问题很简单,输入是非容器对象的列表(
int
,
str
等等),列表中的所有元素都包含在
DataFrame
from pandas import DataFrame
digits = '0123456789abcdef'
df = DataFrame([(a,b) for a, b in zip(digits, range(16))], columns=['hex', 'dec'])
df
df.loc[df.dec == 12, 'hex']
df.loc[df.dec == 12, 'hex'].values[0]
import random
eight = random.sample(range(16), 8)
eight
fun = lambda x: df.loc[df.dec == x, 'hex'].values[0]
''.join(fun(i) for i in eight)
''.join(map(fun, eight))
正如你所看到的,我已经可以做到这一点了,但是我使用了for循环,我知道性能不是很好
pandas
和
numpy
都是关于矢量化的,我想知道有没有一个内置的方法来做到这一点。。。
In [1]: from pandas import DataFrame
In [2]: digits = '0123456789abcdef'
In [3]: df = DataFrame([(a,b) for a, b in zip(digits, range(16))], columns=['hex', 'dec'])
In [4]: df
Out[4]:
hex dec
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 a 10
11 b 11
12 c 12
13 d 13
14 e 14
15 f 15
In [5]: df.loc[df.dec == 12, 'hex']
Out[5]:
12 c
Name: hex, dtype: object
In [6]: df.loc[df.dec == 12, 'hex'].values[0]
Out[6]: 'c'
In [7]: import random
In [8]: eight = random.sample(range(16), 8)
In [9]: eight
Out[9]: [9, 7, 1, 6, 11, 12, 14, 10]
In [10]: fun = lambda x: df.loc[df.dec == x, 'hex'].values[0]
In [11]: ''.join(fun(i) for i in eight)
Out[11]: '9716bcea'
In [12]: ''.join(map(fun, eight))
Out[12]: '9716bcea'
In [13]: %timeit ''.join(fun(i) for i in eight)
2.34 ms ± 136 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [14]: %timeit ''.join(map(fun, eight))
2.34 ms ± 134 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
那么,什么是矢量化的方法来实现与代码中演示的方法相同的结果呢?
