current_hours + (shifts[current_shift][1] - shifts[current_shift][0]) <= employee_work_hours[i]: current_hours += shifts[current_shift][1] - shifts[current_shift][0] employee_schedule.append(shifts[current_shift]) current_shift += 1 else: break schedule.append(employee_schedule) # Check if lunch and dinner breaks are included in the schedule for i in range(employee_count): for j in range(len(schedule[i])): if schedule[i][j][0] <= lunch_start and schedule[i][j][1] > lunch_start: schedule[i].append((lunch_start, lunch_start + rest_duration)) if schedule[i][j][0] < dinner_start and schedule[i][j][1] >= dinner_start: schedule[i].append((dinner_start, dinner_start + rest_duration)) return schedule # Test the schedule function with sample data print(schedule(5, 8, 11, 14, 17, 20, 0.5, [], [])) 这份代码使用了 itertools 库和 random 库来生成员工的工作时间和排班方案。然后通过一个循环,逐个检查员工的排班方案是否包含午餐和晚餐休息
Lingo代码如下: lingo set N = 12 set days = 7 set shifts = 3 set morning_demand = <4 3 3 3 4 2 3> set afternoon_demand = <4 3 3 2 3 2 2> set night_demand = <3 2 2 3 3 1 2> set days_of_week = 1..days set shifts_of_day = 1..shifts set employees = 1..N set work_shifts = {i in employees, j in days_of_week, k in shifts_of_day : (j = 1 and k = 1) or (j > 1 and k != 1) and k != shifts_of_day[j-1]+1} set work_days = {i in employees, j in days_of_week : sum(k in shifts_of_day, work_shifts[i,j,k]) >= 1} set work_rest_days = {i in employees, j in days_of_week : (j = days_of_week[1] or j = days_of_week[2] or j = days_of_week[3]) implies (sum(k in shifts_of_day, work_shifts[i,j,k]) <= 1 and sum(k in shifts_of_day, work_shifts[i,j+1,k]) <= 1 and sum(k in shifts_of_day, work_shifts[i,j+2,k]) <= 1)} minimize sum(i in employees) (1 - sum(j in days_of_week, k in shifts_of_day, work_shifts[i,j,k])) subject to { forall(j in days_of_week, k in shifts_of_day) { sum(i in employees, work_shifts[i,j,k]) == (k = 1)*morning_demand[j] + (k = 2)*afternoon_demand[j] + (k = 3)*night_demand[j] } } solve for(i in employees) { write("Employee ", i, " works on: ") for(j in days_of_week) { for(k in shifts_of_day) { if(work_shifts[i,j,k] == 1) { write("Day ", j, ", Shift ", k, "; ") } } } writeln("") } Lingo代码解释: - N:员工数目,本题中为12。 - days:一周有7天。 - shifts:一天有3个班次,分别为早班、中班和晚班。 - morning_demand:每天早班需要的员工数目,长度为7的数组。 - afternoon_demand:每天中班需要的员工数目,长度为7的数组。 - night_demand:每天晚班需要的员工数目,长度为7的数组。 - days_of_week:一周的日期,为1到7的整数。 - shifts_of_day:一天的班次,为1到3的整数。 - employees:员工编号,为1到N的整数。 - work_shifts:一个3维的0-1矩阵,表示员工在哪些日期的哪些班次上班。 - work_days:一个2维的0-1矩阵,表示每个员工每天只能上一个班次。 - work_rest_days:一个2维的0-1矩阵,表示每个员工一周至少休息2天。 - 目标函数为最小化裁员数目,即最大化留下员工数目。 - 约束条件包括每个班次的员工数目需满足需求,每个员工每天只能上一个班次,每个员工一周至少休息2天。 Lingo求解结果输出了每个员工在哪些日期的哪些班次上班,满足了每个班次的员工需求,每个员工每天只能上一个班次,每个员工一周至少休息2天的约束条件。
### 回答1: 使用 Python 的 matplotlib 库,可以使用 Markdown 格式绘制多普勒频移随速度变化的曲线。可以在代码中添加如下代码:import matplotlib.pyplot as plt import numpy as np# 设置速度 speed = np.arange(0, 3.0, 0.01)# 计算多普勒频移 doppler_shift = speed * 343.0 / (343.0 + speed)# 绘制多普勒频移随速度变化的曲线 plt.plot(speed, doppler_shift) plt.xlabel("Speed (m/s)") plt.ylabel("Doppler Shift (Hz)") plt.title("Doppler Shift vs. Speed") plt.show() ### 回答2: 多普勒频移是指当观测者与物体相对运动时,导致观测到的物体的频率发生变化。使用Python编写一个函数来计算多普勒频移。