英文:
how to convert a string containing characters into date time using strptime in Python
问题
from datetime import datetime
datetime.strptime(x, "%Y%m%d%H%M")
ValueError: unconverted data remains: A1745
英文:
I'm trying to convert a string where there exists a characters between date information into a datetime. I was wondering if this is possible without using replace function. Suppose my string is defined as '20220117A1745' implying Jan. 17th, 2022 at 5:45pm.
If I use strptime, as expected, I receive an error.
from datetime import datetime
datetime.strptime(x,"%Y%m%d%H%M")
ValueError: unconverted data remains: A1745
I was wondering if I can do this without using another method. The reason why I don't want to use antoher method is that, replace has a bad time ecomplexity and will slow the execution time badly (see for reference). I'll do this operation for hundreds of thousans of strings in a for loop.
答案1
得分: 1
以下是翻译好的内容：
from datetime import datetime
x = '20220117A1745'
date_str = x[:8] + x[9:]
date_obj = datetime.strptime(date_str, '%Y%m%d%H%M')
print(date_obj)
输出结果为：
2022-01-17 17:45:00
英文:
What about
from datetime import datetime
x = '20220117A1745'
date_str = x[:8] + x[9:]
date_obj = datetime.strptime(date_str, '%Y%m%d%H%M')
print(date_obj)
which gives
2022-01-17 17:45:00
答案2
得分: 1
如果字符串始终在相同位置包含'A'，则可以这样做：
from datetime import datetime
x = '20220117A1745'
print(datetime.strptime(x, "%Y%m%dA%H%M"))
如果不知道字母是什么：
print(datetime.strptime(x[:8]+x[9:], "%Y%m%d%H%M"))
两种方法的输出结果：
2022-01-17 17:45:00
英文:
If the string always has 'A' in the same position then you can do this:
from datetime import datetime
x = '20220117A1745'
print(datetime.strptime(x,"%Y%m%dA%H%M"))
If you don't know what the letter is then:
print(datetime.strptime(x[:8]+x[9:],"%Y%m%d%H%M"))
Output for both methods:
2022-01-17 17:45:00
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