REVISED： 看完后，似乎我的第一个问题是使用for循环-df.append方法创建数据帧。
使用print语句查看下面的“要附加的行”部分，然后我将变量分配给dict指针，并将dict附加到以空白数据帧开头的内容。
第一行/附加似乎改变了最后一列上第一个条目的类型。
while read line
if xx is not None and yy is not None and xx < yy:
zz=yy-xx
print (yy, xx, zz)
parms['datein'] = xx
parms['dateout'] = yy
parms['time'] = zz
df = df.append(parms,ignore_index=True)
yy=None
xx=None
此时数据框在第一行和最后一列中有一个奇数条目。
下面：
datein dateout time
0 2013-11-01 06:10:00 2013-11-01 12:06:00 21360000000000
1 2013-11-01 12:51:00 2013-11-01 14:53:00 2:02:00
2 2013-11-04 06:02:00 2013-11-04 14:04:00 8:02:00
3 2013-11-05 05:56:00 2013-11-05 12:11:00 6:15:00
所以现在这是一个dataframe.append问题。
尝试将列中的datetime.timedeltas相加会产生错误：
TypeError: unsupported operand type(s) for +: 'long' and 'datetime.timedelta'
显示数据框本身会显示第一行没有出现在数据框附加的行中的问题。
要追加的行：
(datetime.datetime(2013, 11, 1, 12, 6), datetime.datetime(2013, 11, 1, 6, 10),datetime.timedelta(0, 21360))
(datetime.datetime(2013, 11, 1, 14, 53), datetime.datetime(2013, 11, 1, 12, 51),datetime.timedelta(0, 7320))
(datetime.datetime(2013, 11, 4, 14, 4), datetime.datetime(2013, 11, 4, 6, 2),datetime.timedelta(0, 28920))
(datetime.datetime(2013, 11, 5, 12, 11), datetime.datetime(2013, 11, 5, 5, 56),datetime.timedelta(0, 22500))
(datetime.datetime(2013, 11, 5, 14, 42), datetime.datetime(2013, 11, 5, 12, 38),datetime.timedelta(0, 7440))
使用以下命令从dat文件创建带有for循环的数据帧：
parms = dict.fromkeys(keys)
df = pd.DataFrame(columns=keys)
df = df.append(parms,ignore_index=True)
DataFrame输出：
time
dateout
2013-11-01 12:06:00 21360000000000
2013-11-01 14:53:00 2:02:00
2013-11-04 14:04:00 8:02:00
2013-11-05 12:11:00 6:15:00
2013-11-05 14:42:00 2:04:00
修订2
Dat文件：
IN 11/01/2013 14:32
OUT 11/01/2013 18:32
IN 11/01/2013 18:58
OUT 11/01/2013 20:57
IN 11/04/2013 14:33
OUT 11/04/2013 18:30
IN 11/04/2013 18:57
OUT 11/04/2013 23:01
IN 11/05/2013 14:29
OUT 11/05/2013 18:31
IN 11/05/2013 18:58
OUT 11/05/2013 23:01
IN 11/06/2013 14:30
OUT 11/06/2013 18:31
IN 11/06/2013 18:57
OUT 11/06/2013 23:00
IN 11/07/2013 14:30
OUT 11/07/2013 18:31
代码：
import numpy as np
import pandas as pd
import struct
from datetime import datetime
keys = ['datein','dateout','time']
parms = dict.fromkeys(keys)
df = pd.DataFrame(columns=keys)
dat = open(datfile,'r')
for line in dat.readlines():
opt, date, mgr, tim = line[:3], line[6:16], line[18:22], line[24:29]
f = datetime.combine(datetime.strptime(date, '%m/%d/%Y'),datetime.strptime(tim, '%H:%M').time())
if opt == 'IN ' and f > datetime(2013, 11, 1) and f < datetime(2013, 12, 1):
xx = f
if opt == 'OUT' and f > datetime(2013, 11, 1) and f < datetime(2013, 12, 1):
yy = f
if xx is not None and yy is not None and xx < yy:
zz=yy-xx
print (yy, xx, zz) ## <-- Check lines before appending dataframe, output below
parms['datein'] = xx
parms['dateout'] = yy
parms['time'] = zz
df = df.append(parms,ignore_index=True)
if len(df.index) == 1:
print df.dtypes ## <---- This shows 'time' as timedelta64
if len(df.index) == 2:
print df.dtypes ## after next line appended 'time shows 'object' and first line loses type.
