我有一个如下数据框:
dates 0
numbers letters
0 a 2013-01-01 0.261092
2013-01-02 -1.267770
2013-01-03 0.008230
b 2013-01-01 -1.515866
2013-01-02 0.351942
2013-01-03 -0.245463
c 2013-01-01 -0.253103
2013-01-02 -0.385411
2013-01-03 -1.740821
1 a 2013-01-01 -0.108325
2013-01-02 -0.212350
2013-01-03 0.021097
b 2013-01-01 -1.922214
2013-01-02 -1.769003
2013-01-03 -0.594216
c 2013-01-01 -0.419775
2013-01-02 1.511700
2013-01-03 0.994332
2 a 2013-01-01 -0.020299
2013-01-02 -0.749474
2013-01-03 -1.478558
b 2013-01-01 -1.357671
2013-01-02 0.161185
2013-01-03 -0.658246
c 2013-01-01 -0.564796
2013-01-02 -0.333106
2013-01-03 -2.814611
现在我得到了一个列表:
numbers letters
0 0 b
1 1 c
我需要选择索引满足列表的数据。答案如下:
dates 0
numbers letters
0 b 2013-01-01 -1.515866
2013-01-02 0.351942
2013-01-03 -0.245463
1 c 2013-01-01 -0.419775
2013-01-02 1.511700
2013-01-03 0.994332
如何从MultiIndex的Dataframe中选择特定数据?
答案 0 :(得分:3)
您也可以使用索引交集:
In [39]: l
Out[39]:
numbers letters
0 0 b
1 1 c
In [40]: df.loc[df.index.intersection(l.set_index(['numbers','letters']).index)]
Out[40]:
dates 0
numbers letters
0 b 2013-01-01 -1.515866
b 2013-01-02 0.351942
b 2013-01-03 -0.245463
1 c 2013-01-01 -0.108325
c 2013-01-02 -0.212350
c 2013-01-03 0.021097
c 2013-01-01 -0.419775
c 2013-01-02 1.511700
c 2013-01-03 0.994332
或more straightforward and faster solution from @Javier:
In [155]: df.loc[l.set_index(['numbers','letters']).index]
Out[155]:
dates 0
numbers letters
0 b 2013-01-01 -1.515866
b 2013-01-02 0.351942
b 2013-01-03 -0.245463
1 c 2013-01-01 -0.108325
c 2013-01-02 -0.212350
c 2013-01-03 0.021097
c 2013-01-01 -0.419775
c 2013-01-02 1.511700
c 2013-01-03 0.994332
<强>定时:强>
表示27.000行多指数DF
In [156]: df = pd.concat([df.reset_index()] * 10**3, ignore_index=True).set_index(['numbers','letters'])
In [157]: df.shape
Out[157]: (27000, 2)
In [158]: %%timeit
...: q = l.apply(lambda r: "(numbers == {} and letters == '{}')".format(r.numbers, r.letters),
...: axis=1) \
...: .str.cat(sep=' or ')
...: df.query(q)
...:
10 loops, best of 3: 21.3 ms per loop
In [159]: %%timeit
...: df.loc[l.set_index(['numbers','letters']).index]
...:
10 loops, best of 3: 20.2 ms per loop
In [160]: %%timeit
...: df.loc[df.index.intersection(l.set_index(['numbers','letters']).index)]
...:
10 loops, best of 3: 27.2 ms per loop
270.000行多指数DF
In [163]: %%timeit
...: q = l.apply(lambda r: "(numbers == {} and letters == '{}')".format(r.numbers, r.letters),
...: axis=1) \
...: .str.cat(sep=' or ')
...: df.query(q)
...:
10 loops, best of 3: 117 ms per loop
In [164]: %%timeit
...: df.loc[l.set_index(['numbers','letters']).index]
...:
1 loop, best of 3: 142 ms per loop
In [165]: %%timeit
...: df.loc[df.index.intersection(l.set_index(['numbers','letters']).index)]
...:
10 loops, best of 3: 185 ms per loop
结论: df.query()
方法使用numexpr
模块internaly似乎对更大的DF更快
答案 1 :(得分:1)
假设您有以下DF,其中包含您想要获取的值:
In [28]: l
Out[28]:
numbers letters
0 0 b
1 1 c
如果您需要选择numbers
为0
或1
且letters
位于['b','c']
的所有行,您可以使用df.query()
方法如下:
In [29]: df.query("numbers in @l.numbers and letters in @l.letters")
Out[29]:
dates 0
numbers letters
0 b 2013-01-01 -1.515866
b 2013-01-02 0.351942
b 2013-01-03 -0.245463
c 2013-01-01 -0.253103
c 2013-01-02 -0.385411
c 2013-01-03 -1.740821
1 c 2013-01-01 -0.108325
c 2013-01-02 -0.212350
c 2013-01-03 0.021097
b 2013-01-01 -1.922214
b 2013-01-02 -1.769003
b 2013-01-03 -0.594216
c 2013-01-01 -0.419775
c 2013-01-02 1.511700
c 2013-01-03 0.994332
或简单地说:
df.query("numbers in [0,1] and letters in ['b','c']")
更新:如果必须完全匹配,例如(0, 'b')
和(1, 'c')
:
In [14]: q = l.apply(lambda r: "(numbers == {} and letters == '{}')".format(r.numbers, r.letters),
...: axis=1) \
...: .str.cat(sep=' or ')
...:
In [15]: q
Out[15]: "(numbers == 0 and letters == 'b') or (numbers == 1 and letters == 'c')"
In [16]: df.query(q)
Out[16]:
dates 0
numbers letters
0 b 2013-01-01 -1.515866
b 2013-01-02 0.351942
b 2013-01-03 -0.245463
1 c 2013-01-01 -0.108325
c 2013-01-02 -0.212350
c 2013-01-03 0.021097
c 2013-01-01 -0.419775
c 2013-01-02 1.511700
c 2013-01-03 0.994332