【问题标题】:pandas merge dataframes on closest timestamp熊猫在最近的时间戳上合并数据帧
【发布时间】:2016-12-12 23:17:27
【问题描述】:

我想合并三个列上的两个数据框: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接近的匹配时间戳。

下面是一个可重现的示例,它使用为this 问题建议的最接近匹配函数。

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

结果错误,主要是因为没有考虑email & subject,所以最近的日期不正确。

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,因为没有为组对象定义函数。 这样做的最佳方法是什么?

【问题讨论】:

  • 请不要将 png 用于代码或数据。
  • 好的,你想要什么格式?
  • 您的预期输出是文本;将其作为文本而不是图像添加到您的帖子中。

标签: python pandas merge


【解决方案1】:

您希望对每组“电子邮件”和“主题”应用最接近的时间戳逻辑

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'])

【讨论】:

  • 抱歉,这并没有给出预期的结果。
  • 那么它给出了什么?一个错误,还有什么?你能更具体一点吗?请。
  • 我帖子中的图片显示了预期的结果。主要问题是最接近的时间戳是错误的,因为它没有考虑其他两个维度,即电子邮件和主题。如果您查看内部连接的结果,它只包含 5 封电子邮件,但应该显示 10 封。(请参阅我帖子中的图片)。
  • 哦,明白了!您需要给定电子邮件和主题的最接近的时间戳。我正在编辑我的答案,它应该在编辑后工作。
  • 谢谢卡西克!!这很有帮助!输出几乎是正确的。请注意,b@gmail.com 应该只有 0 次点击和 var1。将df3.sort_values(['email_x']) 与预期输出进行比较。电子邮件 B 在所有情况下都是零。
【解决方案2】:

请注意,如果在email 和subject 上合并df1 和df2,则结果 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']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

【讨论】:

    猜你喜欢
    • 1970-01-01
    • 1970-01-01
    • 2021-06-22
    • 1970-01-01
    • 2021-11-20
    • 2013-01-08
    • 2016-03-23
    • 2019-06-23
    • 1970-01-01
    相关资源
    最近更新 更多