yy=None ## <-- reset before next loop
xx=None ## <-- reset before next loop
print dg.dtypes
print dg
在附加到df之前从行打印返回：
(datetime.datetime(2013, 11, 4, 14, 4), datetime.datetime(2013, 11, 4, 6, 2),datetime.timedelta(0, 28920))
(datetime.datetime(2013, 11, 5, 12, 11), datetime.datetime(2013, 11, 5, 5, 56), datetime.timedelta(0, 22500))
(datetime.datetime(2013, 11, 5, 14, 42), datetime.datetime(2013, 11, 5, 12, 38), datetime.timedelta(0, 7440))
(datetime.datetime(2013, 11, 6, 12, 7), datetime.datetime(2013, 11, 6, 5, 49), datetime.timedelta(0, 22680))
(datetime.datetime(2013, 11, 6, 14, 37), datetime.datetime(2013, 11, 6, 12, 24), datetime.timedelta(0, 7980))
(datetime.datetime(2013, 11, 7, 14, 7), datetime.datetime(2013, 11, 7, 6, 8), datetime.timedelta(0, 28740))
(datetime.datetime(2013, 11, 8, 11, 58), datetime.datetime(2013, 11, 8, 5, 53), datetime.timedelta(0, 21900))
(datetime.datetime(2013, 11, 8, 14, 10), datetime.datetime(2013, 11, 8, 12, 21), datetime.timedelta(0, 6540))
(datetime.datetime(2013, 11, 11, 12, 16), datetime.datetime(2013, 11, 11, 6, 6), datetime.timedelta(0, 22200))
(datetime.datetime(2013, 11, 11, 14, 31), datetime.datetime(2013, 11, 11, 12, 49), datetime.timedelta(0, 6120))
和dtypes：
datein datetime64[ns]
dateout datetime64[ns]
time object
dtype: object
显示第一行关闭的数据框：
datein dateout time
0 2013-11-01 06:10:00 2013-11-01 12:06:00 21360000000000
1 2013-11-01 12:51:00 2013-11-01 14:53:00 2:02:00
2 2013-11-04 06:02:00 2013-11-04 14:04:00 8:02:00
3 2013-11-05 05:56:00 2013-11-05 12:11:00 6:15:00
4 2013-11-05 12:38:00 2013-11-05 14:42:00 2:04:00
5 2013-11-06 05:49:00 2013-11-06 12:07:00 6:18:00
6 2013-11-06 12:24:00 2013-11-06 14:37:00 2:13:00
7 2013-11-07 06:08:00 2013-11-07 14:07:00 7:59:00
8 2013-11-08 05:53:00 2013-11-08 11:58:00 6:05:00
9 2013-11-08 12:21:00 2013-11-08 14:10:00 1:49:00
10 2013-11-11 06:06:00 2013-11-11 12:16:00 6:10:00
11 2013-11-11 12:49:00 2013-11-11 14:31:00 1:42:00
12 2013-11-12 06:04:00 2013-11-12 12:24:00 6:20:00
13 2013-11-12 12:40:00 2013-11-12 12:59:00 0:19:00
14 2013-11-13 06:04:00 2013-11-13 12:19:00 6:15:00
15 2013-11-13 12:42:00 2013-11-13 14:35:00 1:53:00
16 2013-11-14 06:05:00 2013-11-14 12:22:00 6:17:00
答案 0 :(得分：0)
这是使用pandas 0.13（现在0.13rc1）。 Timedelta支持在以前的版本中仅为最小。
In [23]: df
Out[23]:
date1 date2 td
0 2013-11-01 12:06:00 2013-11-01 06:10:00 05:56:00
1 2013-11-01 14:53:00 2013-11-01 12:51:00 02:02:00
2 2013-11-04 14:04:00 2013-11-04 06:02:00 08:02:00
3 2013-11-05 12:11:00 2013-11-05 05:56:00 06:15:00
4 2013-11-05 14:42:00 2013-11-05 12:38:00 02:04:00
[5 rows x 3 columns]
In [24]: df2 = df.set_index('date1')
In [25]: df2
Out[25]:
date2 td
date1
2013-11-01 12:06:00 2013-11-01 06:10:00 05:56:00
2013-11-01 14:53:00 2013-11-01 12:51:00 02:02:00
2013-11-04 14:04:00 2013-11-04 06:02:00 08:02:00
2013-11-05 12:11:00 2013-11-05 05:56:00 06:15:00
2013-11-05 14:42:00 2013-11-05 12:38:00 02:04:00
[5 rows x 2 columns]
In [26]: pd.to_timedelta(df2.groupby(df2.index.date)['td'].sum())
Out[26]:
2013-11-01 07:58:00
2013-11-04 08:02:00
2013-11-05 08:19:00
Name: td, dtype: timedelta64[ns]
即使这不是完全支持（因为我必须'将输出'转换回'timedeltas），这当前是groupby的一个未解决的问题，将修复为0.14，请参见此处：https://github.com/pydata/pandas/issues/5724
答案 1 :(得分：0)
这是a bug。在下面，可以看到一个比原始问题更短的例子。
设定：
import datetime
import pandas as pd
parms = {'d': datetime.datetime(2013, 11, 5, 5, 56), 't':datetime.timedelta(0, 22500)}
df = pd.DataFrame(columns=list('dt'))
df = df.append(parms, ignore_index=True)
错误的代码：
>>> df.append(parms, ignore_index=True)
d t
0 2013-11-05 05:56:00 22500000000000
1 2013-11-05 05:56:00 6:15:00