我有四个pandas DataFrames（A，B，C和D）。 A有一系列时间戳和一列引用其他DataFrame之一：
A
Timestamp Source
----------- ------
2012-4-3 B
2013-12-20 C
2012-3-5 C
2014-12-7 D
2012-7-10 B
...
其他DataFrame包含更多数据：
B
Timestamp Foo Bar
----------- ---- ----
2012-1-1 1.5 1.3
2012-1-2 2.3 5.6
2012-1-3 3.4 3.3
...
2014-3-31 0.8 2.1
C
Timestamp Foo Bar
----------- ---- ----
2012-1-1 9.2 5.6
2012-1-2 4.8 7.6
2012-1-3 2.7 6.4
...
2014-3-31 7.0 6.5
D
Timestamp Foo Bar
----------- ---- ----
2012-1-1 6.8 4.2
2012-1-2 4.2 9.3
2012-1-3 5.5 0.7
...
2014-3-31 6.3 2.0
我想从A，B，C和D构建一个包含三列（Timestamp，Foo的数据框架} {和Bar）其中Foo和Bar的值来自Timestamp中列为Source的DataFrame中的相应A 。
并非A中的所有时间戳都显示在其他三个DataFrame中，在这种情况下，我希望Foo和Bar的值为np.nan。并非B，C和D中的所有时间戳都显示在A中，并且不会出现在最终的DataFrame中。
我目前的方法是循环遍历A中的每一行并返回相应的Source DataFrame中的值：
srcs = {'B': B, 'C': C, 'D': D}
A['Foo'] = np.nan
A['Bar'] = np.nan
for i in range(len(A)):
ts = A.iloc[i].Timestamp
src = A.iloc[i].Source
A.iloc[i].Foo = srcs[src][srcs[src].Timestamp == ts].Foo
A.iloc[i].Bar = srcs[src][srcs[src].Timestamp == ts].Bar
必须有更高效，更多的Pandithic（？）方式来执行此操作？
答案 0 :(得分：2)
看起来您可以使用多索引来执行此操作。您的索引将包含时间戳和来源。您可以使用DataFrame上的set_index方法进行此操作。
以下是一些代码，用于创建一些假的DataFrame，每个都有MultiIndex。
# Imports for creating fake data
from random import random
from random import choice
# Setup the sample data
A = pd.DataFrame({'TimeStamp':range(20), 'Source':[choice(others) for i in range(20)]})
# Create the MultiIndex on A
A.set_index(['TimeStamp', 'Source'], inplace=True)
A['Bar'] = [np.nan] * len(A)
A['Foo'] = [np.nan] * len(A)
B = pd.DataFrame({'TimeStamp':range(5),
'Foo':[random()*5+5 for i in range(5)],
'Bar':[random()*5+5 for i in range(5)]})
C = pd.DataFrame({'TimeStamp':range(5,10),
'Foo':[random()*5+5 for i in range(5)],
'Bar':[random()*5+5 for i in range(5)]})
D = pd.DataFrame({'TimeStamp':range(10,15),
'Foo':[random()*5+5 for i in range(5)],
'Bar':[random()*5+5 for i in range(5)]})
sources = {'B':B, 'C':C, 'D':D}
# create the MultiIndex on the Source data sets
for s, df in sources.items():
df['Source'] = [s]*len(df)
df.set_index(['TimeStamp', 'Source'], inplace=True)
现在，您可以使用A上的索引索引源数据集（B，C和D）。
for s, df in sources.items():
temp = df.loc[A.index] # the source data set indexed by A's index
# this will contain NaN's where df does not
# have corresponding index entries
temp.dropna(inplace=True) # dropping the NaN values leaves you with
# only the values in df matching the index in A
if len(temp) > 0:
A.loc[temp.index] = temp # now assign the data to A
print(A)
结果如下：
Bar Foo
TimeStamp Source
0 D NaN NaN
1 C NaN NaN
2 D NaN NaN
3 B 7.927154 8.581380
4 B 7.638422 5.970348
5 D NaN NaN
6 C 6.938001 6.417248
7 B NaN NaN
8 C 5.131940 9.144621
9 B NaN NaN
10 D 9.186963 5.991877
11 D 8.070543 7.735040
12 C NaN NaN
13 B NaN NaN
14 C NaN NaN
15 D NaN NaN
16 C NaN NaN
17 C NaN NaN
18 C NaN NaN
19 B NaN NaN
答案 1 :(得分：1)
import pandas as pd
from StringIO import StringIO
texta = """Timestamp Source
2012-4-3 B
2012-4-2 B
2013-12-20 C
2012-3-5 C
2014-12-7 D
2012-7-10 B"""
A = pd.read_csv(StringIO(texta), delim_whitespace=1, parse_dates=[0])
textb = """Timestamp Foo Bar
2012-1-1 1.5 1.3
2012-4-3 3.1 4.1
2012-1-2 2.3 5.6
2012-1-3 3.4 3.3
2014-3-31 0.8 2.1"""
B = pd.read_csv(StringIO(textb), delim_whitespace=1, parse_dates=[0])
textc = """Timestamp Foo Bar
2012-1-1 9.2 5.6
2012-3-5 4.8 7.6
2012-1-2 4.8 7.6
2012-1-3 2.7 6.4
2014-3-31 7.0 6.5"""
C = pd.read_csv(StringIO(textc), delim_whitespace=1, parse_dates=[0])
textd = """Timestamp Foo Bar
2012-1-1 6.8 4.2
2012-1-2 4.2 9.3
2012-1-3 5.5 0.7
2014-3-31 6.3 2.0"""
D = pd.read_csv(StringIO(textd), delim_whitespace=1, parse_dates=[0])
然后我与pd.concat仅B C和D
bdf = pd.concat([B, C, D], keys=['B', 'C', 'D'])
bdf.reset_index(level=1, inplace=1, drop=1)
bdf.index.name = 'Source'
bdf.reset_index(inplace=1)
print bdf
看起来像这样：
Source Timestamp Foo Bar
0 B 2012-01-01 1.5 1.3
1 B 2012-04-03 3.1 4.1
2 B 2012-01-02 2.3 5.6
3 B 2012-01-03 3.4 3.3
4 B 2014-03-31 0.8 2.1
5 C 2012-01-01 9.2 5.6
6 C 2012-03-05 4.8 7.6
7 C 2012-01-02 4.8 7.6
8 C 2012-01-03 2.7 6.4
9 C 2014-03-31 7.0 6.5
10 D 2012-01-01 6.8 4.2
11 D 2012-01-02 4.2 9.3
12 D 2012-01-03 5.5 0.7
13 D 2014-03-31 6.3 2.0
一个简单的合并
A.merge(bdf, how='left')
看起来像：
Timestamp Source Foo Bar
0 2012-04-03 B 3.1 4.1
1 2012-04-02 B NaN NaN
2 2013-12-20 C NaN NaN
3 2012-03-05 C 4.8 7.6
4 2014-12-07 D NaN NaN
5 2012-07-10 B NaN NaN