考虑以下迷你版本的问题：

from io import StringIO

from pandas import read_csv, to_datetime

# how close do sessions have to be to be considered equal? (in minutes)

threshold = 5

# datetime column (combination of date + start_time)

dtc = [['date', 'start_time']]

# index column (above combination)

ixc = 'date_start_time'

df1 = read_csv(StringIO(u'''

date,start_time,employee_id,session_id

01/01/2016,02:03:00,7261824,871631182

01/01/2016,06:03:00,7261824,871631183

01/01/2016,11:01:00,7261824,871631184

01/01/2016,14:01:00,7261824,871631185

'''), parse_dates=dtc)

df2 = read_csv(StringIO(u'''

date,start_time,employee_id,session_id

01/01/2016,02:03:00,7261824,871631182

01/01/2016,06:05:00,7261824,871631183

01/01/2016,11:04:00,7261824,871631184

01/01/2016,14:10:00,7261824,871631185

'''), parse_dates=dtc)

这使

>>> df1

date_start_time employee_id session_id

0 2016-01-01 02:03:00 7261824 871631182

1 2016-01-01 06:03:00 7261824 871631183

2 2016-01-01 11:01:00 7261824 871631184

3 2016-01-01 14:01:00 7261824 871631185

>>> df2

date_start_time employee_id session_id

0 2016-01-01 02:03:00 7261824 871631182

1 2016-01-01 06:05:00 7261824 871631183

2 2016-01-01 11:04:00 7261824 871631184

3 2016-01-01 14:10:00 7261824 871631185

您希望在合并时将 df2[0:3] 视为 df1[0:3] 的重复项(因为它们分别相隔不到5分钟)，但将 df1[3] 和 df2[3] 视为单独的会话 .

解决方案1：间隔匹配

这基本上就是您在编辑中建议的内容 . 您希望将两个表中的时间戳映射到以时间戳为中心的10分钟间隔，四舍五入到最接近的5分钟 .

每个间隔可以通过其中点唯一地表示，因此您可以将时间戳上的数据帧合并到最接近的5分钟 . 例如：

import numpy as np

# half-threshold in nanoseconds

threshold_ns = threshold * 60 * 1e9

# compute "interval" to which each session belongs

df1['interval'] = to_datetime(np.round(df1.date_start_time.astype(np.int64) / threshold_ns) * threshold_ns)

df2['interval'] = to_datetime(np.round(df2.date_start_time.astype(np.int64) / threshold_ns) * threshold_ns)

# join

cols = ['interval', 'employee_id', 'session_id']

print df1.merge(df2, on=cols, how='outer')[cols]

打印

interval employee_id session_id

0 2016-01-01 02:05:00 7261824 871631182

1 2016-01-01 06:05:00 7261824 871631183

2 2016-01-01 11:00:00 7261824 871631184

3 2016-01-01 14:00:00 7261824 871631185

4 2016-01-01 11:05:00 7261824 871631184

5 2016-01-01 14:10:00 7261824 871631185

请注意，这不完全正确 . 会话 df1[2] 和 df2[2] 不会被视为重复，尽管它们相距仅3分钟 . 这是因为它们位于区间边界的不同侧 .

解决方案2：一对一匹配

这是另一种方法，它取决于 df1 中的会话在 df2 中具有零或一个重复的条件 .

我们将 df1 中的时间戳替换为 df2 中与 employee_id 匹配的最接近的时间戳，并且 session_id and 距离不到5分钟 .

from datetime import timedelta

# get closest match from "df2" to row from "df1" (as long as it's below the threshold)

def closest(row):

matches = df2.loc[(df2.employee_id == row.employee_id) &

(df2.session_id == row.session_id)]

deltas = matches.date_start_time - row.date_start_time

deltas = deltas.loc[deltas <= timedelta(minutes=threshold)]

try:

return matches.loc[deltas.idxmin()]

except ValueError: # no items

return row

# replace timestamps in "df1" with closest timestamps in "df2"

df1 = df1.apply(closest, axis=1)

# join

cols = ['date_start_time', 'employee_id', 'session_id']

print df1.merge(df2, on=cols, how='outer')[cols]

打印

date_start_time employee_id session_id

0 2016-01-01 02:03:00 7261824 871631182

1 2016-01-01 06:05:00 7261824 871631183

2 2016-01-01 11:04:00 7261824 871631184

3 2016-01-01 14:01:00 7261824 871631185

4 2016-01-01 14:10:00 7261824 871631185

这种方法要慢得多，因为你必须为 df1 中的每一行搜索整个 df2 . 我写的内容可能会进一步优化，但这仍需要很长时间才能完成大型数据集 .