基于条件转换DataFrame时间字段为24小时制及错误解决

Let's tackle your problem step by step. First, let's recap your scenario: you have a pandas DataFrame with 12-hour time strings (NewFirst, NewLast as object type) and corresponding hour values (FirstNumberHour, LastNumberHour as float64). You want to convert these to 24-hour format (add 12 when the string contains "p"), then combine the resulting hours with a date column to create datetime objects.

The Error You Encountered #

Your initial code threw a TypeError: can only concatenate str (not "float") to str — this likely stemmed from an accidental string concatenation typo (maybe in how you defined or referenced variables). But regardless, there are cleaner, more reliable ways to achieve your goal.

Solution 1: List Comprehension for Offset Calculation #

This approach uses list comprehensions to generate 12-hour offsets for entries containing "p", then adds those offsets to your existing hour columns:

import pandas as pd
import numpy as np

# Sample DataFrame matching your input
DF = pd.DataFrame({
    'ID': [1,2,3],
    'FirstNumberHour': [7.0,2.0,4.0],
    'NewFirst': ['7a','2a','4p'],
    'LastNumberHour': [4.0,10.0,11.0],
    'NewLast': ['4p','10p','11p']
})

# Calculate offset and update FirstNumberHour
DF["FirstOffset"] = [12 if "p" in item else 0 for item in DF["NewFirst"].astype(str)]
DF["FirstNumberHourUpdate"] = DF["FirstNumberHour"].astype(int) + DF["FirstOffset"]

# Calculate offset and update LastNumberHour
DF["LastOffset"] = [12 if "p" in item else 0 for item in DF["NewLast"].astype(str)]
DF["LastNumberHourUpdate"] = DF["LastNumberHour"].astype(int) + DF["LastOffset"]

# Optional: Clean up temporary offset columns
DF = DF.drop(["FirstOffset", "LastOffset"], axis=1)

This produces your desired output:

Solution 2: Use pd.to_datetime (Simpler, More Robust) #

Instead of manual offset calculations, you can directly parse the NewFirst/NewLast strings to 24-hour hours using pandas' datetime tools. Note we use %I (for 12-hour format hours) instead of %H (24-hour format):

# Generate 24-hour hours directly from the time strings
DF["FirstNumberHourUpdate"] = pd.to_datetime(DF["NewFirst"], format="%I%p").dt.hour
DF["LastNumberHourUpdate"] = pd.to_datetime(DF["NewLast"], format="%I%p").dt.hour

This method automatically handles edge cases like 12a/12p and requires fewer lines of code.

Combine with Date Column to Create Datetime Objects #

Assuming you have a date column (e.g., Date with values like 2024-01-01), here are two ways to merge dates with your 24-hour hours:

Method A: String Concatenation + pd.to_datetime

# Add sample date column for demonstration
DF["Date"] = "2024-01-01"

# Combine date and original time string to create full datetime
DF["FirstDatetime"] = pd.to_datetime(DF["Date"] + " " + DF["NewFirst"], format="%Y-%m-%d %I%p")
DF["LastDatetime"] = pd.to_datetime(DF["Date"] + " " + DF["NewLast"], format="%Y-%m-%d %I%p")

Method B: dt.combine with datetime.time

If you want to use the updated 24-hour hours directly:

from datetime import time

# Convert date column to datetime type first
DF["Date"] = pd.to_datetime(DF["Date"])

# Merge date with time objects created from updated hours
DF["FirstDatetime"] = DF.apply(lambda row: row["Date"].combine(time(row["FirstNumberHourUpdate"])), axis=1)
DF["LastDatetime"] = DF.apply(lambda row: row["Date"].combine(time(row["LastNumberHourUpdate"])), axis=1)

Key Notes #

• You don't need to convert NewFirst/NewLast to float to check for "p" — just ensure they're treated as strings (use .astype(str) if needed) to avoid errors with missing values.
• The pd.to_datetime method is more robust than manual calculations, as it handles edge cases and formatting inconsistencies automatically.

内容的提问来源于stack exchange，提问作者Raven