如何使用Python获取时间限制内的行？

Question

我从 Excel 读取了销售交易 table，我想知道第一件商品售出后 1 小时内的销售数量。设 A 为销售报表，我要创建 B.

A=
item    Location    time
X       Canada      10:03:18
X       Canada      10:08:38
X       Canada      10:24:46
X       Canada      11:16:35
X       US          10:00:16
X       US          11:52:12
Y       Canada      2:08:38
Y       Canada      4:01:48
Y       US          13:32:02
Y       US          14:07:03

B=
item    location    first sale  count
X       Canada      10:03:18    3
X       US          10:00:16    1
Y       Canada      2:08:38     1
Y       US          13:32:02    2

这是我所做的：

A= A.sort('time', ascending=True).reset_index()
sale_loc= pd.DataFrame(A.groupby(['item', 'Location'], sort = False).first()).reset_index()
for i in sale_loc.index:
    sale_cutoff = (A.time[i] + dt.timedelta(hours=1)).time

但是时间操作部分出现错误。我尝试了不同的功能，我也尝试添加一个新列 A (time+1hour) 而不是循环，但类似的问题...

Answer 1

import numpy as np
import pandas as pd

df = pd.DataFrame({'Location': ['Canada', 'Canada', 'Canada', 'Canada', 'US', 'US', 'Canada', 'Canada', 'US', 'US'], 'item': ['X', 'X', 'X', 'X', 'X', 'X', 'Y', 'Y', 'Y', 'Y'], 'time': ['10:03:18', '10:08:38', '10:24:46', '11:16:35', '10:00:16', '11:52:12', '2:08:38', '4:01:48', '13:32:02', '14:07:03']})

df['start'] = pd.to_datetime(df['time'])
grouped = df.groupby(['item', 'Location'])
df['end'] = (grouped['start'].transform(lambda grp: grp.min()+pd.Timedelta(hours=1)))
df['mask'] = (df['start'] < df['end'])

result = grouped['mask'].sum()
print(result)

产量

item  Location
X     Canada      3.0
      US          1.0
Y     Canada      1.0
      US          2.0
Name: mask, dtype: float64

主要思路是按item和Location分组，求每组最短开始时间，然后加1小时：

df['end'] = (grouped['start'].transform(lambda grp: grp.min()+pd.Timedelta(hours=1)))

transform returns 与 df 长度相同的系列，因此每一行都有一个值：

In [319]: df
Out[319]: 
  Location item      time               start                 end
0   Canada    X  10:03:18 2016-05-06 10:03:18 2016-05-06 11:03:18
1   Canada    X  10:08:38 2016-05-06 10:08:38 2016-05-06 11:03:18
2   Canada    X  10:24:46 2016-05-06 10:24:46 2016-05-06 11:03:18
3   Canada    X  11:16:35 2016-05-06 11:16:35 2016-05-06 11:03:18
4       US    X  10:00:16 2016-05-06 10:00:16 2016-05-06 11:00:16
5       US    X  11:52:12 2016-05-06 11:52:12 2016-05-06 11:00:16
6   Canada    Y   2:08:38 2016-05-06 02:08:38 2016-05-06 03:08:38
7   Canada    Y   4:01:48 2016-05-06 04:01:48 2016-05-06 03:08:38
8       US    Y  13:32:02 2016-05-06 13:32:02 2016-05-06 14:32:02
9       US    Y  14:07:03 2016-05-06 14:07:03 2016-05-06 14:32:02

现在您可以轻松识别感兴趣的行。它们是 start 小于 end:

的那些

In [320]: df['mask'] = (df['start'] < df['end'])
In [321]: df
Out[321]: 
  Location item      time               start                 end   mask
0   Canada    X  10:03:18 2016-05-06 10:03:18 2016-05-06 11:03:18   True
1   Canada    X  10:08:38 2016-05-06 10:08:38 2016-05-06 11:03:18   True
2   Canada    X  10:24:46 2016-05-06 10:24:46 2016-05-06 11:03:18   True
3   Canada    X  11:16:35 2016-05-06 11:16:35 2016-05-06 11:03:18  False
4       US    X  10:00:16 2016-05-06 10:00:16 2016-05-06 11:00:16   True
5       US    X  11:52:12 2016-05-06 11:52:12 2016-05-06 11:00:16  False
6   Canada    Y   2:08:38 2016-05-06 02:08:38 2016-05-06 03:08:38   True
7   Canada    Y   4:01:48 2016-05-06 04:01:48 2016-05-06 03:08:38  False
8       US    Y  13:32:02 2016-05-06 13:32:02 2016-05-06 14:32:02   True
9       US    Y  14:07:03 2016-05-06 14:07:03 2016-05-06 14:32:02   True

再次按 item 和 Location 分组，通过对每组 mask 为真的次数求和来找到所需的结果：

result = grouped['mask'].sum()

Answer 2

我没有生成整个代码，而是专注于您所说的引发错误的部分。这是将一个小时添加到您列出的时间的工作示例：

sale_time = ['10:03:18', '10:08:38', '11:16:35', '10:00:16']

import datetime
for i in sale_time:
    sale_time1 = datetime.time(hour = int(i[0:2]), minute=int(i[3:5]), second=int(i[6:8]))
    print(sale_time1)
    sale_cutoff = datetime.time(sale_time1.hour+1, sale_time1.minute, sale_time1.second)
    print(sale_cutoff)

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