Question

我想在三列合并两个数据框：电子邮件，主题和时间戳。数据帧之间的时间戳不同，因此我需要确定一组电子邮件的最接近的匹配时间戳＆amp;学科。

以下是使用针对this问题建议的最接近匹配的函数的可重现示例。

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，最接近的匹配时间戳是10:17:39，而对于b@gmail.com，最接近的匹配是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＆amp;主题。

预期结果是

修改函数以给出给定电子邮件和主题的最接近的时间戳将是有帮助的。

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

但由于没有为组对象定义函数，因此会出错。这样做的最佳方法是什么？

Answer 1

请注意，如果您在df1和df2上合并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

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']查找每个组的差异最小的行。

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

Answer 2

您希望将最接近的时间戳逻辑应用于每组＆＃39;电子邮件＆＃39;和＆＃39;主题＆＃39;

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'])
df2 = df2.set_index(['email', 'subject'])

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
    time_series = time_series.values
    timepoint = np.datetime64(timepoint)
    deltas = np.abs(np.subtract(time_series, timepoint))
    idx_closest_date = np.argmin(deltas)
    res = {"closest_date": time_series[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)

# Then group df1 as needed
grouped = df1.groupby(['email', 'subject'])

# Finally loop over the group items, finding the closest timestamps
join_ts = pd.DataFrame()
for name, group in grouped:
    try:
        join_ts = pd.concat([join_ts, group['timestamp']\
                             .apply(find_closest_date, time_series=df2.loc[name, 'timestamp'])],
                            axis=0)
    except KeyError:
        pass

df3 = pd.merge(pd.concat([df1, join_ts], axis=1), df2, left_on=['closest_date'], right_on=['timestamp'])

pandas在最接近的时间戳

2 个答案: