Question

我有一个像这样的列表

[['a', 'word'],
['University', 'org'],
['of', 'org'],
['Michigan', 'org'], 
['Michigan', 'country']]

我想做的是如果我在第一次迭代时找到密歇根这个词，我将返回标签，我将用＆＃34;删除＆＃34;当我再次迭代时，我应该找到像这样的列表

[['a', 'word'],
['University', 'org'],
['of', 'org'],
['deleted', 'org'],
['Michigan', 'country']]

我尝试的是

for i in range(len(list)):

    if word.lower()==list[i][0].lower():
        found= list[i][1]

        list.pop(i)
        list[i][0] = "deletedword"


        break

Answer 1

使用生成器表达式查找第一个"Michigan"的位置。将其替换为必填字词：

lst = [['a', 'word'], ['University', 'org'], ['of', 'org'], ['Michigan', 'org'], ['Michigan', 'country']]

try:
    pos = next((i, x.index('Michigan')) for i, x in enumerate(lst) if 'Michigan' in x)
    lst[pos[0]][pos[1]] = 'deleted'

except StopIteration:
    print('There is no Michigan in the list')

print(lst)

# [['a', 'word'], 
#  ['University', 'org'], 
#  ['of', 'org'], 
#  ['deleted', 'org'], 
#  ['Michigan', 'country']]

Answer 2

l = [['a', 'word'], ['University', 'org'], ['of', 'org'], ['Michigan', 'org'], ['Michigan', 'country']]

toReplace = 'Michigan'
for i in l:
    if i[0].lower() == toReplace.lower():
        i[0] = "deletedword"    #No need to Pop just replace.
        break
print(l)

<强>输出：

[['a', 'word'], ['University', 'org'], ['of', 'org'], ['deletedword', 'org'], ['Michigan', 'country']]

Answer 3

您希望break退出嵌套for循环。一种可读的方法是使用函数和return语句：

def remove_first_el(A, k='Michigan'):
    for i in range(len(A)):                         # iterate outer lists
        for j in range(len(A[i])):                  # iterate inner lists
            if A[i][j].casefold() == k.casefold():  # ignore case for comparison
                A[i][j] = 'deleted'
                return A

res = remove_first_el(lst)

print(res)

[['a', 'word'],
 ['University', 'org'],
 ['of', 'org'],
 ['deleted', 'org'],
 ['Michigan', 'country']]

如果你需要＆＃34;密歇根＆＃34;在子列表的第一个元素中，您可以相应地修改您的函数：

def remove_first_el_zero_pos(A, k='Michigan'):
    for i in range(len(A)):
        if A[i][0].casefold() == k.casefold():
            A[i][0] = 'deleted'
            return A

删除第二个列表中第一个出现的单词？

3 个答案: