如何解决比较两个时间列,然后在 python 数据帧的新列中返回 hh:mm:ss 中的最小时间差
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 函数将相同的代码应用于整个列时,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
解决方法
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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