如何格式化日期到1900年代? [英] How to format date to 1900's?

问题描述

我正在预处理数据，其中一栏代表日期，例如'6/1/51'

I'm preprocessing data and one column represents dates such as '6/1/51'

我正在尝试将字符串转换为日期对象，到目前为止，我所拥有的是:

I'm trying to convert the string to a date object and so far what I have is:

    date = row[2].strip()
    format = "%m/%d/%y"
    datetime_object = datetime.strptime(date, format)
    date_object = datetime_object.date()
    print(date_object)
    print(type(date_object))

我面临的问题是将2051改为1951.

The problem I'm facing is changing 2051 to 1951.

我尝试写

    format = "%m/%d/19%y"

但是它给了我一个ValueError.

But it gives me a ValueError.

    ValueError: time data '6/1/51' does not match format '%m/%d/19%y'

我无法轻松地在线找到答案，所以我在这里问.谁能帮我这个忙吗?

I couldn't easily find the answer online so I'm asking here. Can anyone please help me with this?

谢谢.

推荐答案

使用'%m/%d/%y'解析没有世纪的日期，然后:

Parse the date without the century using '%m/%d/%y', then:

year_1900 = datetime_object.year - 100
datetime_object = datetime_object.replace(year=year_1900)

您应该对此附加条件，以便仅在1900年代的日期进行设置，例如今天之后的任何日期.

You should put conditionals around that so you only do it on dates that are actually in the 1900's, for example anything later than today.

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