Question

我有一个列表：

List1 = ['M1', 'M2', 'M3', 'M4', 'M5', 'M5', 'M5', 'M6', 'M7', 'M7', 'M8', 'M9', 'M10', 'M10', 'M10', 'M11']

我想查找所有重复项，然后在每个重复项之后添加小写字母字符，如下所示：

List1 = ['M1', 'M2', 'M3', 'M4', 'M5a', 'M5b', 'M5c', 'M6', 'M7a', 'M7b', 'M8', 'M9', 'M10a', 'M10b', 'M10c', 'M11']

我可以找到重复项，但是我不确定该怎么做。

任何帮助将不胜感激！

Answer 1

您可以使用列表推导和string.ascii_lowercase：

import string, collections
def addition(val, _i, l):
   return string.ascii_lowercase[sum(c == val for c in l[:_i])]

List1 = ['M1', 'M2', 'M3', 'M4', 'M5', 'M5', 'M5', 'M6', 'M7', 'M7', 'M8', 'M9', 'M10', 'M10', 'M10', 'M11']
c = collections.Counter(List1)
new_results = ['{}{}'.format(a, '' if c[a] == 1 else addition(a, i, List1)) for i, a in enumerate(List1)]

输出：

['M1', 'M2', 'M3', 'M4', 'M5a', 'M5b', 'M5c', 'M6', 'M7a', 'M7b', 'M8', 'M9', 'M10a', 'M10b', 'M10c', 'M11']

Answer 2

使用Counter和enumerate的另一种简便方法：

from collections import Counter

List1 = ['M1', 'M2', 'M3', 'M4', 'M5', 'M5', 'M5', 'M6', 'M7', 'M7', 'M8', 'M9', 'M10', 'M10', 'M10', 'M11']

c = Counter(List1)

prev = List1[0]
for i, x in enumerate(List1):
    if c[x] > 1 and prev != x:
        l = 'a'
        List1[i] += l
        prev = x
    elif c[x] > 1 and prev == x:
        l = chr(ord(l) + 1)
        List1[i] += l

print(List1)
# ['M1', 'M2', 'M3', 'M4', 'M5a', 'M5b', 'M5c', 'M6', 'M7a', 'M7b', 'M8', 'M9', 'M10a', 'M10b', 'M10c', 'M11']

Answer 3

单线的另一个可能的灵魂：

List2 = [x if List1.count(x) == 1 else x + chr(ord('a') + List1[:i].count(x)) 
         for i, x in enumerate(List1)]

Answer 4

以下解决方案应一次解决您的问题-一次遍历阵列。比Ajax1234，Austin和taras的解决方案性能更高。当然，您可以进一步压缩此代码。

count = -1
newList = []
for indx,ele in enumerate(List1):
    if indx == 0:
        continue
    if List1[indx]==List1[indx-1]:
        count+=1
        newList.append(List1[indx-1]+str(chr(97+count)))
    elif count == -1:
        newList.append(List1[indx-1])
    else:
        count+=1
        newList.append(List1[indx-1]+str(chr(97+count)))
        count = -1
if count == -1:
    newList.append(List1[indx])
else:
    count +=1
    newList.append(List1[indx]+str(chr(97+count)))

在列表中查找重复项，并在每一个之后添加连续的字母字符

4 个答案: