我有两个数据帧，并希望通过时间戳匹配它们。例如：
A
Time X
0 05-01-2017 09:08 3
1 05-01-2017 09:09 6
2 07-01-2017 09:09 5
3 07-01-2017 09:19 4
4 07-01-2017 09:19 8
5 07-02-2017 09:19 7
6 07-02-2017 09:19 5
B
Time Y
0 06-01-2017 14:45 1
1 04-01-2017 03:31 9
2 07-01-2017 03:31 4
3 07-01-2017 14:57 5
4 09-01-2017 14:57 7
有太多数据可以将df_A中的每个项目与df_B中的每个项目进行比较。我希望找到在受控时间阈值内的每个匹配，例如2天。那就是：
dT = Time A – Time B
-2 < dT < 2
结果应为：
C
Index A Time A X Index B Time B Y dT
0 05-01-2017 09:08 3 0 06-01-2017 14:45 1 -1.2
0 05-01-2017 09:08 3 1 04-01-2017 03:31 9 1.2
0 05-01-2017 09:08 3 2 07-01-2017 03:31 4 -1.8
1 05-01-2017 09:09 6 0 06-01-2017 14:45 1 -1.2
1 05-01-2017 09:09 6 1 04-01-2017 03:31 9 1.2
1 05-01-2017 09:09 6 2 07-01-2017 03:31 4 -1.8
2 07-01-2017 09:09 5 0 06-01-2017 14:45 1 0.8
2 07-01-2017 09:09 5 2 07-01-2017 03:31 4 0.2
2 07-01-2017 09:09 5 3 07-01-2017 14:57 5 -0.2
3 07-01-2017 09:19 4 0 06-01-2017 14:45 1 0.8
3 07-01-2017 09:19 4 2 07-01-2017 03:31 4 0.2
3 07-01-2017 09:19 4 3 07-01-2017 14:57 5 -0.2
4 07-01-2017 09:19 8 0 06-01-2017 14:45 1 0.8
4 07-01-2017 09:19 8 2 07-01-2017 03:31 4 0.2
4 07-01-2017 09:19 8 3 07-01-2017 14:57 5 -0.2
5 07-02-2017 09:19 7
6 07-02-2017 09:19 5
4 09-01-2017 14:57 7
我已尝试过以下代码，但它无效：
import pandas as pd
import datetime as dt
from datetime import timedelta
# Data
df_A = pd.DataFrame({'X':[3,6,5,4,8,7,5], 'Time_A': [dt.datetime(2017,1,5,9,8), dt.datetime(2017,1,5,9,9), dt.datetime(2017,1,7,9,19), dt.datetime(2017,1,7,9,19), dt.datetime(2017,1,7,9,19), dt.datetime(2017,2,7,9,19), dt.datetime(2017,2,7,9,19)]})
df_B = pd.DataFrame({'Y':[1,9,4,5,7], 'Time_B': [dt.datetime(2017,1,6,14,45), dt.datetime(2017,1,4,3,31), dt.datetime(2017,1,7,3,31), dt.datetime(2017,1,7,14,57), dt.datetime(2017,1,9,14,57)]})
# Match
def slice_datetime(Time, window):
return (Time + timedelta(hours=window)).strftime('%Y-%m-%d %H:%m')
lst = []
for Time in df_A[['X', 'Time_A']].iterrows():
tmp = df_B.ix[slice_datetime(Time,-48):slice_datetime(Time,48)] # Define the time threshold (hours)
if not tmp.empty:
_match = pd.DataFrame()
for Time_A, (X, Y, Time_B) in tmp.iterrows():
lst.append([X, Y, Time_A, Time_B])
df_C = pd.DataFrame(lst, columns = ['X', 'Y', 'Time_A', 'Time_B'])
答案 0 :(得分：1)
您可以创建两个具有时间边界的新列
df_A["start_date"] = df_A["Time_A"]+datetime.timedelta(days=-2)
df_A["end_date"] = df_A["Time_A"]+datetime.timedelta(days=2)
然后使用条件
连接两个数据帧(df_B.Time_B >= df_A.start_date)&(df_B.Time_B <= df_A.end_date)
希望这有帮助！
答案 1 :(得分：1)
以下是如何在没有循环的情况下执行此操作：
import pandas as pd
df_A = pd.DataFrame({'X':[3,6,5,4,8,7,5],
'Time_A': [pd.datetime(2017,1,5,9,8), pd.datetime(2017,1,5,9,9),
pd.datetime(2017,1,7,9,19), pd.datetime(2017,1,7,9,19),
pd.datetime(2017,1,7,9,19), pd.datetime(2017,2,7,9,19),
pd.datetime(2017,2,7,9,19)]})
df_B = pd.DataFrame({'Y':[1,9,4,5,7],
'Time_B': [pd.datetime(2017,1,6,14,45), pd.datetime(2017,1,4,3,31),
pd.datetime(2017,1,7,3,31), pd.datetime(2017,1,7,14,57),
pd.datetime(2017,1,9,14,57)]})
#first reset_index and rename
df_A = df_A.reset_index().rename(columns = {'index':'index_A'})
df_B = df_B.reset_index().rename(columns = {'index':'index_B'})
#then create a list of index_B where time_B is within 2 days for each time_A
time_delta = pd.Timedelta(days=2) #check the documentation for more parameter
df_A['list_B'] = (df_A['Time_A'].apply(lambda time_A:
df_B.index_B[(time_A - time_delta <= df_B['Time_B']) &
(time_A + time_delta >= df_B['Time_B'])].tolist()))
#now use pd.Series and stack, with reset_index drop and rename
# for finally merge to achieve your goal
df_C = (df_A.set_index(['index_A','Time_A','X'])['list_B']
.apply(pd.Series).stack().astype(int)
.reset_index().drop('level_3',1).rename(columns={0:'index_B'})
.merge(df_B).sort_values('index_A'))
# Create the columns dT
df_C['dT'] = ((df_C['Time_A'] - df_C['Time_B']).dt.total_seconds()/(24.*3600.)).round(1)
#add the time from df_A and df_B without corresponding time in the other df
# using append and ~ with isin
df_C = (df_C.append(df_A[~df_A['Time_A'].isin(df_C['Time_A'])].drop('list_B',1))
.append(df_B[~df_B['Time_B'].isin(df_C['Time_B'])]).fillna(''))
您可能需要重新排序列，但您应该获得所需的输出