## The Question :

*337 people think this question is useful*

I have a `dataframe`

with over 200 columns. The issue is as they were generated the order is

['Q1.3','Q6.1','Q1.2','Q1.1',......]

I need to re-order the columns as follows:

['Q1.1','Q1.2','Q1.3',.....'Q6.1',......]

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

*The Question Comments :*

## The Answer 1

*426 people think this answer is useful*

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.

## The Answer 2

*351 people think this answer is useful*

You can also do more succinctly:

df.sort_index(axis=1)

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)

## The Answer 3

*37 people think this answer is useful*

You can just do:

df[sorted(df.columns)]

Edit: Shorter is

df[sorted(df)]

## The Answer 4

*24 people think this answer is useful*

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:

data
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:

data
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

## The Answer 5

*20 people think this answer is useful*

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

## The Answer 6

*16 people think this answer is useful*

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)

## The Answer 7

*16 people think this answer is useful*

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.

## The Answer 8

*4 people think this answer is useful*

The quickest method is:

df.sort_index(axis=1)

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

sortedDf=df.sort_index(axis=1)

## The Answer 9

*1 people think this answer is useful*

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']

## The Answer 10

*0 people think this answer is useful*

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) ], []) ]

## The Answer 11

*-2 people think this answer is useful*

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