比较两个字符串列表

Question

我熟悉比较 2 个整数和字符串列表；然而，当比较 2 个包含额外字符的字符串列表时，可能会有点挑战。

假设输出包含以下内容，我将其分解为字符串列表。我在我的代码中称它为 diff。

输出

164c164
< Apples = 
---
> Apples = 0
168c168
< Berries = 
---
> Berries = false
218c218
< Cherries = 
---
> Cherries = 20
223c223
< Bananas = 
---
> Bananas = 10
233,234c233,234
< Lemons = 2
< Strawberries = 4
---
> Lemons = 4
> Strawberries = 2
264c264
< Watermelons = 
---
> Watermelons = 524288

第二组字符串包含我希望与第一个列表进行比较的忽略变量。

>>> ignore
['Apples', 'Lemons']

我的代码：

>>> def str_compare (ignore, output):
...     flag = 0
...     diff = output.strip ().split ('\n')
...     if ignore:
...         for line in diff:
...             for i in ignore:
...                 if i in line:
...                     flag = 1
...             if flag:
...                 flag = 0
...             else:
...                 print (line)
... 
>>>

该代码适用于省略了 Apple 和 Lemons 的情况。

>>> str_compare(ignore, output)
164c164
---
168c168
< Berries = 
---
> Berries = false
218c218
< Cherries = 
---
> Cherries = 20
223c223
< Bananas = 
---
> Bananas = 10
233,234c233,234
< Strawberries = 4
---
> Strawberries = 2
264c264
< Watermelons = 
---
> Watermelons = 524288
>>>

必须有更好的方法来比较 2 个不是 O(n^2) 的字符串。如果我的差异列表不包含像 "Apples =" 这样的额外字符，那么比较两个列表可以用 O(n) 来实现。有什么建议或想法可以在不循环遍历每个差异元素上的 "ignore" 变量的情况下进行比较吗？

更新 #1 为了避免混淆和使用建议的评论，我更新了代码。

>>> def str_compare (ignore, output):
...     diff = output.strip ().split ('\n')
...     if ignore:
...         for line in diff:
...             if not any ([i in line for i in ignore]):
...                 print (line)
...                 print ("---")
>>>

无论如何，它仍然会为每个 diff 元素循环忽略两次。

Answer 1

为了提高效率，请忽略集而不是列表。使用拆分从行中获取关键字。

>>> def str_compare (ignore, output):
...     ignore = set (ignore)
...     diff = output.strip ().split ('\n')
...     for line in diff:
...         if line.startswith('<') or line.startswith('>'):
...             var = line.split () [1]
...             if var not in ignore:
...                 print (line)
...         else:
...             print (line)
...

输出

>>> str_compare (ignore, output)
164c164
---
168c168
< Berries = 
---
> Berries = false
218c218
< Cherries = 
---
> Cherries = 20
223c223
< Bananas = 
---
> Bananas = 10
233,234c233,234
< Strawberries = 4
---
> Strawberries = 2
264c264
< Watermelons = 
---
> Watermelons = 524288

您可以通过拆分和连接“---\n”来消除对标志的需求（比标志或打字 ---- 更通用的解决方案）

请注意，字符串包含 s1 在 s2 最坏情况下应约为 len(s1) * len(2)，而相等约为 max(len(s1),len(s2)。而 python 实现是相当不错（对于一般情况），线性复杂度算法似乎存在 http://monge.univ-mlv.fr/~mac/Articles-PDF/CP-1991-jacm.pdf 另见 Algorithm to find multiple string matches

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