python - How to change the order of DataFrame columns?

1146
votes

I have the following DataFrame (df):

import numpy as np
import pandas as pd

df = pd.DataFrame(np.random.rand(10, 5))

I add more column(s) by assignment:

df['mean'] = df.mean(1)

How can I move the column mean to the front, i.e. set it as first column leaving the order of the other columns untouched?

30
pythonpandasdataframe
possible duplicate of Python Pandas - Re-ordering columns in a dataframe based on column name – Laurence
For a generalized NumPy-based solution see How to move a column in a pandas dataframe, assumes one column level only, i.e. no MultiIndex. – jpp
After searching enough, I got this best link for columns re-arranging multiple logics in pretty simple terms [columns re-arrange logic for pandas] [datasciencemadesimple.com/… – ravibeli

30 Answers

1058
votes

One easy way would be to reassign the dataframe with a list of the columns, rearranged as needed.

This is what you have now: 

In [6]: df
Out[6]:
          0         1         2         3         4      mean
0  0.445598  0.173835  0.343415  0.682252  0.582616  0.445543
1  0.881592  0.696942  0.702232  0.696724  0.373551  0.670208
2  0.662527  0.955193  0.131016  0.609548  0.804694  0.632596
3  0.260919  0.783467  0.593433  0.033426  0.512019  0.436653
4  0.131842  0.799367  0.182828  0.683330  0.019485  0.363371
5  0.498784  0.873495  0.383811  0.699289  0.480447  0.587165
6  0.388771  0.395757  0.745237  0.628406  0.784473  0.588529
7  0.147986  0.459451  0.310961  0.706435  0.100914  0.345149
8  0.394947  0.863494  0.585030  0.565944  0.356561  0.553195
9  0.689260  0.865243  0.136481  0.386582  0.730399  0.561593

In [7]: cols = df.columns.tolist()

In [8]: cols
Out[8]: [0L, 1L, 2L, 3L, 4L, 'mean']

Rearrange cols in any way you want. This is how I moved the last element to the first position: 

In [12]: cols = cols[-1:] + cols[:-1]

In [13]: cols
Out[13]: ['mean', 0L, 1L, 2L, 3L, 4L]

Then reorder the dataframe like this: 

In [16]: df = df[cols]  #    OR    df = df.ix[:, cols]

In [17]: df
Out[17]:
       mean         0         1         2         3         4
0  0.445543  0.445598  0.173835  0.343415  0.682252  0.582616
1  0.670208  0.881592  0.696942  0.702232  0.696724  0.373551
2  0.632596  0.662527  0.955193  0.131016  0.609548  0.804694
3  0.436653  0.260919  0.783467  0.593433  0.033426  0.512019
4  0.363371  0.131842  0.799367  0.182828  0.683330  0.019485
5  0.587165  0.498784  0.873495  0.383811  0.699289  0.480447
6  0.588529  0.388771  0.395757  0.745237  0.628406  0.784473
7  0.345149  0.147986  0.459451  0.310961  0.706435  0.100914
8  0.553195  0.394947  0.863494  0.585030  0.565944  0.356561
9  0.561593  0.689260  0.865243  0.136481  0.386582  0.730399
614
votes

You could also do something like this:

df = df[['mean', '0', '1', '2', '3']]

You can get the list of columns with:

cols = list(df.columns.values)

The output will produce:

['0', '1', '2', '3', 'mean']

...which is then easy to rearrange manually before dropping it into the first function

355
votes

Just assign the column names in the order you want them:

In [39]: df
Out[39]: 
          0         1         2         3         4  mean
0  0.172742  0.915661  0.043387  0.712833  0.190717     1
1  0.128186  0.424771  0.590779  0.771080  0.617472     1
2  0.125709  0.085894  0.989798  0.829491  0.155563     1
3  0.742578  0.104061  0.299708  0.616751  0.951802     1
4  0.721118  0.528156  0.421360  0.105886  0.322311     1
5  0.900878  0.082047  0.224656  0.195162  0.736652     1
6  0.897832  0.558108  0.318016  0.586563  0.507564     1
7  0.027178  0.375183  0.930248  0.921786  0.337060     1
8  0.763028  0.182905  0.931756  0.110675  0.423398     1
9  0.848996  0.310562  0.140873  0.304561  0.417808     1

In [40]: df = df[['mean', 4,3,2,1]]

Now, 'mean' column comes out in the front:

In [41]: df
Out[41]: 
   mean         4         3         2         1
0     1  0.190717  0.712833  0.043387  0.915661
1     1  0.617472  0.771080  0.590779  0.424771
2     1  0.155563  0.829491  0.989798  0.085894
3     1  0.951802  0.616751  0.299708  0.104061
4     1  0.322311  0.105886  0.421360  0.528156
5     1  0.736652  0.195162  0.224656  0.082047
6     1  0.507564  0.586563  0.318016  0.558108
7     1  0.337060  0.921786  0.930248  0.375183
8     1  0.423398  0.110675  0.931756  0.182905
9     1  0.417808  0.304561  0.140873  0.310562
158
votes

In your case,

df = df.reindex(columns=['mean',0,1,2,3,4])

will do exactly what you want.

In my case (general form):

df = df.reindex(columns=sorted(df.columns))
df = df.reindex(columns=(['opened'] + list([a for a in df.columns if a != 'opened']) ))
80
votes

You need to create a new list of your columns in the desired order, then use df = df[cols] to rearrange the columns in this new order.

cols = ['mean']  + [col for col in df if col != 'mean']
df = df[cols]

You can also use a more general approach. In this example, the last column (indicated by -1) is inserted as the first column.

cols = [df.columns[-1]] + [col for col in df if col != df.columns[-1]]
df = df[cols]

You can also use this approach for reordering columns in a desired order if they are present in the DataFrame.

inserted_cols = ['a', 'b', 'c']
cols = ([col for col in inserted_cols if col in df] 
        + [col for col in df if col not in inserted_cols])
df = df[cols]
77
votes
import numpy as np
import pandas as pd
df = pd.DataFrame()
column_names = ['x','y','z','mean']
for col in column_names: 
    df[col] = np.random.randint(0,100, size=10000)

You can try out the following solutions :

Solution 1:

df = df[ ['mean'] + [ col for col in df.columns if col != 'mean' ] ]

Solution 2:

df = df[['mean', 'x', 'y', 'z']]

Solution 3:

col = df.pop("mean")
df = df.insert(0, col.name, col)

Solution 4:

df.set_index(df.columns[-1], inplace=True)
df.reset_index(inplace=True)

Solution 5:

cols = list(df)
cols = [cols[-1]] + cols[:-1]
df = df[cols]

solution 6:

order = [1,2,3,0] # setting column's order
df = df[[df.columns[i] for i in order]]

Time Comparison:

Solution 1:

CPU times: user 1.05 ms, sys: 35 µs, total: 1.08 ms Wall time: 995 µs

Solution 2:

CPU times: user 933 µs, sys: 0 ns, total: 933 µs Wall time: 800 µs

Solution 3:

CPU times: user 0 ns, sys: 1.35 ms, total: 1.35 ms Wall time: 1.08 ms

Solution 4:

CPU times: user 1.23 ms, sys: 45 µs, total: 1.27 ms Wall time: 986 µs

Solution 5:

CPU times: user 1.09 ms, sys: 19 µs, total: 1.11 ms Wall time: 949 µs

Solution 6:

CPU times: user 955 µs, sys: 34 µs, total: 989 µs Wall time: 859 µs

55
votes

Suppose you have df with columns A B C.

The most simple way is:

df = df.reindex(['B','C','A'], axis=1)
51
votes

If your column names are too-long-to-type then you could specify the new order through a list of integers with the positions:

Data:

          0         1         2         3         4      mean
0  0.397312  0.361846  0.719802  0.575223  0.449205  0.500678
1  0.287256  0.522337  0.992154  0.584221  0.042739  0.485741
2  0.884812  0.464172  0.149296  0.167698  0.793634  0.491923
3  0.656891  0.500179  0.046006  0.862769  0.651065  0.543382
4  0.673702  0.223489  0.438760  0.468954  0.308509  0.422683
5  0.764020  0.093050  0.100932  0.572475  0.416471  0.389390
6  0.259181  0.248186  0.626101  0.556980  0.559413  0.449972
7  0.400591  0.075461  0.096072  0.308755  0.157078  0.207592
8  0.639745  0.368987  0.340573  0.997547  0.011892  0.471749
9  0.050582  0.714160  0.168839  0.899230  0.359690  0.438500

Generic example:

new_order = [3,2,1,4,5,0]
print(df[df.columns[new_order]])  

          3         2         1         4      mean         0
0  0.575223  0.719802  0.361846  0.449205  0.500678  0.397312
1  0.584221  0.992154  0.522337  0.042739  0.485741  0.287256
2  0.167698  0.149296  0.464172  0.793634  0.491923  0.884812
3  0.862769  0.046006  0.500179  0.651065  0.543382  0.656891
4  0.468954  0.438760  0.223489  0.308509  0.422683  0.673702
5  0.572475  0.100932  0.093050  0.416471  0.389390  0.764020
6  0.556980  0.626101  0.248186  0.559413  0.449972  0.259181
7  0.308755  0.096072  0.075461  0.157078  0.207592  0.400591
8  0.997547  0.340573  0.368987  0.011892  0.471749  0.639745
9  0.899230  0.168839  0.714160  0.359690  0.438500  0.050582

Although it might seem like I'm just explicitly typing the column names in a different order, the fact that there's a column 'mean' should make it clear that new_order relates to actual positions and not column names.