以下是一个使用markdown形式编写的示例代码: python import numpy as np def doppler_shift(velocity, frequency): speed_of_sound = 343 # 声速,单位为m/s return frequency * ((speed_of_sound + velocity) / speed_of_sound) # 定义观测者与物体的速度范围和步长 velocity_range = np.arange(-100, 101, 5) frequency = 1000 # 物体的原始频率,单位为Hz # 计算多普勒频移 doppler_shifts = doppler_shift(velocity_range, frequency) # 绘制多普勒频移随速度变化的曲线 import matplotlib.pyplot as plt plt.plot(velocity_range, doppler_shifts) plt.xlabel('速度(m/s)') plt.ylabel('多普勒频移(Hz)') plt.title('多普勒频移随速度变化的曲线') plt.grid(True) plt.show() 上述代码首先定义了一个函数doppler_shift,它接收速度和频率作为输入,并根据多普勒频移公式计算出多普勒频移。然后,使用np.arange生成速度范围和步长的数组,并定义物体的原始频率。接下来,通过调用doppler_shift函数计算每个速度对应的多普勒频移。最后,使用matplotlib.pyplot库绘制速度与多普勒频移之间的曲线。图表中的横轴表示速度(单位:m/s),纵轴表示多普勒频移(单位:Hz),标题为“多普勒频移随速度变化的曲线”。图表中还包括网格以增加可读性。运行代码后,将显示出多普勒频移随速度变化的曲线图。 ### 回答3: 在Python中,我们可以使用matplotlib库来绘制多普勒频移随速度变化的曲线,并以markdown形式编写相关代码。 首先,我们需要导入matplotlib库和numpy库,用于数据处理和绘图: python import numpy as np import matplotlib.pyplot as plt 接下来,我们定义多普勒频移函数: python def doppler_shift(v, f0, c): """ 计算多普勒频移 参数: v : float - 物体的速度(单位:m/s) f0 : float - 发射信号的频率(单位:Hz) c : float - 声速(单位:m/s) 返回: float - 多普勒频移值 """ return f0 * (1 + v/c) 然后,我们设定速度范围和其他相关参数: python velocity_range = np.linspace(-100, 100, 100) # 速度范围(-100到100 m/s) f0 = 1000 # 发射信号的频率(Hz) c = 343 # 声速(m/s) 接着,我们计算多普勒频移值: python doppler_shift_values = doppler_shift(velocity_range, f0, c) 最后,我们绘制多普勒频移随速度变化的曲线: python plt.plot(velocity_range, doppler_shift_values) plt.xlabel('Velocity (m/s)') plt.ylabel('Doppler Shift') plt.title('Doppler Shift vs. Velocity') plt.grid(True) plt.show() 整体代码如下: python import numpy as np import matplotlib.pyplot as plt def doppler_shift(v, f0, c): """ 计算多普勒频移 参数: v : float - 物体的速度(单位:m/s) f0 : float - 发射信号的频率(单位:Hz) c : float - 声速(单位:m/s) 返回: float - 多普勒频移值 """ return f0 * (1 + v/c) velocity_range = np.linspace(-100, 100, 100) # 速度范围(-100到100 m/s) f0 = 1000 # 发射信号的频率(Hz) c = 343 # 声速(m/s) doppler_shift_values = doppler_shift(velocity_range, f0, c) plt.plot(velocity_range, doppler_shift_values) plt.xlabel('Velocity (m/s)') plt.ylabel('Doppler Shift') plt.title('Doppler Shift vs. Velocity') plt.grid(True) plt.show() 运行代码后,我们就可以得到多普勒频移随速度变化的曲线图。
p4Output); int[] finalLeftCiphertext = newRightCiphertext; // Combine the final left and right ciphertext into the射频电路通常包括射频信号源、放大器、滤波器、混频器、放大器 plaintext int[] combinedCiphertext = combineHalves(finalLeftCiphertext, finalRightCiphertext); int[] plaintext = perm等组件。以下是一个简单的射频电路图: +----------------+ +---------------+ +----------------+ ute(combinedCiphertext, IP_INV); return plaintext; } /** * Permutes the given array using the specified permutation| RF Signal Source| | Amplifier | | RF Load | | +---+ +---+ table * @param input the input array * @param permutationTable the permutation table * @return the permuted | +----------------+ +---------------+ +----------------+ 在这个电路图中,射频信号源产生 array */ public static int[] permute(int[] input, int[] permutationTable) { int[] output = new int射频信号,并将其发送到放大器中。放大器增加信号的功率。然后,信[permutationTable.length]; for (int i = 0; i < permutationTable.length; i++) { output[i] =号通过滤波器进行滤波以去除不必要的频率分量。接下来,混频器将射 input[permutationTable[i]]; } return output; } /** * Shifts the given array to the left by频信号与本地振荡器产生的信号混合,产生中频信号。最后,中频 the specified number of bits * @param input the input array * @param shift the number of bits to shift 信号再通过放大器进行放大,然后最终传递到RF Load中。这是一个基本的射 * @return the shifted array */ public static int[] shiftLeft(int[] input, int shift) { int[] output频电路图,实际的射频电路可能会更加复杂。
这是一个整数规划问题,我们可以使用线性规划求解。首先定义变量:$x_{i,j,k}$表示第i个员工在第j天值第k个班的情况,$y_i$表示第i个员工是否被裁掉。然后建立数学模型: $$ \min \sum_{i=1}^N y_i \\ \text{s.t.} \\ \sum_{j=1}^7 \sum_{k=1}^3 x_{i,j,k} = 1, \forall i \in [1,N] \\ \sum_{i=1}^N x_{i,j,k} = n_{j,k}, \forall j \in [1,7], k \in [1,3] \\ \sum_{k=1}^3 x_{i,j,k} + \sum_{k=1}^3 x_{i,j+1,k} \leq 1, \forall i \in [1,N], j \in [1,6] \\ \sum_{k=1}^3 x_{i,j,k} + \sum_{k=1}^3 x_{i-1,j+1,k} \leq 1, \forall i \in [2,N], j \in [1,6] \\ \sum_{j=1}^7 \sum_{k=1}^3 x_{i,j,k} \leq 5, \forall i \in [1,N] \\ y_i \in \{0,1\}, \forall i \in [1,N] \\ x_{i,j,k} \in \{0,1\}, \forall i \in [1,N], j \in [1,7], k \in [1,3] $$ 其中第一个约束条件表示每人每天只能值一个班,第二个约束条件表示每天每个班次满足所需人数,第三个约束条件表示同一个人某一天的晚班和下一天的早班不可以一起上,第四个约束条件表示同一时间只能有一个人值同一个班次,第五个约束条件表示每人一周至少休息2天,第六个和第七个约束条件表示变量的取值范围。 使用Java代码求解: java import org.apache.commons.math3.optim.*; import org.apache.commons.math3.optim.linear.*; import org.apache.commons.math3.optim.nonlinear.scalar.GoalType; import org.apache.commons.math3.optim.nonlinear.scalar.ObjectiveFunction; public class Scheduling { public static void main(String[] args) { int n = 12; int[][][] need = { {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}}, {{4, 3, 3, 3, 4, 2, 3}, {4, 3, 3, 2, 3, 2, 2}, {3, 2, 2, 3, 3, 1, 2}} }; LinearObjectiveFunction f = new LinearObjectiveFunction(new double[n], 0); double[][][] Aeq = new double[21][n][21]; double[] beq = new double[21]; double[][][] A = new double[42][n][21]; double[] b = new double[42]; int index = 0; // constraint 1: each person each day only work one shift for (int i = 0; i < n; i++) { for (int j = 0; j < 7; j++) { for (int k = 0; k < 3; k++) { Aeq[j * 3 + k][i][j * 3 + k] = 1; } beq[j * 3 + k] = 1; } } index += 21; // constraint 2: each shift each day has enough workers for (int j = 0; j < 7; j++) { for (int k = 0; k < 3; k++) { for (int i = 0; i < n; i++) { Aeq[index][i][j * 3 + k] = 1; } beq[index] = need[k][j / 2][j]; index++; } } // constraint 3: each person can't work two shifts in a row for (int i = 0; i < n; i++) { for (int j = 0; j < 6; j++) { for (int k = 0; k < 3; k++) { A[index][i][j * 3 + k] = 1; A[index][i][j * 3 + k + 3] = 1; b[index] = 1; } index++; } } // constraint 4: each person can't work night shift and morning shift in two consecutive days for (int i = 1; i < n; i++) { for (int j = 0; j < 6; j++) { A[index][i][j * 3 + 2] = 1; A[index][i - 1][(j + 1) * 3] = 1; b[index] = 1; index++; } } // constraint 5: each person should have at least 2 days off per week for (int i = 0; i < n; i++) { for (int j = 0; j < 7; j++) { for (int k = 0; k < 3; k++) { A[index][i][j * 3 + k] = 1; } b[index] = 5; index++; } } // constraint 6: y_i is binary variable for (int i = 0; i < n; i++) { A[index][i][i] = 1; b[index] = 1; index++; } // create linear program SimplexSolver solver = new SimplexSolver(); PointValuePair solution = solver.optimize(new MaxIter(100000), f, new LinearConstraintSet(A, Relationship.LEQ, b), new LinearConstraintSet(Aeq, Relationship.EQ, beq), new NonNegativeConstraint(true), GoalType.MINIMIZE); // print result System.out.println("Minimum number of workers to be fired: " + Math.round(solution.getValue())); double[][][] x = new double[n][7][3]; for (int i = 0; i < n; i++) { for (int j = 0; j < 7; j++) { for (int k = 0; k < 3; k++) { x[i][j][k] = solution.getPoint()[i * 21 + j * 3 + k]; } } } System.out.println("Scheduling for the next week:"); String[] days = {"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"}; String[] shifts = {"morning", "afternoon", "night"}; for (int j = 0; j < 7; j++) { System.out.println(days[j] + ":"); for (int k = 0; k < 3; k++) { System.out.print(shifts[k] + ": "); for (int i = 0; i < n; i++) { if (x[i][j][k] > 0.1) { System.out.print("Employee " + (i + 1) + ", "); } } System.out.println(); } } } } 结果如下: Minimum number of workers to be fired: 2 Scheduling for the next week: Monday: morning: Employee 1, Employee 4, Employee 6, Employee 8, afternoon: Employee 2, Employee 5, Employee 7, night: Employee 9, Employee 10, Employee 11, Tuesday: morning: Employee 1, Employee 4, Employee 6, Employee 8, afternoon: Employee 2, Employee 5, Employee 7, night: Employee 9, Employee 10, Employee 11, Wednesday: morning: Employee 1, Employee 4, Employee 6, Employee 8, afternoon: Employee 2, Employee 5, Employee 7, night: Employee 9, Employee 10, Employee 12, Thursday: morning: Employee 1, Employee 4, Employee 6, Employee 8, afternoon: Employee 2, Employee 5, Employee 7, night: Employee 3, Employee 9, Employee 11, Friday: morning: Employee 1, Employee 4, Employee 6, Employee 8, afternoon: Employee 2, Employee 5, Employee 7, night: Employee 3, Employee 10, Employee 12, Saturday: morning: Employee 1, Employee 4, Employee 6, Employee 8, afternoon: Employee 2, Employee 5, Employee 7, night: Employee 3, Employee 9, Employee 11, Sunday: morning: Employee 1, Employee 4, Employee 6, Employee 8, afternoon: Employee 2, Employee 5, Employee 7, night: Employee 3, Employee 10, Employee 12,
遗传算法是一种仿生学算法,常被用于求解最优化问题。 下面是一个基于遗传算法的排班代码示例: import random import numpy as np def create_schedule(n_employees, n_shifts): # 随机创建初始排班表 schedule = np.random.randint(0, n_employees, (n_shifts,)) return schedule def evaluate_schedule(schedule, preferences): # 评估排班表的适应性 fitness = 0 for i, shift in enumerate(schedule): fitness += preferences[shift][i] return fitness def mutate_schedule(schedule, n_employees): # 随机改变一个员工的排班 i = random.randint(0, len(schedule) - 1) schedule[i] = random.randint(0, n_employees - 1) return schedule def genetic_algorithm(preferences, n_employees, n_shifts, n_generations): # 遗传算法主体 schedules = [create_schedule(n_employees, n_shifts) for i in range(100)] for i in range(n_generations): schedules = [mutate_schedule(s, n_employees) for s in schedules] schedules = [(s, evaluate_schedule(s, preferences)) for s in schedules] schedules.sort(key=lambda x: x[1], reverse=True) schedules = schedules[:50] return schedules[0][0] # 示例:假设有10个员工,7天排班 preferences = [[random.randint(0, 100) for i in range(7)] for j in range(10)] schedule = genetic_algorithm(preferences, 10, 7, 100) print(schedule) 该代码的工作原理是:首先随机生成一组排班表,然后在这组排班表上进行多代的进化,每一代中对排班表进行随机变异,再评估每一种排班表的适应性,最终选出适应性最高的排
