performance - Python - Speed up for converting a categorical variable to it's numerical index

Question

I need to convert a column of categorical variables in a Pandas data frame into a numerical value that corresponds to the index into an array of the unique categorical variables in the column (long story !) and here's a code snippet that accomplishes that:

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)

which converts the data frame:

    col
 0  baked
 1  beans
 2  baked
 3  baked
 4  beans

into the data frame:

    col
 0  0.0
 1  1.0
 2  0.0
 3  0.0
 4  1.0

as desired. But my problem is that my dumb little for loop (the only way I've thought of to do this) is slow as molasses when I try to run similar code on big data files. I was just curious as to whether anyone had any thoughts on whether there were any ways to do this more efficiently. Thanks in advance for any thoughts.

Answer 1

Use factorize:

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

Docs

EDIT:

As Jeff mentioned in comment, then the best is convert column to categorical mainly because less memory usage:

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

Timings:

It is interesting, in large df pandas is faster as numpy. I cant believe it.

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

Code for testing:

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))