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python - 比较两个时间列，然后在 python 数据帧的新列中返回 hh:mm:ss 中的最小时间差

问题描述

from datetime import datetime
start='12:25:03'
format = '%H:%M:%S'
startDateTime = datetime.strptime(start, format)
end='12:30:40'
endDateTime = datetime.strptime(end, format)
diff = endDateTime - startDateTime
print(diff)
0:05:37

上面的代码工作正常，但是当我使用 lambda 函数将其应用于整个列时，我得到不同格式的结果，我想以 hh:mm:ss 格式获取 Diff 列的值。

t1 - Object type 
t2 - Object type 
Diff - timedelta64[ns] type



 df["Diff"] = df.apply(lambda x: datetime.strptime(x["t1"], format) - datetime.strptime(x["t2"], format), axis = 1)
 df.head()
           t1          t2             Diff
    0   01:27:19    01:28:58    -1 days +23:58:21
    1   01:49:57    01:50:40    -1 days +23:59:17
    2   03:35:24    03:36:14    -1 days +23:59:10

标签： pythonpandaspython-datetimetimedelta

解决方案

相关：将 timedelta 格式化为字符串

您可以为 timedelta 对象编写自己的格式化程序，例如

def formatTimedelta(td):
    """
    format a timedelta object to string, in HH:MM:SS format (seconds floored).
    negative timedeltas will be prefixed with a minus, '-'.
    """
    total = td.total_seconds()
    prefix, total = ('-', total*-1) if total < 0 else ('', total)
    h, r = divmod(total, 3600)
    m, s = divmod(r, 60)
    return f"{prefix}{int(h):02d}:{int(m):02d}:{int(s):02d}"

这将为您提供示例 df

df
         t1        t2
0  01:27:19  01:28:58
1  01:49:57  01:50:40
2  03:35:24  03:36:14

# to datetime
df['t1'] = pd.to_datetime(df['t1'])
df['t2'] = pd.to_datetime(df['t2'])

# calculate timedeltas and format
df['diff0'] = (df['t1']-df['t2']).apply(formatTimedelta)
df['diff1'] = (df['t2']-df['t1']).apply(formatTimedelta)

df['diff0']
0    -00:01:39
1    -00:00:43
2    -00:00:50
Name: diff0, dtype: object

df['diff1']
0    00:01:39
1    00:00:43
2    00:00:50
Name: diff1, dtype: object

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