计算熊猫中另一列上连续日期列与groupby之间的差异？答案

【问题标题】：Calculate difference between successive date column with groupby on another column in pandas?计算熊猫中另一列上连续日期列与groupby之间的差异？
【发布时间】：2020-05-07 03:01:36
【问题描述】：

我有一个熊猫数据框，

data = pd.DataFrame([['Car','2019-01-06T21:44:09Z'],
                     ['Train','2019-01-06T19:44:09Z'],
                     ['Train','2019-01-02T19:44:09Z'],
                     ['Car','2019-01-08T06:44:09Z'],
                     ['Car','2019-01-06T18:44:09Z'],
                     ['Train','2019-01-04T19:44:09Z'],
                     ['Car','2019-01-05T16:34:09Z'],
                     ['Train','2019-01-08T19:44:09Z'],
                     ['Car','2019-01-07T14:44:09Z'],
                     ['Car','2019-01-06T11:44:09Z'],
                     ['Train','2019-01-10T19:44:09Z'],
                     ], 
                    columns=['Type', 'Date'])

在按日期排序后，需要找出每种类型的连续日期之间的差异

最终数据看起来像

data = pd.DataFrame([['Car','2019-01-06T21:44:09Z',1],
                     ['Train','2019-01-06T19:44:09Z',4],
                     ['Train','2019-01-02T19:44:09Z',0],
                     ['Car','2019-01-08T06:44:09Z',3],
                     ['Car','2019-01-06T18:44:09Z',1],
                     ['Train','2019-01-04T19:44:09Z',2],
                     ['Car','2019-01-05T16:34:09Z',0],
                     ['Train','2019-01-08T19:44:09Z',6],
                     ['Car','2019-01-07T14:44:09Z',2],
                     ['Car','2019-01-06T11:44:09Z',1],
                     ['Train','2019-01-10T19:44:09Z',8],
                     ], 
                    columns=['Type', 'Date','diff'])

这里，Type Car min(Date) 是 2019-01-05T16:34:09Z，所以差异从 0 开始，然后第二个日期是 2019-01-06T18:44:09Z 和 2019-01-06T11:44 :09Z，所以 diff 是 1 天（这里不确定是否可以包括时间）等等.. 对于 Type Train min(Date) 是 2019-01-02T19:44:09Z，所以 diff 是 0 然后是 2019-01-04T19:44:09Z 所以 2 天 diff

我尝试了 groupby，但不确定如何包含按日期排序

data['diff'] = data.groupby('Type')['Date'].diff() / np.timedelta64(1, 'D')

【问题讨论】：

标签： python pandas

【解决方案1】：

将pandas.DataFrame.groupby 与dt.date 一起使用：

df['diff'] = df.groupby('Type')['Date'].apply(lambda x: x.dt.date - x.min().date())

输出：

     Type                      Date   diff
0     Car 2019-01-06 21:44:09+00:00 1 days
1   Train 2019-01-06 19:44:09+00:00 4 days
2   Train 2019-01-02 19:44:09+00:00 0 days
3     Car 2019-01-08 06:44:09+00:00 3 days
4     Car 2019-01-06 18:44:09+00:00 1 days
5   Train 2019-01-04 19:44:09+00:00 2 days
6     Car 2019-01-05 16:34:09+00:00 0 days
7   Train 2019-01-08 19:44:09+00:00 6 days
8     Car 2019-01-07 14:44:09+00:00 2 days
9     Car 2019-01-06 11:44:09+00:00 1 days
10  Train 2019-01-10 19:44:09+00:00 8 days

如果您希望他们成为int，请添加dt.days：

df['diff'] = df.groupby('Type')['Date'].apply(lambda x: x.dt.date - x.min().date()).dt.days

输出：

     Type                      Date  diff
0     Car 2019-01-06 21:44:09+00:00     1
1   Train 2019-01-06 19:44:09+00:00     4
2   Train 2019-01-02 19:44:09+00:00     0
3     Car 2019-01-08 06:44:09+00:00     3
4     Car 2019-01-06 18:44:09+00:00     1
5   Train 2019-01-04 19:44:09+00:00     2
6     Car 2019-01-05 16:34:09+00:00     0
7   Train 2019-01-08 19:44:09+00:00     6
8     Car 2019-01-07 14:44:09+00:00     2
9     Car 2019-01-06 11:44:09+00:00     1
10  Train 2019-01-10 19:44:09+00:00     8

【讨论】：

【解决方案2】：

首先将日期转换为日期到其他列中
使用 lambda 函数减去日期的最小值并使用 dt.days 查找天数
然后删除额外的日期列

data['Date_date'] = pd.to_datetime(data['Date']).dt.date
data['diff'] = data.groupby(['Type'])['Date_date'].apply(lambda x:(x-x.min()).dt.days)
data.drop(['Date_date'],axis=1,inplace=True,errors='ignore')
print(data)

     Type                  Date  diff
0     Car  2019-01-06T21:44:09Z     1
1   Train  2019-01-06T19:44:09Z     4
2   Train  2019-01-02T19:44:09Z     0
3     Car  2019-01-08T06:44:09Z     3
4     Car  2019-01-06T18:44:09Z     1
5   Train  2019-01-04T19:44:09Z     2
6     Car  2019-01-05T16:34:09Z     0
7   Train  2019-01-08T19:44:09Z     6
8     Car  2019-01-07T14:44:09Z     2
9     Car  2019-01-06T11:44:09Z     1
10  Train  2019-01-10T19:44:09Z     8

【讨论】：

【解决方案3】：

从transform直接减法

s = pd.to_datetime(data['Date']).dt.date
data['diff'] = (s - s.groupby(data.Type).transform('min')).dt.days

Out[36]:
     Type                  Date  diff
0     Car  2019-01-06T21:44:09Z     1
1   Train  2019-01-06T19:44:09Z     4
2   Train  2019-01-02T19:44:09Z     0
3     Car  2019-01-08T06:44:09Z     3
4     Car  2019-01-06T18:44:09Z     1
5   Train  2019-01-04T19:44:09Z     2
6     Car  2019-01-05T16:34:09Z     0
7   Train  2019-01-08T19:44:09Z     6
8     Car  2019-01-07T14:44:09Z     2
9     Car  2019-01-06T11:44:09Z     1
10  Train  2019-01-10T19:44:09Z     8

【讨论】：