For the specific case of OP's question:

new_order = [-1,0,1,2,3,4]
df = df[df.columns[new_order]]
print(df)

       mean         0         1         2         3         4
0  0.500678  0.397312  0.361846  0.719802  0.575223  0.449205
1  0.485741  0.287256  0.522337  0.992154  0.584221  0.042739
2  0.491923  0.884812  0.464172  0.149296  0.167698  0.793634
3  0.543382  0.656891  0.500179  0.046006  0.862769  0.651065
4  0.422683  0.673702  0.223489  0.438760  0.468954  0.308509
5  0.389390  0.764020  0.093050  0.100932  0.572475  0.416471
6  0.449972  0.259181  0.248186  0.626101  0.556980  0.559413
7  0.207592  0.400591  0.075461  0.096072  0.308755  0.157078
8  0.471749  0.639745  0.368987  0.340573  0.997547  0.011892
9  0.438500  0.050582  0.714160  0.168839  0.899230  0.359690

The main problem with this approach is that calling the same code multiple times will create different results each time, so one needs to be careful :)

30
votes

This question has been answered before but reindex_axis is deprecated now so I would suggest to use:

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

For those who want to specify the order they want instead of just sorting them, here's the solution spelled out:

df = df.reindex(['the','order','you','want'], axis=1)

Now, how you want to sort the list of column names is really not a pandas question, that's a Python list manipulation question. There are many ways of doing that, and I think this answer has a very neat way of doing it.

21
votes

I think this is a slightly neater solution:

df.insert(0, 'mean', df.pop("mean"))

This solution is somewhat similar to @JoeHeffer 's solution but this is one liner.

Here we remove the column "mean" from the dataframe and attach it to index 0 with the same column name.

19
votes

I ran into a similar question myself, and just wanted to add what I settled on. I liked the reindex_axis() method for changing column order. This worked:

df = df.reindex_axis(['mean'] + list(df.columns[:-1]), axis=1)

An alternate method based on the comment from @Jorge:

df = df.reindex(columns=['mean'] + list(df.columns[:-1]))

Although reindex_axis seems to be slightly faster in micro benchmarks than reindex, I think I prefer the latter for its directness.

18
votes

This function avoids you having to list out every variable in your dataset just to order a few of them. 

def order(frame,var):
    if type(var) is str:
        var = [var] #let the command take a string or list
    varlist =[w for w in frame.columns if w not in var]
    frame = frame[var+varlist]
    return frame

It takes two arguments, the first is the dataset, the second are the columns in the data set that you want to bring to the front.

So in my case I have a data set called Frame with variables A1, A2, B1, B2, Total and Date. If I want to bring Total to the front then all I have to do is: 

frame = order(frame,['Total'])

If I want to bring Total and Date to the front then I do:

frame = order(frame,['Total','Date'])

EDIT:

Another useful way to use this is, if you have an unfamiliar table and you're looking with variables with a particular term in them, like VAR1, VAR2,... you may execute something like: 

frame = order(frame,[v for v in frame.columns if "VAR" in v])
17
votes

Simply do,

df = df[['mean'] + df.columns[:-1].tolist()]
12
votes

You could do the following (borrowing parts from Aman's answer):

cols = df.columns.tolist()
cols.insert(0, cols.pop(-1))

cols
>>>['mean', 0L, 1L, 2L, 3L, 4L]

df = df[cols]
10
votes

Just type the column name you want to change, and set the index for the new location.

def change_column_order(df, col_name, index):
    cols = df.columns.tolist()
    cols.remove(col_name)
    cols.insert(index, col_name)
    return df[cols]

For your case, this would be like:

df = change_column_order(df, 'mean', 0)
9
votes

Moving any column to any position: 

import pandas as pd
df = pd.DataFrame({"A": [1,2,3], 
                   "B": [2,4,8], 
                   "C": [5,5,5]})

cols = df.columns.tolist()
column_to_move = "C"
new_position = 1

cols.insert(new_position, cols.pop(cols.index(column_to_move)))
df = df[cols]
8
votes

