Question

我有一个字符串列表，需要删除其他项目中包含的项目，如图所示：

a = ["one", "one single", "one single trick", "trick", "trick must", "trick must get", "one single trick must", "must get", "must get the job done"]

我只需要删除同一列表中另一个字符串中包含的每个字符串，例如：“一个”包含在“单个”中，因此必须将其删除，然后“单个”包含在“单个技巧”中所以也需要删除

我尝试过：

b=a
for item in a:
    for element in b:
        if item in element:
            b.remove(element)

预期结果：

a = ["trick must get", "one single trick must", "must get the job done"]

任何帮助将不胜感激！预先感谢！

Answer 1

结合Python的any函数，列表解析应该可以很好地做到这一点：

a = [phrase for phrase in a if not any([phrase2 != phrase and phrase in phrase2 for phrase2 in a])]

结果：

>>> a = ["one", "one single", "one single trick", "trick", "trick must", "trick must get", "one single trick must", "must get", "must get the job done"]
>>> a = [phrase for phrase in a if not any([phrase2 != phrase and phrase in phrase2 for phrase2 in a])]
>>> a
['trick must get', 'one single trick must', 'must get the job done']

Answer 2

解决O O（n）时间复杂度问题的有效方法是使用一个集合，该集合跟踪给定短语的所有子短语，从最长的字符串迭代到最短的字符串，并且仅将字符串添加到输出中（如果该字符串尚未在子短语集中）：

seen = set()
output = []
for s in sorted(a, key=len, reverse=True):
    words = tuple(s.split())
    if words not in seen:
        output.append(s)
    seen.update({words[i: i + n] for i in range(len(words)) for n in range(len(words) - i + 1)})

output变为：

['one single trick must', 'must get the job done', 'trick must get']

Answer 3

这不是一种有效的解决方案，但是通过将最长到最小排序并删除最后一个元素，我们可以检查每个元素是否在任何地方都显示为子字符串。

a = ['one', 'one single', 'one single trick', 'trick', 'trick must', 'trick must get', 
     'one single trick must', 'must get', 'must get the job done']
a = sorted(a, key=len, reverse=True)
b = []
for i in range(len(a)):
    x = a.pop()
    if x not in "\t".join(a):
        b.append(x)

# ['trick must get', 'must get the job done', 'one single trick must']

如何删除包含在同一字符串列表中的其他字符串中的字符串？

3 个答案: