Question

我在下面有字符串。如何在python中不使用for循环替换最后一个空格而不是一个空格。

"814409014860", "BOA ", "604938XXXXXX5410 ",,"ADOM ","ADU SAVBOSS SVGIDIIADOM0001 Int. charge ","24/05/18 09:39:08 0.70 ",0.00

我想避免for循环，因为数据很大。

在分钟时间戳

之后需要逗号

09:39:08 0.70 ",0.00 =＆gt; 09:39:08, 0.70 ",0.00

Python，bash，c＃是首选。

Answer 1

您可以使用rfind和rsplit：

# First find the occurrence of the last space
last_space_index = big_string.rfind(" ")

# then split from the right the substring that ends in that index
new_big_string = ", ".join(big_string[:last_space_index].rsplit(" ", 1)) + big_string[last_space_index:]

Answer 2

Python正则表达式替换？

import re

s = '"814409014860", "BOA ", "604938XXXXXX5410 ",,"ADOM ","ADU SAVBOSS SVGIDIIADOM0001 Int. charge ","24/05/18 09:39:08 0.70 ",0.00'

pattern = r'(\d{2}\:\d{2}\:\d{2})'

output = re.sub(pattern,r'\1,',s)

print(output)

>'"814409014860", "BOA ", "604938XXXXXX5410 ",,"ADOM ","ADU SAVBOSS SVGIDIIADOM0001 Int. charge ","24/05/18 09:39:08, 0.70 ",0.00'

在python中使用逗号替换最后一个空格

2 个答案: