解释这段代码data = xlrd.open_workbook(r'data/data.xlsx') table = data.sheet_by_name('data_en')

这段代码使用了`xlrd`库中的`open_workbook`函数打开了名为"data.xlsx"的Excel文件，并将其存储在`data`变量中。

接下来，使用`sheet_by_name`函数从`data`变量中选择名为"data_en"的工作表，并将其存储在`table`变量中。

因此，这段代码的作用是打开Excel文件"data.xlsx"中名为"data_en"的工作表，以便对其进行操作。

相关问题

将以下代码转化为matlab代码表示：import xlrd import sympy import numpy as np from scipy import linalg #%% queue = [ 0, 29, 17, 2, 1, 20, 19, 26, 18, 25, 14, 6, 11, 7, 15, 9, 8, 12, 27, 16, 10, 13, 5, 4, 3, 22, 28, 24, 23, 21, 0] def read_data_model(): data = xlrd.open_workbook("/Users/lzs/Downloads/2020szcupc/data/C2.xlsx") table = data.sheet_by_name("Sheet1") rowNum = table.nrows colNum = table.ncols consumes = [] for i in range(1, rowNum): # 忽略DC的消耗 if i == 1: pass else: consumes.append(0 if table.cell_value(i, 3) == '/' else table.cell_value(i, 3)) return consumes #%% 获得矩阵A def get_A_matrix(data): A = np.ones([29,29], dtype = float) diagonal = np.eye(29) for i in range(29): for j in range(29): A[i][j] = data['consumes'][j] / data['r'] A = A - diagonal return A #%% def get_b_maatrix(data): b = np.ones([29,1], dtype=float) for i in range(29): b[i][0] = -data['dst']*data['consumes'][i]/data['velocity']+data['f'] for j in range(29): b[i][0] = b[i][0] + data['f']*data['consumes'][i]/data['r'] return b #%% 数值解 def numerical(data): data['velocity'] = 50 data['dst'] = 11469 data['r'] = 200 data['f'] = 10 A = get_A_matrix(data) b = get_b_maatrix(data) x = linalg.solve(A, b) return x #%% 符号解决方案 def symbolic(data): data['velocity'] = sympy.symbols("v", integer = True) data['dst'] = 12100 data['r'] = sympy.symbols("r", integer = True) data['f'] = sympy.symbols("f", integer = True) # 获取矩阵A并转移到符号矩阵M A = np.ones([29,29], dtype = float).tolist() diagonal = np.eye(29).tolist() for i in range(29): for j in range(29): A[i][j] = data['consumes'][j] / data['r'] - diagonal[i][j] M = sympy.Matrix(A) # 得到矩阵b并转移到符号矩阵b b = np.ones([29,1], dtype=float).tolist() for i in range(29): b[i][0] = -data['dst']*data['consumes'][i]/data['velocity']+data['f'] for j in range(29): b[i][0] = b[i][0] + data['f']*data['consumes'][i]/data['r'] b = sympy.Matrix(b) # LU solver x = M.LUsolve(b) return x #%% 主功能 if name == 'main': data = {} data['consumes'] = read_data_model() options = {"numerical":1, "symbolic":2} option = 1 if option == options['numerical']: x = numerical(data) print(x) elif option == options['symbolic']: x = symbolic(data) print(x) else: print("WARN!!!")

以下是将给定代码转化为 Matlab 代码的结果：

%% 导入所需库
import matlab.io.*
import sympy.*
import numpy as np
from scipy import linalg

%% 读取数据模型
function consumes = read_data_model()
    data = xlsread('/Users/lzs/Downloads/2020szcupc/data/C2.xlsx');
    consumes = [];
    for i = 2:size(data,1)
        if i == 2
            continue % 忽略DC的消耗
        end
        consumes = [consumes, 0];
        if data(i,4) ~= '/'
            consumes(end) = data(i,4);
        end
    end
end

%% 获取矩阵A
function A = get_A_matrix(data)
    A = ones(29,29);
    diagonal = eye(29);
    for i = 1:29
        for j = 1:29
            A(i,j) = data.consumes(j) / data.r;
        end
        A(i,i) = A(i,i) - 1;
    end
end

%% 获取矩阵b
function b = get_b_maatrix(data)
    b = ones(29,1);
    for i = 1:29
        b(i) = -data.dst*data.consumes(i)/data.velocity + data.f;
        for j = 1:29
            b(i) = b(i) + data.f*data.consumes(j)/data.r;
        end
    end
end

%% 数值解
function x = numerical(data)
    data.velocity = 50;
    data.dst = 11469;
    data.r = 200;
    data.f = 10;
    A = get_A_matrix(data);
    b = get_b_maatrix(data);
    x = linalg.solve(A, b);
end

%% 符号解决方案
function x = symbolic(data)
    syms v r f
    data.velocity = v;
    data.dst = 12100;
    data.r = r;
    data.f = f;
    % 获取矩阵A并转移到符号矩阵M
    A = ones(29,29);
    diagonal = eye(29);
    for i = 1:29
        for j = 1:29
            A(i,j) = data.consumes(j) / data.r;
        end
        A(i,i) = A(i,i) - 1;
    end
    M = sym(A);
    % 得到矩阵b并转移到符号矩阵b
    b = ones(29,1);
    for i = 1:29
        b(i) = -data.dst*data.consumes(i)/data.velocity + data.f;
        for j = 1:29
            b(i) = b(i) + data.f*data.consumes(j)/data.r;
        end
    end
    b = sym(b);
    % LU求解
    x = M\LUsolve(b);
end

%% 主功能
data = struct();
data.consumes = read_data_model();
option = 1;
if option == 1
    x = numerical(data);
    disp(x);
elseif option == 2
    x = symbolic(data);
    disp(x);
else
    disp('WARN!!!');
end

需要注意的是，Matlab 中没有 `xlrd` 库，需要使用 `xlsread` 函数来读取 Excel 文件。另外，Matlab 中的 `eye` 函数可以直接得到单位矩阵，无需转化为列表再使用 `diag` 函数。

AttributeError: 'Worksheet' object has no attribute 'row_values'

根据你提供的引用和引用，你遇到的错误是`AttributeError: 'Worksheet' object has no attribute 'row_values'`。这个错误表明在你的代码中使用了`row_values`方法，但是Worksheet对象并没有该属性。

解决这个问题，你可以使用`sheet_by_index`或者`sheet_by_name`方法来获取Worksheet对象，然后使用`row_values`方法获取行的值。下面是一个示例代码：

import xlrd

data = xlrd.open_workbook('F:\\message\\test.xlsx')
table = data.sheet_by_name(u'Sheet1')

nrows = table.nrows

for i in range(nrows):
    row_values = table.row_values(i)
    print(row_values)

在这个示例中，我们首先使用`sheet_by_name`方法获取名为'Sheet1'的Worksheet对象。然后，我们使用`nrows`属性获取行数，然后使用`row_values`方法获取每一行的值并打印出来。

希望这个解决方案对你有帮助。如果你仍然遇到问题，请提供更多的代码和错误信息。<em>1</em><em>2</em><em>3</em>
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