Pandas. How to sort a DataFrame without changing index?

Question

df2 = pd.DataFrame({
    "A": [26, 2, 3],
    "B": [0, 7, 1],
    "C": [7, 5, 4]
},
    index=list('abc'))
df2

Output:

    A  B  C
a  26  0  7
b   2  7  5
c   3  1  4

df2.sort_values(['B', 'A'], ascending=[False, True]) gives:

    A  B  C
b   2  7  5
c   3  1  4
a  26  0  7

The column with indexes is now shuffled in new order, but I want it to be the same even after sorting. Parameter ignore_index just sets indexes from 0 to n-1. And the sort_index function isn't helpful too, because indexes can be not in lexicographical order.

Its not randomly shuffled. It retains the row index after sorting. — NYC Coder
– NYC Coder, Commented Nov 4, 2020 at 20:02

NYC Coder · Accepted Answer · 2020-11-04 20:11:43Z

4

You can add the index back after sorting:

df2 = df2.sort_values(['B', 'A'], ascending=[False, True]).reset_index(drop=True)
df2['index'] = ['a', 'b', 'c']
df2.set_index('index', inplace=True)

print(df2)

        A  B  C
index          
a       2  7  5
b       3  1  4
c      26  0  7

answered Nov 4, 2020 at 20:11

NYC Coder

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Comments

Scott Boston · Accepted Answer · 2020-11-04 20:35:23Z

2

Use dataframe constructor:

df2 = pd.DataFrame({
    "A": [26, 2, 3],
    "B": [0, 7, 1],
    "C": [7, 5, 4]
},
    index=list('abc'))
print(df2)

Output:

    A  B  C
a  26  0  7
b   2  7  5
c   3  1  4

Create new dataframe with constructor:

df2 = pd.DataFrame(df2.sort_values(['B', 'A'], ascending=[False, True]).to_numpy(), 
                   index=df2.index, columns=df2.columns)
print(df2)

Output:

    A  B  C
a   2  7  5
b   3  1  4
c  26  0  7

answered Nov 4, 2020 at 20:35

Scott Boston

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