导出到 excel 时将 class 'pandas.tslib.Timedelta' 转换为字符串

Convert class 'pandas.tslib.Timedelta' to string when export to excel

初始数据帧:

arrivalTime        
0   2016-01-12 06:35:42
2   2016-01-12 06:54:02
3   2016-01-12 07:01:43
4   2016-01-12 07:02:28
5   2016-01-12 07:12:29
6   2016-01-12 07:18:41 

在数据上我应用了这个函数:

def function(df):
    df['arrivalTime_cal'] = pd.to_datetime(df['arrivalTime'], format='%Y-%m-%d %H:%M:%S')
    df['diff_time'] = df['arrivalTime_cal'].diff().fillna(0)
    del df['arrivalTime_cal']
    return df

我得到了这些结果(更正 ipython):

diff_time  
0  00:00:00  
1  00:04:37  
2  00:13:43  
3  00:07:41  
4  00:00:45

当导出到 excel 时,结果会更改格式:

arrivalTime             diff_time
0   2016-01-12 06:35:42 0
1   2016-01-12 06:40:19 0,003206019
2   2016-01-12 06:54:02 0,009525463
3   2016-01-12 07:01:43 0,005335648
4   2016-01-12 07:02:28 0,000520833

如何在 Excel 中保留字符串格式?

提前致谢

IIUC 然后你可以将类型转换为 str 然后 split str:

In [53]:
df['diff_time'].astype(str).str.split().str[-1].str.rsplit('.').str[0]

Out[53]:
index
0    00:00:00
2    00:18:20
3    00:07:41
4    00:00:45
5    00:10:01
6    00:06:12
dtype: object

将以上内容分解为多个步骤,使用 astype:

转换为 str
In [54]:
df['diff_time'].astype(str)

Out[54]:
index
0    0 days 00:00:00.000000000
2    0 days 00:18:20.000000000
3    0 days 00:07:41.000000000
4    0 days 00:00:45.000000000
5    0 days 00:10:01.000000000
6    0 days 00:06:12.000000000
Name: diff_time, dtype: object

现在拆分(默认字符为空格)并只取最后一个拆分元素,即时间分量:

In [55]:
df['diff_time'].astype(str).str.split().str[-1]

Out[55]:
index
0    00:00:00.000000000
2    00:18:20.000000000
3    00:07:41.000000000
4    00:00:45.000000000
5    00:10:01.000000000
6    00:06:12.000000000
dtype: object

现在 rsplit 减去 hte 微秒

In [56]:
df['diff_time'].astype(str).str.split().str[-1].str.rsplit('.')

Out[56]:
index
0    [00:00:00, 000000000]
2    [00:18:20, 000000000]
3    [00:07:41, 000000000]
4    [00:00:45, 000000000]
5    [00:10:01, 000000000]
6    [00:06:12, 000000000]
dtype: object

可以看到转换后的值确实是str:

In [57]:
df['diff_time'].astype(str).str.split().str[-1].str.rsplit('.').str[0][0]

Out[57]:
'00:00:00'