Python - 加速将分类变量转换为其数字索引

Question

我需要将 Pandas 数据框中的一列分类变量转换为对应于索引的数值，转换为列中唯一分类变量的数组（长话短说！），这是一个代码 sn-p实现：

import pandas as pd
import numpy as np

d = {'col': ["baked","beans","baked","baked","beans"]}
df = pd.DataFrame(data=d)
uniq_lab = np.unique(df['col'])

for lab in uniq_lab:
    df['col'].replace(lab,np.where(uniq_lab == lab)[0][0].astype(float),inplace=True)

转换数据框：

    col
 0  baked
 1  beans
 2  baked
 3  baked
 4  beans

进入数据框：

    col
 0  0.0
 1  1.0
 2  0.0
 3  0.0
 4  1.0

根据需要。但我的问题是，当我尝试在大数据文件上运行类似代码时，我的愚蠢的小 for 循环（我想到的唯一方法）就像糖蜜一样慢。我只是好奇是否有人对是否有任何方法可以更有效地做到这一点有任何想法。提前感谢您的任何想法。

Answer 1

使用factorize:

df['col'] = pd.factorize(df.col)[0]
print (df)
   col
0    0
1    1
2    0
3    0
4    1

Docs

编辑：

正如评论中提到的Jeff，那么最好将列转换为categorical，主要是因为更少memory usage：

df['col'] = df['col'].astype("category")

时间安排：

有趣的是，在大 df 中 pandas 比 numpy 更快。我不敢相信。

len(df)=500k:

In [29]: %timeit (a(df1))
100 loops, best of 3: 9.27 ms per loop

In [30]: %timeit (a1(df2))
100 loops, best of 3: 9.32 ms per loop

In [31]: %timeit (b(df3))
10 loops, best of 3: 24.6 ms per loop

In [32]: %timeit (b1(df4))
10 loops, best of 3: 24.6 ms per loop  

len(df)=5k：

In [38]: %timeit (a(df1))
1000 loops, best of 3: 274 µs per loop

In [39]: %timeit (a1(df2))
The slowest run took 6.71 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 273 µs per loop

In [40]: %timeit (b(df3))
The slowest run took 5.15 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 295 µs per loop

In [41]: %timeit (b1(df4))
1000 loops, best of 3: 294 µs per loop

len(df)=5:

In [46]: %timeit (a(df1))
1000 loops, best of 3: 206 µs per loop

In [47]: %timeit (a1(df2))
1000 loops, best of 3: 204 µs per loop

In [48]: %timeit (b(df3))
The slowest run took 6.30 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 164 µs per loop

In [49]: %timeit (b1(df4))
The slowest run took 6.44 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 164 µs per loop

测试代码：

d = {'col': ["baked","beans","baked","baked","beans"]}
df = pd.DataFrame(data=d)
print (df)
df = pd.concat([df]*100000).reset_index(drop=True)
#test for 5k
#df = pd.concat([df]*1000).reset_index(drop=True)


df1,df2,df3, df4 = df.copy(),df.copy(),df.copy(),df.copy()

def a(df):
    df['col'] = pd.factorize(df.col)[0]
    return df

def a1(df):
    idx,_ = pd.factorize(df.col)
    df['col'] = idx
    return df

def b(df):
    df['col'] = np.unique(df['col'],return_inverse=True)[1]
    return df

def b1(df):
    _,idx = np.unique(df['col'],return_inverse=True)
    df['col'] = idx    
    return df

print (a(df1))    
print (a1(df2))   
print (b(df3))   
print (b1(df4))  

Answer 2

您可以使用np.unique 的可选参数return_inverse 根据每个字符串的唯一性来标识每个字符串，并在输入数据框中设置这些字符串，如下所示 -

_,idx = np.unique(df['col'],return_inverse=True)
df['col'] = idx

请注意，IDs 对应于唯一按字母顺序排序的字符串数组。如果你必须得到那个唯一的数组，你可以用它替换_，就像这样-

uniq_lab,idx = np.unique(df['col'],return_inverse=True)

示例运行 -

>>> d = {'col': ["baked","beans","baked","baked","beans"]}
>>> df = pd.DataFrame(data=d)
>>> df
     col
0  baked
1  beans
2  baked
3  baked
4  beans
>>> _,idx = np.unique(df['col'],return_inverse=True)
>>> df['col'] = idx
>>> df
   col
0    0
1    1
2    0
3    0
4    1