如何将日期转换为1900年代？数据预处理日期转换问题

The issue you’re hitting is that Python’s datetime.strptime uses a default cutoff for two-digit years—typically, years >= 70 get mapped to 19xx, while years <70 become 20xx. That’s why '51' ends up as 2051 instead of 1951. Your attempt to use %m/%d/19%y fails because the input string doesn’t include the '19' prefix—this format expects a full four-digit year like '6/1/1951', not the two-digit version you have.

Here are two reliable fixes using either the standard library or a flexible third-party tool:

Method 1: Adjust the Year Post-Conversion (Standard Library Only) #

This is the simplest approach with no extra dependencies. After parsing the date, check if the year falls in the 2000s and subtract 100 to shift it to the 1900s:

from datetime import datetime

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

# Tweak the cutoff value to match your dataset's context
if datetime_object.year >= 2000:
    datetime_object = datetime_object.replace(year=datetime_object.year - 100)

date_object = datetime_object.date()
print(date_object)  # Outputs 1951-06-01 for '6/1/51'
print(type(date_object))  # <class 'datetime.date'>

If your dataset has a mix of 19xx and 20xx two-digit years, adjust the cutoff (e.g., use 1950 instead of 2000) to ensure correct mapping for all entries.

Method 2: Use python-dateutil for Custom Century Cutoffs #

If you’re open to a third-party library, python-dateutil lets you define exactly how two-digit years should be interpreted. First install it with pip install python-dateutil, then use:

from dateutil import parser

date_str = row[2].strip()
# Set cutoff_year to 1950: two-digit years >=50 become 19xx, <50 become 20xx
date_object = parser.parse(date_str, yearfirst=False, cutoff_year=1950).date()
print(date_object)  # Outputs 1951-06-01 for '6/1/51'

This is great if you have inconsistent date formats, but it does require adding an external dependency.

Quick Note #

Never modify the format string to hardcode '19'—it won’t work because your input strings don’t contain those characters. Always adjust the parsed datetime object or use a tool that supports custom century rules.

内容的提问来源于stack exchange，提问作者ekim420