Question

我想比较两个字符串，以便比较应忽略特殊字符的差异。也就是说，

Hai，这是一个测试

应与

匹配

海！这是一个测试“或”海这是一个测试

有没有办法在不修改原始字符串的情况下执行此操作？

Answer 1

这会在进行比较之前删除标点符号和空格：

In [32]: import string

In [33]: def compare(s1, s2):
    ...:     remove = string.punctuation + string.whitespace
    ...:     return s1.translate(None, remove) == s2.translate(None, remove)

In [34]: compare('Hai, this is a test', 'Hai ! this is a test')
Out[34]: True

Answer 2

>>> def cmp(a, b):
...     return [c for c in a if c.isalpha()] == [c for c in b if c.isalpha()]
... 
>>> cmp('Hai, this is a test', 'Hai ! this is a test')
True
>>> cmp('Hai, this is a test', 'Hai this is a test')
True
>>> cmp('Hai, this is a test', 'other string')
False

这会创建两个临时列表，但不会以任何方式修改原始字符串。

Answer 3

比较字母等价的任意数量的字符串，

def samealphabetic(*args):
    return len(set(filter(lambda s: s.isalpha(), arg) for arg in args)) <= 1

print samealphabetic('Hai, this is a test',
                     'Hai ! this is a test',
                     'Hai this is a test')

打印True。应该更改<=，具体取决于您想要返回的参数。

Answer 4

通常，你要替换你想忽略的字符，然后比较它们：

import re
def equal(a, b):
    # Ignore non-space and non-word characters
    regex = re.compile(r'[^\s\w]')
    return regex.sub('', a) == regex.sub('', b)

>>> equal('Hai, this is a test', 'Hai this is a test')
True
>>> equal('Hai, this is a test', 'Hai this@#)($! i@#($()@#s a test!!!')
True

Answer 5

也许你可以先删除两个字符串中的特殊字符，然后比较它们。

在您的示例中，特殊字符为'，'，'！'和空间。

所以你的字符串：

a='Hai, this is a test'
b='Hai ! this is a test'
tempa=a.translate(None,',! ')
tempb=b.translate(None,',! ')

然后你可以比较tempa和tempb。

Answer 6

使用Levenshtein metric测量两个字符串之间的距离。按分数对字符串比较进行排名。选择顶部 n 匹配。

Answer 7

由于您提到您不想修改原始字符串，您还可以执行就地操作，而无需任何额外空间。

>>> import string
>>> first = "Hai, this is a test"
>>> second = "Hai ! this is a test"
>>> third = "Hai this is a test"
>>> def my_match(left, right):
    i, j = 0, 0
    ignored = set(string.punctuation + string.whitespace)
    while i < len(left) and j < len(right):
        if left[i] in ignored:
            i += 1
        elif right[j] in ignored:
            j += 1
        elif left[i] != right[j]:
            return False
        else:
            i += 1
            j += 1
    if i != len(left) or j != len(right):
        return False
    return True

>>> my_match(first, second)
True
>>> my_match(first, third)
True
>>> my_match("test", "testing")
False

Python比较字符串忽略特殊字符

7 个答案: