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我正在尝试使用列表理解来做到这一点。我正在使用 python 2.7 的一个子集，它不允许使用命令any或all

string_list1 = ['James Dean', 'Mr. James Dean', 'Jon Sparrow', 'Timothy Hook', 'Captain Jon Sparrow']
string_list2 = []

# Get elements that are a substring of other elements
for str1 in string_list1:
    for str2 in string_list1:
        if str1 in str2 and str1 != str2:
            string_list2.append(str1)
print('Substrings: ', string_list2)

# remove element if another element is within it
for str2 in string_list2:
    for str1 in string_list1:
        if str2 in str1 and str2 != str1:
            string_list1.remove(str1)
print('Desired: ', string_list1) # all elements that are unique

结果应该是 ['James Dean', 'Jon Sparrow', 'Timothy Hook'] 基本上是子串和非子串元素

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1 回答 1

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您可以将相同的算法应用于列表理解，如下所示：

lst = ['James Dean', 'Mr. James Dean', 'Jon Sparrow', 'Timothy Hook', 'Captain Jon Sparrow']

res = [primitive for primitive in lst 
          if primitive not in (superstr for superstr in lst
              if [substr for substr in lst if substr in superstr and substr != superstr]
          )
      ]

print(res)

但是解释器不会看到内部表达式(superstr ...)只需要计算一次，而不是外部循环的每次迭代。所以我宁愿分两步来做：

lst = ['James Dean', 'Mr. James Dean', 'Jon Sparrow', 'Timothy Hook', 'Captain Jon Sparrow']

exclude = [superstr for superstr in lst
              if [substr for substr in lst if substr in superstr and substr != superstr]
          ]
res = [primitive for primitive in lst if primitive not in exclude]

print(res)
于 2021-07-30T09:12:09.943 回答