CODEWITHSUNDEEP
pythonperformancepandas
Luke
Luke

Reputation: 7099

Fastest way to sort each row in a pandas dataframe

I need to find the quickest way to sort each row in a dataframe with millions of rows and around a hundred columns.

So something like this:

A   B   C   D
3   4   8   1
9   2   7   2

Needs to become:

A   B   C   D
8   4   3   1
9   7   2   2

Right now I'm applying sort to each row and building up a new dataframe row by row. I'm also doing a couple of extra, less important things to each row (hence why I'm using pandas and not numpy). Could it be quicker to instead create a list of lists and then build the new dataframe at once? Or do I need to go cython?

Upvotes: 23

Views: 27299

Answers (5)

Pradeep Vairamani
Pradeep Vairamani

Reputation: 4312

You could use pd.apply.

Eg:

A = pd.DataFrame(np.random.randint(0,100,(4,5)),
        columns=['one','two','three','four','five']) 
print (A)

   one  two  three  four  five
0    2   75     44    53    46
1   18   51     73    80    66
2   35   91     86    44    25
3   60   97     57    33    79

B = A.apply(np.sort, axis = 1) 
print(B)

   one  two  three  four  five
0    2   44     46    53    75
1   18   51     66    73    80
2   25   35     44    86    91
3   33   57     60    79    97

Since you want it in descending order, you can simply multiply the dataframe with -1 and sort it.

A = pd.DataFrame(np.random.randint(0,100,(4,5)),
        columns=['one','two','three','four','five'])
A = A * -1
A = A.apply(np.sort, axis = 1)
A = A * -1

Upvotes: 7

abdelkrim naji
abdelkrim naji

Reputation: 11

One could try this approach to preserve the integrity of the df:

import pandas as pd 
import numpy as np

A = pd.DataFrame(np.random.randint(0,100,(4,5)), columns=['one','two','three','four','five']) 
print (A) 
print(type(A))

   one  two  three  four  five
0   85   27     64    50    55
1    3   90     65    22     8
2    0    7     64    66    82
3   58   21     42    27    30
<class 'pandas.core.frame.DataFrame'>

B = A.apply(lambda x: np.sort(x), axis=1, raw=True) 
print(B) 
print(type(B))

   one  two  three  four  five
0   27   50     55    64    85
1    3    8     22    65    90
2    0    7     64    66    82
3   21   27     30    42    58
<class 'pandas.core.frame.DataFrame'>

Upvotes: 1

Erfan
Erfan

Reputation: 42916

Instead of using pd.DataFrame constructor, an easier way to assign the sorted values back is to use double brackets:

original dataframe:

A   B   C   D
3   4   8   1
9   2   7   2

df[['A', 'B', 'C', 'D']] = np.sort(df)[:, ::-1]

   A  B  C  D
0  8  4  3  1
1  9  7  2  2

This way you can also sort a part of the columns:

df[['B', 'C']] = np.sort(df[['B', 'C']])[:, ::-1]

   A  B  C  D
0  3  8  4  1
1  9  7  2  2

Upvotes: 2

Andy Hayden
Andy Hayden

Reputation: 375685

I think I would do this in numpy:

In [11]: a = df.values

In [12]: a.sort(axis=1)  # no ascending argument

In [13]: a = a[:, ::-1]  # so reverse

In [14]: a
Out[14]:
array([[8, 4, 3, 1],
       [9, 7, 2, 2]])

In [15]: pd.DataFrame(a, df.index, df.columns)
Out[15]:
   A  B  C  D
0  8  4  3  1
1  9  7  2  2

I had thought this might work, but it sorts the columns:

In [21]: df.sort(axis=1, ascending=False)
Out[21]:
   D  C  B  A
0  1  8  4  3
1  2  7  2  9

Ah, pandas raises:

In [22]: df.sort(df.columns, axis=1, ascending=False)

ValueError: When sorting by column, axis must be 0 (rows)

Upvotes: 29

SpmP
SpmP

Reputation: 557

To Add to the answer given by @Andy-Hayden, to do this inplace to the whole frame... not really sure why this works, but it does. There seems to be no control on the order.

    In [97]: A = pd.DataFrame(np.random.randint(0,100,(4,5)), columns=['one','two','three','four','five'])

    In [98]: A
    Out[98]: 
    one  two  three  four  five
    0   22   63     72    46    49
    1   43   30     69    33    25
    2   93   24     21    56    39
    3    3   57     52    11    74

    In [99]: A.values.sort
    Out[99]: <function ndarray.sort>

    In [100]: A
    Out[100]: 
    one  two  three  four  five
    0   22   63     72    46    49
    1   43   30     69    33    25
    2   93   24     21    56    39
    3    3   57     52    11    74

    In [101]: A.values.sort()

    In [102]: A
    Out[102]: 
    one  two  three  four  five
    0   22   46     49    63    72
    1   25   30     33    43    69
    2   21   24     39    56    93
    3    3   11     52    57    74
    In [103]: A = A.iloc[:,::-1]

    In [104]: A
    Out[104]: 
    five  four  three  two  one
    0    72    63     49   46   22
    1    69    43     33   30   25
    2    93    56     39   24   21
    3    74    57     52   11    3

I hope someone can explain the why of this, just happy that it works 8)

Upvotes: 6

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