You can reorder the dataframe columns using a list of names with:

df = df.filter(list_of_col_names)

7
votes

Here's a way to move one existing column that will modify the existing data frame in place.

my_column = df.pop('column name')
df.insert(3, my_column.name, my_column)
7
votes

I wanted to bring two columns in front from a dataframe where I do not know exactly the names of all columns, because they are generated from a pivot statement before. So, if you are in the same situation: To bring columns in front that you know the name of and then let them follow by "all the other columns", I came up with the following general solution:

df = df.reindex_axis(['Col1','Col2'] + list(df.columns.drop(['Col1','Col2'])), axis=1)
6
votes

Here is a very simple answer to this(only one line).

You can do that after you added the 'n' column into your df as follows.

import numpy as np
import pandas as pd

df = pd.DataFrame(np.random.rand(10, 5))
df['mean'] = df.mean(1)
df
           0           1           2           3           4        mean
0   0.929616    0.316376    0.183919    0.204560    0.567725    0.440439
1   0.595545    0.964515    0.653177    0.748907    0.653570    0.723143
2   0.747715    0.961307    0.008388    0.106444    0.298704    0.424512
3   0.656411    0.809813    0.872176    0.964648    0.723685    0.805347
4   0.642475    0.717454    0.467599    0.325585    0.439645    0.518551
5   0.729689    0.994015    0.676874    0.790823    0.170914    0.672463
6   0.026849    0.800370    0.903723    0.024676    0.491747    0.449473
7   0.526255    0.596366    0.051958    0.895090    0.728266    0.559587
8   0.818350    0.500223    0.810189    0.095969    0.218950    0.488736
9   0.258719    0.468106    0.459373    0.709510    0.178053    0.414752


### here you can add below line and it should work 
# Don't forget the two (()) 'brackets' around columns names.Otherwise, it'll give you an error.

df = df[list(('mean',0, 1, 2,3,4))]
df

        mean           0           1           2           3           4
0   0.440439    0.929616    0.316376    0.183919    0.204560    0.567725
1   0.723143    0.595545    0.964515    0.653177    0.748907    0.653570
2   0.424512    0.747715    0.961307    0.008388    0.106444    0.298704
3   0.805347    0.656411    0.809813    0.872176    0.964648    0.723685
4   0.518551    0.642475    0.717454    0.467599    0.325585    0.439645
5   0.672463    0.729689    0.994015    0.676874    0.790823    0.170914
6   0.449473    0.026849    0.800370    0.903723    0.024676    0.491747
7   0.559587    0.526255    0.596366    0.051958    0.895090    0.728266
8   0.488736    0.818350    0.500223    0.810189    0.095969    0.218950
9   0.414752    0.258719    0.468106    0.459373    0.709510    0.178053
6
votes

You can use a set which is an unordered collection of unique elements to do keep the "order of the other columns untouched":

other_columns = list(set(df.columns).difference(["mean"])) #[0, 1, 2, 3, 4]

Then, you can use a lambda to move a specific column to the front by:

In [1]: import numpy as np                                                                               

In [2]: import pandas as pd                                                                              

In [3]: df = pd.DataFrame(np.random.rand(10, 5))                                                         

In [4]: df["mean"] = df.mean(1)                                                                          

In [5]: move_col_to_front = lambda df, col: df[[col]+list(set(df.columns).difference([col]))]            

In [6]: move_col_to_front(df, "mean")                                                                    
Out[6]: 
       mean         0         1         2         3         4
0  0.697253  0.600377  0.464852  0.938360  0.945293  0.537384
1  0.609213  0.703387  0.096176  0.971407  0.955666  0.319429
2  0.561261  0.791842  0.302573  0.662365  0.728368  0.321158
3  0.518720  0.710443  0.504060  0.663423  0.208756  0.506916
4  0.616316  0.665932  0.794385  0.163000  0.664265  0.793995
5  0.519757  0.585462  0.653995  0.338893  0.714782  0.305654
6  0.532584  0.434472  0.283501  0.633156  0.317520  0.994271
7  0.640571  0.732680  0.187151  0.937983  0.921097  0.423945
8  0.562447  0.790987  0.200080  0.317812  0.641340  0.862018
9  0.563092  0.811533  0.662709  0.396048  0.596528  0.348642

In [7]: move_col_to_front(df, 2)                                                                         
Out[7]: 
          2         0         1         3         4      mean
0  0.938360  0.600377  0.464852  0.945293  0.537384  0.697253
1  0.971407  0.703387  0.096176  0.955666  0.319429  0.609213
2  0.662365  0.791842  0.302573  0.728368  0.321158  0.561261
3  0.663423  0.710443  0.504060  0.208756  0.506916  0.518720
4  0.163000  0.665932  0.794385  0.664265  0.793995  0.616316
5  0.338893  0.585462  0.653995  0.714782  0.305654  0.519757
6  0.633156  0.434472  0.283501  0.317520  0.994271  0.532584
7  0.937983  0.732680  0.187151  0.921097  0.423945  0.640571
8  0.317812  0.790987  0.200080  0.641340  0.862018  0.562447
9  0.396048  0.811533  0.662709  0.596528  0.348642  0.563092
5
votes

Just flipping helps often.

df[df.columns[::-1]]

Or just shuffle for a look.

import random
cols = list(df.columns)
random.shuffle(cols)
df[cols]
4
votes

Hackiest method in the book

df.insert(0, "test", df["mean"])
df = df.drop(columns=["mean"]).rename(columns={"test": "mean"})
4
votes

A pretty straightforward solution that worked for me is to use .reindex on df.columns:

df = df[df.columns.reindex(['mean', 0, 1, 2, 3, 4])[0]]
3
votes

You can use reindex which can be used for both axis:

df
#           0         1         2         3         4      mean
# 0  0.943825  0.202490  0.071908  0.452985  0.678397  0.469921
# 1  0.745569  0.103029  0.268984  0.663710  0.037813  0.363821
# 2  0.693016  0.621525  0.031589  0.956703  0.118434  0.484254
# 3  0.284922  0.527293  0.791596  0.243768  0.629102  0.495336
# 4  0.354870  0.113014  0.326395  0.656415  0.172445  0.324628
# 5  0.815584  0.532382  0.195437  0.829670  0.019001  0.478415
# 6  0.944587  0.068690  0.811771  0.006846  0.698785  0.506136
# 7  0.595077  0.437571  0.023520  0.772187  0.862554  0.538182
# 8  0.700771  0.413958  0.097996  0.355228  0.656919  0.444974
# 9  0.263138  0.906283  0.121386  0.624336  0.859904  0.555009

df.reindex(['mean', *range(5)], axis=1)

#        mean         0         1         2         3         4
# 0  0.469921  0.943825  0.202490  0.071908  0.452985  0.678397
# 1  0.363821  0.745569  0.103029  0.268984  0.663710  0.037813
# 2  0.484254  0.693016  0.621525  0.031589  0.956703  0.118434
# 3  0.495336  0.284922  0.527293  0.791596  0.243768  0.629102
# 4  0.324628  0.354870  0.113014  0.326395  0.656415  0.172445
# 5  0.478415  0.815584  0.532382  0.195437  0.829670  0.019001
# 6  0.506136  0.944587  0.068690  0.811771  0.006846  0.698785
# 7  0.538182  0.595077  0.437571  0.023520  0.772187  0.862554
# 8  0.444974  0.700771  0.413958  0.097996  0.355228  0.656919
# 9  0.555009  0.263138  0.906283  0.121386  0.624336  0.859904
3
votes

A simple approach is using set(), in particular when you have a long list of columns and do not want to handle them manually:

cols = list(set(df.columns.tolist()) - set(['mean']))
cols.insert(0, 'mean')
df = df[cols]
3
votes

How about using T?

df = df.T.reindex(['mean', 0, 1, 2, 3, 4]).T
2
votes

I believe @Aman's answer is the best if you know the location of the other column.

If you don't know the location of mean, but only have its name, you cannot resort directly to cols = cols[-1:] + cols[:-1]. Following is the next-best thing I could come up with:

meanDf = pd.DataFrame(df.pop('mean'))
# now df doesn't contain "mean" anymore. Order of join will move it to left or right:
meanDf.join(df) # has mean as first column
df.join(meanDf) # has mean as last column
2
votes

I liked Shoresh's answer to use set functionality to remove columns when you don't know the location, however this didn't work for my purpose as I need to keep the original column order (which has arbitrary column labels).

I got this to work though by using IndexedSet from the boltons package.

I also needed to re-add multiple column labels, so for a more general case I used the following code:

from boltons.setutils import IndexedSet
cols = list(IndexedSet(df.columns.tolist()) - set(['mean', 'std']))
cols[0:0] =['mean', 'std']
df = df[cols]

Hope this is useful to anyone searching this thread for a general solution.