Python日期时间处理问题：负时长行程下车时间修正失败

需求：为所有时长为负的行程，给dropoff_datetime列增加12小时。

按照提示要求使用带三个参数的where函数：条件对比df['duration']与timedelta(0)，设置inplace=True，other参数设为dropoff_datetime列加12小时的结果，但编写的代码输出仍不正确，怀疑other语句存在问题。

原代码 #

# Load libraries
import pandas as pd
from datetime import timedelta

# Loading dataset, creating duration column, and filtering to negative durations
url = 'https://drive.google.com/uc?id=1YV5bKobzYxVAWyB7VlxNH6dmfP4tHBui'
df = pd.read_csv(url, parse_dates = ['pickup_datetime', 'dropoff_datetime', 'dropoff_calculated'])
df["duration"] = pd.to_timedelta(df["duration"])

# Task 1: add 12 hours to dropoff duration for negative durations
df['duration'].where(~(df['duration'] < timedelta(0)), other = df['dropoff_datetime'] + timedelta(12), inplace = True)

# Task 2: recalculate duration column
df['duration'] = df['dropoff_datetime'] - df['pickup_datetime']

# Task 3: inspect first 10 rows with negative duration
print(df[df['duration'] < timedelta(0)][["pickup_datetime", "dropoff_datetime", "trip_duration", "dropoff_calculated"]].head(5))

原输出结果 #

pickup_datetime        dropoff_datetime       trip_duration  
34   2016-09-19 11:47:23   2016-09-19 02:21:19   0 days 02:33:56
66   2016-09-20 12:11:43   2016-09-20 02:15:55   0 days 02:04:13
74   2016-09-20 12:55:00   2016-09-20 01:03:36   0 days 00:08:36
132  2017-04-22 12:38:41   2017-04-22 01:20:13   0 days 00:41:32
231  2017-04-24 12:56:31   2017-04-24 01:06:18   0 days 00:09:47
    dropoff_calculated  
34   2016-09-19 14:21:19
66   2016-09-20 14:15:56
74   2016-09-20 13:03:36
132  2017-04-22 13:20:13
231  2017-04-24 13:06:18

问题根源与修正方案 #

核心错误 #

1. 操作对象错误：原代码修改的是duration列，但需求是修改dropoff_datetime列，完全偏离目标。
2. 时间增量参数错误：timedelta(12)代表12天，而非12小时，需明确指定hours=12。
3. 逻辑无效：因为第一步没改对目标列，后续重新计算duration自然还是负的。

修正后的代码 #

# Load libraries
import pandas as pd
from datetime import timedelta

# Loading dataset, creating duration column
url = 'https://drive.google.com/uc?id=1YV5bKobzYxVAWyB7VlxNH6dmfP4tHBui'
df = pd.read_csv(url, parse_dates = ['pickup_datetime', 'dropoff_datetime', 'dropoff_calculated'])
df["duration"] = pd.to_timedelta(df["duration"])

# Task 1: 为负时长的行程给dropoff_datetime加12小时
df['dropoff_datetime'].where(
    ~(df['duration'] < timedelta(0)), 
    other = df['dropoff_datetime'] + timedelta(hours=12), 
    inplace = True
)

# Task 2: 重新计算duration列
df['duration'] = df['dropoff_datetime'] - df['pickup_datetime']

# Task 3: 检查修正后的负时长行（此时应无负时长数据）
print(df[df['duration'] < timedelta(0)][["pickup_datetime", "dropoff_datetime", "trip_duration", "dropoff_calculated"]].head(5))

修正说明 #

• 修改dropoff_datetime列后，重新计算的duration会变为正值，与输出中dropoff_calculated的预期结果一致（比如第34行的2016-09-19 14:21:19就是原下车时间加12小时后的结果）。
• 修正后执行Task3，不会再输出负时长的行，因为所有异常行程已被修正。

内容的提问来源于stack exchange，提问作者Nick Tsougy