t = 0 for i in new_arr: if i[0] == '0': continue elif i[0] == '6': continue else: new_arr[t] = 0 print(i) t += 1
时间: 2024-02-14 12:20:09 浏览: 22
这段代码的作用是遍历一个名为`new_arr`的列表中的子列表,如果子列表的第一个元素是'0'或'6',则跳过该子列表,否则将其在`new_arr`列表中的位置(即索引)上的值设为0,并打印出该子列表。
其中`t`是一个计数器,用于记录遍历到`new_arr`列表的第几个子列表。在遍历时,如果当前子列表的第一个元素为'0'或'6',则使用`continue`语句跳过这个子列表;否则,将该子列表在`new_arr`列表中的位置的值设为0,并打印出该子列表。
需要注意的是,这段代码将`new_arr`列表中符合条件的子列表的值设为了0,并没有删除它们。如果需要删除这些子列表,可以使用列表的`remove()`方法。
相关问题
for index1, itme1 in enumerate(list_col): for index, itme in enumerate(list_row): if itme1 == '总计': count = count_conditions( teacher_arr, lambda row: row[9] == '学前' or row[9] == '小学', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '女': count = count_conditions( teacher_arr, lambda row: row[2] == '女', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '少数民族': count = count_conditions( teacher_arr, lambda row: row[11] != '汉族', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '在编人员': count = count_conditions( teacher_arr, lambda row: row[12] == '是', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count代码优化
对于提供的代码,以下是一些可能的优化:
1. 将 lambda 表达式的使用替换为一个独立的函数。这将更容易读取和理解代码,并可以避免使用复杂的 lambda 表达式。
2. 循环嵌套可以考虑优化。例如,可以使用列表理解式或其他Python内置函数来简化代码。
3. 考虑更好的变量命名。例如,将 `itme1` 改为 `col_name` 可以更好地描述变量的用途,从而提高代码可读性。
4. 使用参数进行函数调用,以避免代码重复。例如,`count_conditions` 函数可以使用参数,而不是多次调用相同的lambda表达式。
5. 将常量值放入变量中进行引用。例如,将 `"学前"` 和 `"小学"` 放入变量中以进行引用,而不是在lambda表达式中硬编码值。
综上所述,代码可以进行以下优化:
```
def count_total_by_grade(teacher_arr, col_name, col_value):
count = count_conditions(teacher_arr,
lambda row: row[9] == '学前' or row[9] == '小学',
lambda row: row[8] == col_value)
if count != 0:
sh_4149.cell(list_col.index(col_name), list_row.index(col_value)).value = count
def count_female_by_grade(teacher_arr, col_name, col_value):
count = count_conditions(teacher_arr,
lambda row: row[2] == '女',
lambda row: row[8] == col_value)
if count != 0:
sh_4149.cell(list_col.index(col_name), list_row.index(col_value)).value = count
def count_non_han_by_grade(teacher_arr, col_name, col_value):
count = count_conditions(teacher_arr,
lambda row: row[11] != '汉族',
lambda row: row[8] == col_value)
if count != 0:
sh_4149.cell(list_col.index(col_name), list_row.index(col_value)).value = count
def count_in_teacher_by_grade(teacher_arr, col_name, col_value):
count = count_conditions(teacher_arr,
lambda row: row[12] == '是',
lambda row: row[8] == col_value)
if count != 0:
sh_4149.cell(list_col.index(col_name), list_row.index(col_value)).value = count)
for (col_index, col_name) in enumerate(list_col):
for (row_index, row_value) in enumerate(list_row):
if col_name == '总计':
count_total_by_grade(teacher_arr, col_name, row_value)
elif col_name == '女':
count_female_by_grade(teacher_arr, col_name, row_value)
elif col_name == '少数民族':
count_non_han_by_grade(teacher_arr, col_name, row_value)
elif col_name == '在编人员':
count_in_teacher_by_grade(teacher_arr, col_name, row_value)
```
注:此处为未测试代码,若有错误或需要改进的地方,请指出。
import collections def solve_method(n: int) -> None: asc = True list_ = collections.deque() x = 1 for i in range(1, n + 1): arr = [0] * i if asc: for j in range(i): arr[j] = x x += 1 else: for j in range(i - 1, -1, -1): arr[j] = x x += 1 list_.appendleft(arr) asc = not asc res = [] head = "" for ints in list_: content = head for j in range(len(ints)): num = ints[j] content += str(num) content += "*" * (4 - len(str(num))) if j != len(ints) - 1: content += " " res.append(content) head += " " for s in reversed(res): print(s) if __name__ == "__main__": n = int(input()) solve_method(n) 改写以上代码,python
import collections
def solve_method(n):
asc = True
list_ = collections.deque()
x = 1
for i in range(1, n + 1):
arr = [0] * i
if asc:
for j in range(i):
arr[j] = x
x += 1
else:
for j in range(i - 1, -1, -1):
arr[j] = x
x += 1
list_.appendleft(arr)
asc = not asc
res = []
head = ""
for ints in list_:
content = head
for j in range(len(ints)):
num = ints[j]
content += str(num)
content += "*" * (4 - len(str(num)))
if j != len(ints) - 1:
content += " "
res.append(content)
head += " "
for s in reversed(res):
print(s)
if __name__ == "__main__":
n = int(input())
solve_method(n)