[英]How to replace string using regular expression in Python
I need help with replacing characters in a string using regular expressions. 我需要使用正则表达式替换字符串中的字符的帮助。
Input : 输入:
s3 = ['March/21/2019' , 'Mar/23/2019']
Desired Output : 所需输出:
s3 = ['03/21/2019' , '03/23/2019']
I've tried a few things, but none of them seem to make any impact on the input: 我已经尝试了一些方法,但是它们似乎都不会对输入产生任何影响:
s3[i] = s3[i].replace(r'Mar[az]*', '03')
s3[i] = s3[i].replace(r'(?:Mar[az]*)', '03')
Could someone please help me and tell me what I'm doing wrong. 有人可以帮助我,告诉我我做错了什么。
This works. 这可行。
import re
s3 = ['March/21/2019' , 'Mar/23/2019']
s3 = [re.sub(r'Mar[a-z]*', '03', item) for item in s3]
# ['03/21/2019', '03/23/2019']
Of course, you can also use a for loop for better readability. 当然,您也可以使用for循环以提高可读性。
import re
s3 = ['March/21/2019' , 'Mar/23/2019']
for i in range(len(s3)):
s3[i] = re.sub(r'Mar[a-z]*', '03', s3[i])
# ['03/21/2019', '03/23/2019']
If you're only working with dates, try something like this: 如果只使用日期,请尝试以下操作:
import datetime
s3 = ['March/21/2019' , 'Mar/23/2019']
for i in range(0, len(s3)):
try:
newformat = datetime.datetime.strptime(s3[i], '%B/%d/%Y')
except ValueError:
newformat = datetime.datetime.strptime(s3[i], '%b/%d/%Y')
s3[i] = newformat.strftime('%m/%d/%Y')
s3
now contains ['03/21/2019', '03/23/2019']
s3
现在包含['03/21/2019', '03/23/2019']
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