Sorting columns in pandas dataframe based on column name


I have a dataframe with over 200 columns. The issue is as they were generated the order is


I need to sort the columns as follows:


Is there some way for me to do this within Python?

Asked By: pythOnometrist



df = df.reindex(sorted(df.columns), axis=1)

This assumes that sorting the column names will give the order you want. If your column names won’t sort lexicographically (e.g., if you want column Q10.3 to appear after Q9.1), you’ll need to sort differently, but that has nothing to do with pandas.

Answered By: BrenBarn

The sort method and sorted function allow you to provide a custom function to extract the key used for comparison:

>>> ls = ['Q1.3', 'Q6.1', 'Q1.2']
>>> sorted(ls, key=lambda x: float(x[1:]))
['Q1.2', 'Q1.3', 'Q6.1']
Answered By: tweet

You can also do more succinctly:


Make sure you assign the result back:

df = df.sort_index(axis=1)

Or, do it in-place:

df.sort_index(axis=1, inplace=True)
Answered By: Wes McKinney

Tweet’s answer can be passed to BrenBarn’s answer above with

data.reindex_axis(sorted(data.columns, key=lambda x: float(x[1:])), axis=1)

So for your example, say:

vals = randint(low=16, high=80, size=25).reshape(5,5)
cols = ['Q1.3', 'Q6.1', 'Q1.2', 'Q9.1', 'Q10.2']
data = DataFrame(vals, columns = cols)

You get:


    Q1.3    Q6.1    Q1.2    Q9.1    Q10.2
0   73      29      63      51      72
1   61      29      32      68      57
2   36      49      76      18      37
3   63      61      51      30      31
4   36      66      71      24      77

Then do:

data.reindex_axis(sorted(data.columns, key=lambda x: float(x[1:])), axis=1)

resulting in:


     Q1.2    Q1.3    Q6.1    Q9.1    Q10.2
0    2       0       1       3       4
1    7       5       6       8       9
2    2       0       1       3       4
3    2       0       1       3       4
4    2       0       1       3       4
Answered By: Jeremy Low

You can just do:


Edit: Shorter is

Answered By: Ivelin

Don’t forget to add “inplace=True” to Wes’ answer or set the result to a new DataFrame.

df.sort_index(axis=1, inplace=True)
Answered By: burkesquires

The quickest method is:


Be aware that this creates a new instance. Therefore you need to store the result in a new variable:

Answered By: multigoodverse
print df.sort_index(by='Frequency',ascending=False)

where by is the name of the column,if you want to sort the dataset based on column

If you need an arbitrary sequence instead of sorted sequence, you could do:

sequence = ['Q1.1','Q1.2','Q1.3',.....'Q6.1',......]
your_dataframe = your_dataframe.reindex(columns=sequence)

I tested this in 2.7.10 and it worked for me.

Answered By: M.Z

For several columns, You can put columns order what you want:

#['A', 'B', 'C'] <-this is your columns order
df = df[['C', 'B', 'A']]

This example shows sorting and slicing columns:

d = {'col1':[1, 2, 3], 'col2':[4, 5, 6], 'col3':[7, 8, 9], 'col4':[17, 18, 19]}
df = pandas.DataFrame(d)

You get:

col1  col2  col3  col4
 1     4     7    17
 2     5     8    18
 3     6     9    19

Then do:

df = df[['col3', 'col2', 'col1']]

Resulting in:

col3  col2  col1
7     4     1
8     5     2
9     6     3     
Answered By: Myeongsik Joo

One use-case is that you have named (some of) your columns with some prefix, and you want the columns sorted with those prefixes all together and in some particular order (not alphabetical).

For example, you might start all of your features with Ft_, labels with Lbl_, etc, and you want all unprefixed columns first, then all features, then the label. You can do this with the following function (I will note a possible efficiency problem using sum to reduce lists, but this isn’t an issue unless you have a LOT of columns, which I do not):

def sortedcols(df, groups = ['Ft_', 'Lbl_'] ):
    return df[ sum([list(filter(re.compile(r).search, list(df.columns).copy())) for r in (lambda l: ['^(?!(%s))' % '|'.join(l)] + ['^%s' % i  for i in l ] )(groups)   ], [])  ]
Answered By: Roko Mijic
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