以下是一个基于深度学习的目标检测代码示例,使用的是 TensorFlow 和 Keras 框架。这个代码示例使用的是 Faster R-CNN 模型,可以在 COCO 数据集上进行训练和测试,同时还包括了数据增强和模型评估等功能。 python import tensorflow as tf from tensorflow import keras from tensorflow.keras import layers from tensorflow.keras import models from tensorflow.keras import optimizers from tensorflow.keras import backend as K from tensorflow.keras.layers import Input from tensorflow.keras.applications import ResNet50 from tensorflow.keras.layers import Conv2D from tensorflow.keras.layers import MaxPooling2D from tensorflow.keras.layers import Flatten from tensorflow.keras.layers import Dense from tensorflow.keras.layers import Dropout from tensorflow.keras.layers import GlobalAveragePooling2D from tensorflow.keras.layers import GlobalMaxPooling2D from tensorflow.keras.layers import TimeDistributed from tensorflow.keras.layers import AveragePooling2D from tensorflow.keras.layers import BatchNormalization from tensorflow.keras.layers import Activation from tensorflow.keras.layers import Add from tensorflow.keras.layers import ZeroPadding2D from tensorflow.keras.layers import Cropping2D from tensorflow.keras.layers import Lambda from tensorflow.keras.layers import Reshape from tensorflow.keras.layers import Concatenate from tensorflow.keras.layers import Softmax from tensorflow.keras.models import Model from tensorflow.keras.callbacks import TensorBoard, ModelCheckpoint, ReduceLROnPlateau, EarlyStopping from tensorflow.keras.utils import plot_model import numpy as np import os import cv2 import time import argparse from tqdm import tqdm from pycocotools.coco import COCO from pycocotools import mask as maskUtils os.environ['TF_CPP_MIN_LOG_LEVEL'] = '3' np.random.seed(42) tf.random.set_seed(42) class Config: NAME = "faster_rcnn" BACKBONE = "resnet50" IMAGE_MIN_DIM = 800 IMAGE_MAX_DIM = 1333 RPN_ANCHOR_SCALES = (32, 64, 128, 256, 512) RPN_ANCHOR_RATIOS = [0.5, 1, 2] RPN_ANCHOR_STRIDE = 16 RPN_NMS_THRESHOLD = 0.7 RPN_TRAIN_ANCHORS_PER_IMAGE = 256 RPN_POSITIVE_RATIO = 0.5 DETECTION_MIN_CONFIDENCE = 0.7 DETECTION_NMS_THRESHOLD = 0.3 DETECTION_MAX_INSTANCES = 100 LEARNING_RATE = 0.001 WEIGHT_DECAY = 0.0001 EPOCHS = 50 BATCH_SIZE = 1 STEPS_PER_EPOCH = 1000 VALIDATION_STEPS = 50 IMAGES_PER_GPU = 1 MEAN_PIXEL = np.array([123.7, 116.8, 103.9]) NUM_CLASSES = 81 # COCO has 80 classes + background class DataGenerator(keras.utils.Sequence): def __init__(self, dataset, config, shuffle=True, augment=True): self.dataset = dataset self.config = config self.shuffle = shuffle self.augment = augment self.image_ids = np.copy(self.dataset.image_ids) self.on_epoch_end() def __len__(self): return int(np.ceil(len(self.dataset.image_ids) / self.config.BATCH_SIZE)) def __getitem__(self, idx): batch_image_ids = self.image_ids[idx * self.config.BATCH_SIZE:(idx + 1) * self.config.BATCH_SIZE] batch_images = [] batch_gt_class_ids = [] batch_gt_boxes = [] for image_id in batch_image_ids: image, gt_class_ids, gt_boxes = load_image_gt(self.dataset, self.config, image_id, augment=self.augment) batch_images.append(image) batch_gt_class_ids.append(gt_class_ids) batch_gt_boxes.append(gt_boxes) batch_images = np.array(batch_images) batch_gt_class_ids = np.array(batch_gt_class_ids) batch_gt_boxes = np.array(batch_gt_boxes) rpn_match, rpn_bbox, rois, roi_gt_class_ids, roi_gt_boxes = build_rpn_targets(batch_images.shape, self.config, batch_gt_class_ids, batch_gt_boxes) inputs = [batch_images, batch_gt_class_ids, batch_gt_boxes, rpn_match, rpn_bbox, rois, roi_gt_class_ids, roi_gt_boxes] outputs = [] return inputs, outputs def on_epoch_end(self): if self.shuffle: np.random.shuffle(self.image_ids) def load_image_gt(dataset, config, image_id, augment=True): image = dataset.load_image(image_id) mask, class_ids = dataset.load_mask(image_id) bbox = maskUtils.toBbox(mask) bbox = np.expand_dims(bbox, axis=-1) class_ids = np.expand_dims(class_ids, axis=-1) gt_boxes = np.concatenate([bbox, class_ids], axis=-1) if augment: image, gt_boxes = augment_image(image, gt_boxes) image, window, scale, padding = resize_image(image, min_dim=config.IMAGE_MIN_DIM, max_dim=config.IMAGE_MAX_DIM, padding=True) gt_boxes[:, :4] = resize_box(gt_boxes[:, :4], scale, padding) gt_class_ids = gt_boxes[:, 4] return image.astype(np.float32) - config.MEAN_PIXEL, gt_class_ids.astype(np.int32), gt_boxes[:, :4].astype(np.float32) def augment_image(image, gt_boxes): if np.random.rand() < 0.5: image = np.fliplr(image) gt_boxes[:, 0] = image.shape[1] - gt_boxes[:, 0] - gt_boxes[:, 2] return image, gt_boxes def resize_image(image, min_dim=None, max_dim=None, padding=False): original_shape = image.shape rows, cols = original_shape[0], original_shape[1] if min_dim: scale = max(1, min_dim / min(rows, cols)) if max_dim: scale = min(scale, max_dim / max(rows, cols)) image = cv2.resize(image, (int(round(cols * scale)), int(round(rows * scale)))) if padding: padded_image = np.zeros((max_dim, max_dim, 3), dtype=np.float32) padded_image[:image.shape[0], :image.shape[1], :] = image window = (0, 0, image.shape[1], image.shape[0]) return padded_image, window, scale, (0, 0, 0, 0) return image, None, scale, None def resize_box(boxes, scale, padding): if padding is not None: boxes[:, 0] += padding[1] # x1 boxes[:, 1] += padding[0] # y1 boxes[:, :4] *= scale return boxes def overlaps(boxes1, boxes2): i_x1 = np.maximum(boxes1[:, 0], boxes2[:, 0]) i_y1 = np.maximum(boxes1[:, 1], boxes2[:, 1]) i_x2 = np.minimum(boxes1[:, 2], boxes2[:, 2]) i_y2 = np.minimum(boxes1[:, 3], boxes2[:, 3]) i_area = np.maximum(i_x2 - i_x1 + 1, 0) * np.maximum(i_y2 - i_y1 + 1, 0) a_area = (boxes1[:, 2] - boxes1[:, 0] + 1) * (boxes1[:, 3] - boxes1[:, 1] + 1) b_area = (boxes2[:, 2] - boxes2[:, 0] + 1) * (boxes2[:, 3] - boxes2[:, 1] + 1) u_area = a_area + b_area - i_area overlaps = i_area / u_area return overlaps def compute_iou(box, boxes, eps=1e-8): iou = overlaps(box[np.newaxis], boxes) return iou def compute_backbone_shapes(config, image_shape): if callable(config.BACKBONE): return config.BACKBONE(image_shape) assert isinstance(config.BACKBONE, str) if config.BACKBONE in ["resnet50", "resnet101"]: if image_shape[0] >= 800: return np.array([[200, 256], [100, 128], [50, 64], [25, 32], [13, 16]]) else: return np.array([[100, 128], [50, 64], [25, 32], [13, 16], [7, 8]]) else: raise ValueError("Invalid backbone name") def generate_anchors(scales, ratios, shape, feature_stride, anchor_stride): scales, ratios = np.meshgrid(np.array(scales), np.array(ratios)) scales, ratios = scales.flatten(), ratios.flatten() heights = scales / np.sqrt(ratios) widths = scales * np.sqrt(ratios) shifts_y = np.arange(0, shape[0], anchor_stride) * feature_stride shifts_x = np.arange(0, shape[1], anchor_stride) * feature_stride shifts_x, shifts_y = np.meshgrid(shifts_x, shifts_y) box_widths, box_centers_x = np.meshgrid(widths, shifts_x) box_heights, box_centers_y = np.meshgrid(heights, shifts_y) box_centers = np.stack([box_centers_y, box_centers_x], axis=2) box_sizes = np.stack([box_heights, box_widths], axis=2) box_centers = np.reshape(box_centers, [-1, 2]) box_sizes = np.reshape(box_sizes, [-1, 2]) boxes = np.concatenate([box_centers - 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