Question

我想创建一个新列，它将显示两个日期之间的timedelta，如下面的pandas数据帧所示：

>>> hg[['not inc','date']]
   not inc                date
0    False 2012-02-29 00:00:00
1    False 2012-03-16 00:00:00
2    False 2012-04-04 00:00:00
3     True 2012-05-08 00:00:00
4    False 2012-05-12 00:00:00
5    False 2012-05-26 00:00:00
6    False 2012-06-09 00:00:00
7    False 2012-10-13 00:00:00
8    False 2012-11-10 00:00:00
9     True 2013-03-19 00:00:00
10   False 2013-04-01 00:00:00
11   False 2013-04-25 00:00:00
12   False 2013-05-04 00:00:00
13   False 2013-05-18 00:00:00
14   False 2013-06-01 00:00:00
15    True 2013-08-20 00:00:00
16   False 2013-08-31 00:00:00
17   False 2013-09-21 00:00:00
18   False 2013-10-05 00:00:00
19   False 2013-10-19 00:00:00
20   False 2013-11-09 00:00:00
21    True 2014-01-21 00:00:00
22   False 2014-02-08 00:00:00
23   False 2014-02-22 00:00:00
24   False 2014-03-08 00:00:00
25   False 2014-03-29 00:00:00
26   False 2014-04-19 00:00:00
27    True 2014-07-21 00:00:00
28    True 2014-08-01 00:00:00
29   False 2014-08-09 00:00:00
30   False 2014-08-30 00:00:00
31   False 2014-09-13 00:00:00
32    True 2014-09-26 00:00:00
33   False 2014-10-04 00:00:00
34    True 2015-01-08 00:00:00
35    True 2015-01-20 00:00:00
36   False 2015-01-31 00:00:00
37   False 2015-02-14 00:00:00

我希望通过减去2012-01-02开始日期差异并成为一个int。

这是我尝试过但没有成功的原因因为prevdate没有更新到新行的日期，而是继续引用datetime的原始起始位置（2012,01,02）。我在数据帧的行中使用iterrows。

>>>for index, row in hg.iterrows():
    prevdate = datetime(2012,01,02)
    dsince = row['date']-prevdate
    prevdate = row['date']
    print dsince

结果（我也不知道如何将值修改为int）：

58 days, 0:00:00
74 days, 0:00:00
93 days, 0:00:00
127 days, 0:00:00
131 days, 0:00:00
145 days, 0:00:00
159 days, 0:00:00
285 days, 0:00:00
313 days, 0:00:00
442 days, 0:00:00
455 days, 0:00:00
479 days, 0:00:00
488 days, 0:00:00
502 days, 0:00:00
516 days, 0:00:00
596 days, 0:00:00
607 days, 0:00:00
628 days, 0:00:00
642 days, 0:00:00
656 days, 0:00:00
677 days, 0:00:00
750 days, 0:00:00
768 days, 0:00:00
782 days, 0:00:00
796 days, 0:00:00
817 days, 0:00:00
838 days, 0:00:00
931 days, 0:00:00
942 days, 0:00:00
950 days, 0:00:00
971 days, 0:00:00
985 days, 0:00:00
998 days, 0:00:00
1006 days, 0:00:00
1102 days, 0:00:00
1114 days, 0:00:00
1125 days, 0:00:00
1139 days, 0:00:00

为了使事情变得更复杂一点，我想只创建另一个列，其中行之间的日期差异为False，而不是＆＃39; not inc＆＃39;柱。

谢谢。

Answer 1

假设您的日期列已投放为datetime64：

In [61]: hg = pd.DataFrame({"not inc":[False , False, False, True, False],"date":pd.to_datetime(pd.Series(["2012-02-29", "2012-03-16", "2012-04-04", "2012-05-08", "2012-05-12"]))})

In [63]: hg.dtypes
Out[63]:
date       datetime64[ns]
not inc              bool
dtype: object

暂时过滤掉您不想包含的行：

In [64]: included = hg[hg["not inc"] == False]

使用shift获取一系列要减去的日期，在开头填写开始日期：

In [66]: prev_dates = included.date.shift().fillna(pd.datetime(2012,1,2))

In [67]: prev_dates
Out[67]:
0   2012-01-02
1   2012-02-29
2   2012-03-16
4   2012-04-04
Name: date, dtype: datetime64[ns]

减去日期并将timedelta重新设为int：

In [68]: delta = included.date - prev_dates

In [69]: delta = delta.astype("timedelta64[D]")

In [70]: delta
Out[70]:
0    58
1    16
2    19
4    38
Name: date, dtype: float64

然后concat将新系列添加到原始数据框中。

In [71]: delta.name = "delta"

In [72]: hg = pd.concat((hg, delta), axis=1)

In [73]: hg
Out[73]:
        date not inc  delta
0 2012-02-29   False     58
1 2012-03-16   False     16
2 2012-04-04   False     19
3 2012-05-08    True    NaN
4 2012-05-12   False     38

Answer 2

将第prevdate = datetime(2012,01,02)行放在循环之前。

prevdate = datetime(2012,01,02)
for index, row in hg.iterrows():
    dsince = (row['date'] - prevdate).days
    prevdate = row['date']
    print dsince

如果不起作用，请将prevdate和row['date']转换为日期。

大熊猫在一个列之间的日子

2 个答案: