python3、两个字符串的区别

Question

我想记录列表中两个字符串的差异位置（以删除它们）...最好记录每个部分的最高分隔点，因为这些区域将具有动态内容。

比较这些

总字符数 178。两个独特的部分

t1 = 'WhereTisthetotalnumberofght5y5wsjhhhhjhkmhm Thethreemethodsthatreturntheratioofmatchingtototalcharacterscangivedifferentresultsduetodifferinglevelsofapxxxxxxxproximation,although'

和

总字符数 211。两个独特的部分

t2 = 'WhereTisthetotalnumberofdofodfgjnjndfgu><rgregw><sssssuguyguiygis>gggs<GS,Gs Thethreemethodsthatreturntheratioofmatchingtototalcharacterscangivedifferentrexxxxxxxsultsduetodifferinglevelsofapproximation,although'

我知道 difflib 可以做到这一点，但输出很糟糕。

我想存储（在列表中）字符位置，最好是较大的分隔值。

模式位置

t1 = 'WhereTisthetotalnumberof  24  ght5y5wsjhhhhjhkmhm  43  Thethreemethodsthatreturntheratioofmatchingtototalcharacterscangivedifferentresultsduetodifferinglevelsofap  151  xxxxxxx  158  proximation,although'
t2 = 'WhereTisthetotalnumberof  24  dofodfgjnjndfgu><rgregw><sssssuguyguiygis>gggs<GS,Gs  76  Thethreemethodsthatreturntheratioofmatchingtototalcharacterscangivedifferentre  155  xxxxxxx  162  sultsduetodifferinglevelsofapproximation,although'

输出：

output list = [24, 76, 151, 162]

更新

回应post @Olivier

由 *** 分隔的所有 Y 的位置

t1
WhereTisthetotalnumberofght5***y***5wsjhhhhjhkmhm Thethreemethodsthatreturntheratioofmatchingtototalcharacterscangivedifferentresultsduetodifferinglevelsofapxxxxxxxproximation,although

t2 WhereTisthetotalnumberofdofodfgjnjndfgu><rgregw><sssssugu***y***gui***y***gis>gggs<GS,Gs Thethreemethodsthatreturntheratioofmatchingtototalcharacterscangivedifferentrexxxxxxxsultsduetodifferinglevelsofapproximation,although

matcher.get_matching_blocks()后的输出和 string = ''.join([t1[a:a+n] for a, _, n in blocks])

WhereTisthetotalnumberof***y*** Thethreemethodsthatreturntheratioofmatchingtototalcharacterscangivedifferentresultsduetodifferinglevelsofapproximation,although

Answer 1

使用 difflib 可能是您最好的选择，因为您不太可能想出比它提供的算法更有效的解决方案。你想要的是使用SequenceMatcher.get_matching_blocks。这是它根据 doc.

输出的内容

Return list of triples describing matching subsequences. Each triple is of the form (i, j, n), and means that a[i:i+n] == b[j:j+n]. The triples are monotonically increasing in i and j.

这是一种方法，您可以使用它来重建已删除增量的字符串。

from difflib import SequenceMatcher

x = "abc_def"
y = "abc--ef"

matcher = SequenceMatcher(None, x, y)
blocks = matcher.get_matching_blocks()

# blocks: [Match(a=0, b=0, size=4), Match(a=5, b=5, size=2), Match(a=7, b=7, size=0)]

string = ''.join([x[a:a+n] for a, _, n in blocks])

# string: "abcef"

编辑：还指出，如果您有两个这样的字符串。

t1 = 'WordWordaayaaWordWord'
t2 = 'WordWordbbbybWordWord'

那么上面的代码会return'WordWordyWordWord。这是因为 get_matching_blocks 将捕获出现在预期块之间的两个字符串中的 'y'。一个解决方案是按长度过滤 returned 块。

string = ''.join([x[a:a+n] for a, _, n in blocks if n > 1])

如果您想对 returned 块进行更复杂的分析，您还可以执行以下操作。

def block_filter(substring):
    """Outputs True if the substring is to be merged, False otherwise"""
    ...


string = ''.join([x[a:a+n] for a, _, n in blocks if block_filter(x[a:a+n])])

python3、两个字符串的区别

python 3, differences between two strings

string

difflib

python-3.x