删除字符串中的每个逗号，除了最后一个带有python

发布于 2025-01-22 13:37:21 字数 255 浏览 1 评论 0原文

我有那些字符串，

s1 = "1,395,54"
s2 = "1,5,6,75"

除了最后获得以下字符串外，我想要所有逗号：

s1 = "1395,54"
s2 = "156,75"

解决这个简单问题的最有效方法是什么？

原文

I have those strings

s1 = "1,395,54"
s2 = "1,5,6,75"

I would like all commas except the last to obtain the following strings:

s1 = "1395,54"
s2 = "156,75"

What is the most efficient way to solve this simple problem?

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不弃不离 2025-01-29 13:37:21

这是str.replace的解决方案。 替换函数接受第三个参数，该参数指定替换的最大事件数量。因此，您可以传递比字符串中，出现数量的数字（your_string.count（“，”，“”） - 1）。

>>> s1 = "1,395,54"
>>> s1.replace(",", "", s1.count(",") - 1)
'1395,54'
>>> s2 = "1,5,6,75"
>>> s2.replace(",", "", s2.count(",") - 1)
'156,75'

Here is the solution with str.replace. replace function accepts a third argument which specifies Maximum number of occurrences to replace. So you can pass a number which is one less than the number of occurrences of , in your string(your_string.count(",") - 1).

>>> s1 = "1,395,54"
>>> s1.replace(",", "", s1.count(",") - 1)
'1395,54'
>>> s2 = "1,5,6,75"
>>> s2.replace(",", "", s2.count(",") - 1)
'156,75'

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放飞的风筝 2025-01-29 13:37:21

您可以用f -strings拆分并重新加入 -

*rest, last = s1.split(',')
f'{"".join(rest)},{last}'

您可以将它们组合到一个衬里 -

''。join（s1.split（'，'，'，'）[： - 1]） +” ，“ + s1.split（'，'）[ - 1]

You could split and rejoin with f-strings -

*rest, last = s1.split(',')
f'{"".join(rest)},{last}'

You can combine those into a one liner -

''.join(s1.split(',')[:-1]) + "," + s1.split(',')[-1]

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星星的軌跡 2025-01-29 13:37:21

您可能还会使用一种模式，断言右侧有一个逗号，只有数字介于两者之间：

import re

s1 = "1,395,54"
s2 = "1,5,6,75"

pattern = r",(?=\d*,)"

print(re.sub(pattern, "", s1))  # 1395,54
print(re.sub(pattern, "", s2))  # 156,75

更广泛的变体接受除了之间的逗号以外的任何char：

,(?=[^,]*,)

You might also use a pattern asserting that there is a comma present to the right with only digits in between:

import re

s1 = "1,395,54"
s2 = "1,5,6,75"

pattern = r",(?=\d*,)"

print(re.sub(pattern, "", s1))  # 1395,54
print(re.sub(pattern, "", s2))  # 156,75

The broader variant accepting any char except a comma inbetween: