python - Pandas 在最近的时间戳上合并数据帧

Question

我想在三列上合并两个数据框:电子邮件、主题和时间戳。
数据帧之间的时间戳不同，因此我需要为一组电子邮件和主题确定最匹配的时间戳。

以下是使用为 this 建议的最接近匹配函数的可重现示例题。

import numpy as np
import pandas as pd
from pandas.io.parsers import StringIO

def find_closest_date(timepoint, time_series, add_time_delta_column=True):
   # takes a pd.Timestamp() instance and a pd.Series with dates in it
   # calcs the delta between `timepoint` and each date in `time_series`
   # returns the closest date and optionally the number of days in its time delta
   deltas = np.abs(time_series - timepoint)
   idx_closest_date = np.argmin(deltas)
   res = {"closest_date": time_series.ix[idx_closest_date]}
   idx = ['closest_date']
   if add_time_delta_column:
      res["closest_delta"] = deltas[idx_closest_date]
      idx.append('closest_delta')
   return pd.Series(res, index=idx)


a = """timestamp,email,subject
2016-07-01 10:17:00,a@gmail.com,subject3
2016-07-01 02:01:02,a@gmail.com,welcome
2016-07-01 14:45:04,a@gmail.com,subject3
2016-07-01 08:14:02,a@gmail.com,subject2
2016-07-01 16:26:35,a@gmail.com,subject4
2016-07-01 10:17:00,b@gmail.com,subject3
2016-07-01 02:01:02,b@gmail.com,welcome
2016-07-01 14:45:04,b@gmail.com,subject3
2016-07-01 08:14:02,b@gmail.com,subject2
2016-07-01 16:26:35,b@gmail.com,subject4
"""

b = """timestamp,email,subject,clicks,var1
2016-07-01 02:01:14,a@gmail.com,welcome,1,1
2016-07-01 08:15:48,a@gmail.com,subject2,2,2
2016-07-01 10:17:39,a@gmail.com,subject3,1,7
2016-07-01 14:46:01,a@gmail.com,subject3,1,2
2016-07-01 16:27:28,a@gmail.com,subject4,1,2
2016-07-01 10:17:05,b@gmail.com,subject3,0,0
2016-07-01 02:01:03,b@gmail.com,welcome,0,0
2016-07-01 14:45:05,b@gmail.com,subject3,0,0
2016-07-01 08:16:00,b@gmail.com,subject2,0,0
2016-07-01 17:00:00,b@gmail.com,subject4,0,0
"""

请注意，对于 a@gmail.com，最匹配的时间戳是 10:17:39，而对于 b@gmail.com，最匹配的时间戳是 10:17:05。

a = """timestamp,email,subject
2016-07-01 10:17:00,a@gmail.com,subject3
2016-07-01 10:17:00,b@gmail.com,subject3
"""

b = """timestamp,email,subject,clicks,var1
2016-07-01 10:17:39,a@gmail.com,subject3,1,7
2016-07-01 10:17:05,b@gmail.com,subject3,0,0
"""
df1 = pd.read_csv(StringIO(a), parse_dates=['timestamp'])
df2 = pd.read_csv(StringIO(b), parse_dates=['timestamp'])

df1[['closest', 'time_bt_x_and_y']] = df1.timestamp.apply(find_closest_date, args=[df2.timestamp])
df1

df3 = pd.merge(df1, df2, left_on=['email','subject','closest'], right_on=['email','subject','timestamp'],how='left')

df3
timestamp_x        email   subject             closest  time_bt_x_and_y         timestamp_y  clicks  var1
  2016-07-01 10:17:00  a@gmail.com  subject3 2016-07-01 10:17:05         00:00:05                 NaT     NaN   NaN
  2016-07-01 02:01:02  a@gmail.com   welcome 2016-07-01 02:01:03         00:00:01                 NaT     NaN   NaN
  2016-07-01 14:45:04  a@gmail.com  subject3 2016-07-01 14:45:05         00:00:01                 NaT     NaN   NaN
  2016-07-01 08:14:02  a@gmail.com  subject2 2016-07-01 08:15:48         00:01:46 2016-07-01 08:15:48     2.0   2.0
  2016-07-01 16:26:35  a@gmail.com  subject4 2016-07-01 16:27:28         00:00:53 2016-07-01 16:27:28     1.0   2.0
  2016-07-01 10:17:00  b@gmail.com  subject3 2016-07-01 10:17:05         00:00:05 2016-07-01 10:17:05     0.0   0.0
  2016-07-01 02:01:02  b@gmail.com   welcome 2016-07-01 02:01:03         00:00:01 2016-07-01 02:01:03     0.0   0.0
  2016-07-01 14:45:04  b@gmail.com  subject3 2016-07-01 14:45:05         00:00:01 2016-07-01 14:45:05     0.0   0.0
  2016-07-01 08:14:02  b@gmail.com  subject2 2016-07-01 08:15:48         00:01:46                 NaT     NaN   NaN
  2016-07-01 16:26:35  b@gmail.com  subject4 2016-07-01 16:27:28         00:00:53                 NaT     NaN   NaN

结果是错误的，主要是因为最近的日期不正确，因为它没有考虑电子邮件和主题。

预期的结果是



修改函数以提供给定电子邮件和主题的最接近的时间戳会很有帮助。

df1.groupby(['email','subject'])['timestamp'].apply(find_closest_date, args=[df1.timestamp])

但这会产生错误，因为该函数未为组对象定义。
这样做的最佳方法是什么？

Answer 1

请注意，如果您合并 df1和 df2在 email和 subject ，那么结果
具有所有可能的相关时间戳配对:

In [108]: result = pd.merge(df1, df2, how='left', on=['email','subject'], suffixes=['', '_y']); result
Out[108]: 
             timestamp        email   subject         timestamp_y  clicks  var1
0  2016-07-01 10:17:00  a@gmail.com  subject3 2016-07-01 10:17:39       1     7
1  2016-07-01 10:17:00  a@gmail.com  subject3 2016-07-01 14:46:01       1     2
2  2016-07-01 02:01:02  a@gmail.com   welcome 2016-07-01 02:01:14       1     1
3  2016-07-01 14:45:04  a@gmail.com  subject3 2016-07-01 10:17:39       1     7
4  2016-07-01 14:45:04  a@gmail.com  subject3 2016-07-01 14:46:01       1     2
5  2016-07-01 08:14:02  a@gmail.com  subject2 2016-07-01 08:15:48       2     2
6  2016-07-01 16:26:35  a@gmail.com  subject4 2016-07-01 16:27:28       1     2
7  2016-07-01 10:17:00  b@gmail.com  subject3 2016-07-01 10:17:05       0     0
8  2016-07-01 10:17:00  b@gmail.com  subject3 2016-07-01 14:45:05       0     0
9  2016-07-01 02:01:02  b@gmail.com   welcome 2016-07-01 02:01:03       0     0
10 2016-07-01 14:45:04  b@gmail.com  subject3 2016-07-01 10:17:05       0     0
11 2016-07-01 14:45:04  b@gmail.com  subject3 2016-07-01 14:45:05       0     0
12 2016-07-01 08:14:02  b@gmail.com  subject2 2016-07-01 08:16:00       0     0
13 2016-07-01 16:26:35  b@gmail.com  subject4 2016-07-01 17:00:00       0     0

您现在可以获取每行时间戳差异的绝对值:

result['diff'] = (result['timestamp_y'] - result['timestamp']).abs()

然后使用

idx = result.groupby(['timestamp','email','subject'])['diff'].idxmin()
result = result.loc[idx]

根据 ['timestamp','email','subject'] 为每个组找到差异最小的行.

import numpy as np
import pandas as pd
from pandas.io.parsers import StringIO

a = """timestamp,email,subject
2016-07-01 10:17:00,a@gmail.com,subject3
2016-07-01 02:01:02,a@gmail.com,welcome
2016-07-01 14:45:04,a@gmail.com,subject3
2016-07-01 08:14:02,a@gmail.com,subject2
2016-07-01 16:26:35,a@gmail.com,subject4
2016-07-01 10:17:00,b@gmail.com,subject3
2016-07-01 02:01:02,b@gmail.com,welcome
2016-07-01 14:45:04,b@gmail.com,subject3
2016-07-01 08:14:02,b@gmail.com,subject2
2016-07-01 16:26:35,b@gmail.com,subject4
"""

b = """timestamp,email,subject,clicks,var1
2016-07-01 02:01:14,a@gmail.com,welcome,1,1
2016-07-01 08:15:48,a@gmail.com,subject2,2,2
2016-07-01 10:17:39,a@gmail.com,subject3,1,7
2016-07-01 14:46:01,a@gmail.com,subject3,1,2
2016-07-01 16:27:28,a@gmail.com,subject4,1,2
2016-07-01 10:17:05,b@gmail.com,subject3,0,0
2016-07-01 02:01:03,b@gmail.com,welcome,0,0
2016-07-01 14:45:05,b@gmail.com,subject3,0,0
2016-07-01 08:16:00,b@gmail.com,subject2,0,0
2016-07-01 17:00:00,b@gmail.com,subject4,0,0
"""

df1 = pd.read_csv(StringIO(a), parse_dates=['timestamp'])
df2 = pd.read_csv(StringIO(b), parse_dates=['timestamp'])

result = pd.merge(df1, df2, how='left', on=['email','subject'], suffixes=['', '_y'])
result['diff'] = (result['timestamp_y'] - result['timestamp']).abs()
idx = result.groupby(['timestamp','email','subject'])['diff'].idxmin()
result = result.loc[idx].drop(['timestamp_y','diff'], axis=1)
result = result.sort_index()
print(result)

产量

             timestamp        email   subject  clicks  var1
0  2016-07-01 10:17:00  a@gmail.com  subject3       1     7
2  2016-07-01 02:01:02  a@gmail.com   welcome       1     1
4  2016-07-01 14:45:04  a@gmail.com  subject3       1     2
5  2016-07-01 08:14:02  a@gmail.com  subject2       2     2
6  2016-07-01 16:26:35  a@gmail.com  subject4       1     2
7  2016-07-01 10:17:00  b@gmail.com  subject3       0     0
9  2016-07-01 02:01:02  b@gmail.com   welcome       0     0
11 2016-07-01 14:45:04  b@gmail.com  subject3       0     0
12 2016-07-01 08:14:02  b@gmail.com  subject2       0     0
13 2016-07-01 16:26:35  b@gmail.com  subject4       0     0