Rearrange the columns of a pandas DataFrame based on row number

Question

There is a dataframe:

import pandas as pd
df = pd.DataFrame(data = {'a':[3,0,2,1],'b':[4,3,2,1],'c':[3,2,1,0],'d':[4,3,2,0]})
print(df)
>>> df
   a  b  c  d
0  3  4  3  4
1  0  3  2  3
2  2  2  1  2
3  1  1  0  0

How to rearrange(sort?) entire df based on the distribution of numbers in the cells in each column? for example, the content of a column is [3,2,1], then it should be arranged before other columns that elements is [3,2,0]. When comparing two columns, the numbers in each row are compared in sequence. If the first row is the same, the next row is compared.

The desired results are as follows:

>>> dd
   b  d  c  a
0  4  4  3  3
1  3  3  2  0
2  2  2  1  2
3  1  0  0  1

Asked By: skywave1980

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Source

Answer 1

Code:

import pandas as pd

df = pd.DataFrame(data={'a': [3, 0, 2, 1], 'b': [4, 3, 2, 1], 'c': [3, 2, 1, 0], 'd': [4, 3, 2, 0]})

def custom_sort(col):
    return tuple(col.values)

sorted_columns = sorted(df.columns, key=lambda col: custom_sort(df[col]), reverse=True)
df_sorted = df[sorted_columns]
print(df_sorted)

Output:

Answered By: Ömer Sezer

Answer 2

I also found a solution by weighting the elements of each column.

>>> df
   a  b  c  d
0  3  4  3  4
1  0  3  2  3
2  2  2  1  2
3  1  1  0  0
df[df.apply(lambda x:sum(x * pow(10, np.arange(len(x))[::-1]))).sort_values(ascending=False).index]
   b  d  c  a
0  4  4  3  3
1  3  3  2  0
2  2  2  1  2
3  1  0  0  1

Answered By: skywave1980

Answer 3

For an efficient solution, don’t reinvent the wheel, use sort_values directly:

out = df.sort_values(by=list(df.index), axis=1, ascending=False)

Or eventually numpy.lexsort:

out = df.iloc[:, np.lexsort(-df[::-1].to_numpy())]

Output:

   b  d  c  a
0  4  4  3  3
1  3  3  2  0
2  2  2  1  2
3  1  0  0  1

Answered By: mozway

Rearrange the columns of a pandas DataFrame based on row number

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