如何转换含未补零、空格的不规范字符串日期列？

Got it, let's work through this date formatting headache you're facing. The problem here is twofold: your dates have random extra spacing (like 2/9 /2018 with that stray space after the day) and single-digit months/days that don't play nice with strict strptime formatting rules. Here's how to fix it step by step:

1. First, Clean Up the Messy Strings #

Before parsing, we need to eliminate those random spaces that are throwing off the format. Using pandas, we can strip all whitespace from each date string with a simple regex replace:

import pandas as pd

# Remove all spaces from the date column
df['start'] = df['start'].str.replace('\s+', '', regex=True)

This turns 2/9 /2018 into 2/9/2018, which is already way easier to work with.

2. Use Flexible Date Parsing (No Strict Formats Needed) #

Your original strptime call failed because it was expecting a %m %d %Y format (month-day-year separated by spaces), but your dates use slashes. Even if you switched to %m/%d/%Y, strptime is strict—it requires two-digit months/days, so 1/26/2018 would still throw an error because the month is a single digit.

Instead, use one of these more forgiving approaches:

Option A: Pandas' Built-in to_datetime (Simplest) #

Pandas has a fantastic to_datetime function that automatically handles most common date variations, including single-digit months/days and (funnily enough) even the original spaced-out dates without pre-cleaning:

# Parse the date column directly, even with leftover spaces
df['start_datetime'] = pd.to_datetime(df['start'], errors='coerce')

The errors='coerce' flag will turn any unparseable dates into NaT (Not a Time) instead of crashing your code—super useful for messy real-world data.

Option B: dateutil.parser (More Control) #

If you need more customization, the dateutil library's parser can auto-detect almost any date format. First install it if you haven't (pip install python-dateutil), then use it with apply:

from dateutil import parser

# Parse cleaned date strings
df['start_datetime'] = df['start'].apply(lambda x: parser.parse(x))

This works great for edge cases that pandas might miss, though for your specific example, pandas alone is more than enough.

Why Your Original Code Failed #

Just to clarify: datetime.datetime.strptime(df['start'][1], '%m %d %Y') was looking for a date like 01 26 2018 (month, day, year separated by spaces), but your data uses slashes (1/26/2018). Even if you corrected the format to %m/%d/%Y, strptime requires two-digit months/days—so 1/26/2018 would still fail because the month is a single digit. Flexible parsers avoid this strictness entirely.

内容的提问来源于stack exchange，提问作者mitch