pandas 用值交换索引的最快方法
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时间：2020-09-14 02:15:04 来源:igfitidea点击:
quickest way to swap index with values
提问by piRSquared
consider the pd.Seriess
考虑 pd.Seriess
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
What is the quickest way to swap index and values and get the following
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
回答by jezrael
One posible solution is swap keys and values by:
一种可能的解决方案是通过以下方式交换键和值：
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
Another the fastest:
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
Timings:
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
If length of Seriesis 1M:
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
