用值交换索引的最快方法

Question

考虑一下 pd.Series s

s = pd.Series(list('abcdefghij'), list('ABCDEFGHIJ'))
s

A    a
B    b
C    c
D    d
E    e
F    f
G    g
H    h
I    i
J    j
dtype: object

交换索引和值并获得以下内容的最快方法是什么

a    A
b    B
c    C
d    D
e    E
f    F
g    G
h    H
i    I
j    J
dtype: object

Answer 1

一个可行的解决方案是交换键和值:

s1 = pd.Series(dict((v,k) for k,v in s.iteritems()))
print (s1)
a    A
b    B
c    C
d    D
e    E
f    F
g    G
h    H
i    I
j    J
dtype: object

另一个最快的:

print (pd.Series(s.index.values, index=s ))
a    A
b    B
c    C
d    D
e    E
f    F
g    G
h    H
i    I
j    J
dtype: object

时间:

In [63]: %timeit pd.Series(dict((v,k) for k,v in s.iteritems()))
The slowest run took 6.55 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 146 µs per loop

In [71]: %timeit (pd.Series(s.index.values, index=s ))
The slowest run took 7.42 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 102 µs per loop

如果长度Series是1M:

s = pd.Series(list('abcdefghij'), list('ABCDEFGHIJ'))
s = pd.concat([s]*1000000).reset_index(drop=True)
print (s)

In [72]: %timeit (pd.Series(s.index, index=s ))
10000 loops, best of 3: 106 µs per loop

In [229]: %timeit pd.Series(dict((v,k) for k,v in s.iteritems()))
1 loop, best of 3: 1.77 s per loop

In [230]: %timeit (pd.Series(s.index, index=s ))
10 loops, best of 3: 130 ms per loop

In [231]: %timeit (pd.Series(s.index.values, index=s ))
10 loops, best of 3: 26.5 ms per loop