Redefining the Index in a Pandas DataFrame object

Question

I am trying to re-index a pandas DataFrame object, like so,

From:
            a   b   c
        0   1   2   3
        1  10  11  12
        2  20  21  22

To :
           b   c
       1   2   3
      10  11  12
      20  21  22

I am going about this as shown below and am getting the wrong answer. Any clues on how to do this?

>>> col = ['a','b','c']
>>> data = DataFrame([[1,2,3],[10,11,12],[20,21,22]],columns=col)
>>> data
    a   b   c
0   1   2   3
1  10  11  12
2  20  21  22
>>> idx2 = data.a.values
>>> idx2
array([ 1, 10, 20], dtype=int64)
>>> data2 = DataFrame(data,index=idx2,columns=col[1:])
>>> data2
     b   c
1   11  12
10 NaN NaN
20 NaN NaN

Any idea why this is happening?

because you are using the 1st DF to build the second, this will only get the rows where idx2 intersect the data.index, ie. row 1 — RuiDC
– RuiDC, Commented Jun 27, 2012 at 12:56

Avaris · Accepted Answer · 2012-05-05 02:44:12Z

192

Why don't you simply use set_index method?

In : col = ['a','b','c']

In : data = DataFrame([[1,2,3],[10,11,12],[20,21,22]],columns=col)

In : data
Out:
    a   b   c
0   1   2   3
1  10  11  12
2  20  21  22

In : data2 = data.set_index('a')

In : data2
Out:
     b   c
a
1    2   3
10  11  12
20  21  22

answered May 5, 2012 at 2:44

Avaris

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1 Comment

Daniele Over a year ago

In order to remove the index name, as in the original example: data2.index.name = None

Aakash Saxena · Accepted Answer · 2017-05-10 12:39:42Z

5

If you don't want 'a' in the index

In :

col = ['a','b','c']

data = DataFrame([[1,2,3],[10,11,12],[20,21,22]],columns=col)

data

Out:

    a   b   c
0   1   2   3
1  10  11  12
2  20  21  22

In :

data2 = data.set_index('a')

Out:

In :

data2.index.name = None

Out:

answered May 10, 2017 at 12:39

Aakash Saxena

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1 Comment

Yster Over a year ago

The second Out is not correct. There is still an unnamed column on the left containing 0, 1, 2.

Zero · Accepted Answer · 2021-03-24 09:17:48Z

0

To avoid index name with a single-liner, you can use set_index('a') with rename_axis(None)

In [8]: data.set_index('a').rename_axis(None)
Out[8]: 
     b   c
1    2   3
10  11  12
20  21  22

Details

In [9]: data
Out[9]: 
    a   b   c
0   1   2   3
1  10  11  12
2  20  21  22

answered Mar 24, 2021 at 9:17

Zero

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