Renaming columns in pandas

Question

I have a DataFrame using pandas and column labels that I need to edit to replace the original column labels.

I'd like to change the column names in a DataFrame A where the original column names are:

['$a', '$b', '$c', '$d', '$e'] 

to

['a', 'b', 'c', 'd', 'e'].

I have the edited column names stored it in a list, but I don't know how to replace the column names.

Answer 1

Column names vs Names of Series

I would like to explain a bit what happens behind the scenes.

Dataframes are a set of Series.

Series in turn are an extension of a numpy.array

numpy.arrays have a property .name

This is the name of the series. It is seldom that pandas respects this attribute, but it lingers in places and can be used to hack some pandas behaviors.

Naming the list of columns

A lot of answers here talks about the df.columns attribute being a list when in fact it is a Series. This means it has a .name attribute.

This is what happens if you decide to fill in the name of the columns Series:

df.columns = ['column_one', 'column_two']
df.columns.names = ['name of the list of columns']
df.index.names = ['name of the index']

name of the list of columns     column_one  column_two
name of the index       
0                                    4           1
1                                    5           2
2                                    6           3

Note that the name of the index always comes one column lower.

Artifacts that linger

The .name attribute lingers on sometimes. If you set df.columns = ['one', 'two'] then the df.one.name will be 'one'.

If you set df.one.name = 'three' then df.columns will still give you ['one', 'two'], and df.one.name will give you 'three'

BUT

pd.DataFrame(df.one) will return

    three
0       1
1       2
2       3

Because pandas reuses the .name of the already defined Series.

Multi level column names

Pandas has ways of doing multi layered column names. There is not so much magic involved but I wanted to cover this in my answer too since I don't see anyone picking up on this here.

    |one            |
    |one      |two  |
0   |  4      |  1  |
1   |  5      |  2  |
2   |  6      |  3  |

This is easily achievable by setting columns to lists, like this:

df.columns = [['one', 'one'], ['one', 'two']]

Answer 2

One line or Pipeline solutions

I'll focus on two things:

1. OP clearly states

   I have the edited column names stored it in a list, but I don't know how to replace the column names.

   I do not want to solve the problem of how to replace '$' or strip the first character off of each column header. OP has already done this step. Instead I want to focus on replacing the existing columns object with a new one given a list of replacement column names.

2. df.columns = new where new is the list of new columns names is as simple as it gets. The drawback of this approach is that it requires editing the existing dataframe's columns attribute and it isn't done inline. I'll show a few ways to perform this via pipelining without editing the existing dataframe.

---

Setup 1
To focus on the need to rename of replace column names with a pre-existing list, I'll create a new sample dataframe df with initial column names and unrelated new column names.

df = pd.DataFrame({'Jack': [1, 2], 'Mahesh': [3, 4], 'Xin': [5, 6]})
new = ['x098', 'y765', 'z432']

df

   Jack  Mahesh  Xin
0     1       3    5
1     2       4    6

---

Solution 1
pd.DataFrame.rename

It has been said already that if you had a dictionary mapping the old column names to new column names, you could use pd.DataFrame.rename.

d = {'Jack': 'x098', 'Mahesh': 'y765', 'Xin': 'z432'}
df.rename(columns=d)

   x098  y765  z432
0     1     3     5
1     2     4     6

However, you can easily create that dictionary and include it in the call to rename. The following takes advantage of the fact that when iterating over df, we iterate over each column name.

# given just a list of new column names
df.rename(columns=dict(zip(df, new)))

   x098  y765  z432
0     1     3     5
1     2     4     6

This works great if your original column names are unique. But if they are not, then this breaks down.

---

Setup 2
non-unique columns

df = pd.DataFrame(
    [[1, 3, 5], [2, 4, 6]],
    columns=['Mahesh', 'Mahesh', 'Xin']
)
new = ['x098', 'y765', 'z432']

df

   Mahesh  Mahesh  Xin
0       1       3    5
1       2       4    6

---

Solution 2
pd.concat using the keys argument

First, notice what happens when we attempt to use solution 1:

df.rename(columns=dict(zip(df, new)))

   y765  y765  z432
0     1     3     5
1     2     4     6

We didn't map the new list as the column names. We ended up repeating y765. Instead, we can use the keys argument of the pd.concat function while iterating through the columns of df.

pd.concat([c for _, c in df.items()], axis=1, keys=new) 

   x098  y765  z432
0     1     3     5
1     2     4     6

---

Solution 3
Reconstruct. This should only be used if you have a single dtype for all columns. Otherwise, you'll end up with dtype object for all columns and converting them back requires more dictionary work.

Single dtype

pd.DataFrame(df.values, df.index, new)

   x098  y765  z432
0     1     3     5
1     2     4     6

Mixed dtype

pd.DataFrame(df.values, df.index, new).astype(dict(zip(new, df.dtypes)))

   x098  y765  z432
0     1     3     5
1     2     4     6

---

Solution 4
This is a gimmicky trick with transpose and set_index. pd.DataFrame.set_index allows us to set an index inline but there is no corresponding set_columns. So we can transpose, then set_index, and transpose back. However, the same single dtype versus mixed dtype caveat from solution 3 applies here.

Single dtype

df.T.set_index(np.asarray(new)).T

   x098  y765  z432
0     1     3     5
1     2     4     6

Mixed dtype

df.T.set_index(np.asarray(new)).T.astype(dict(zip(new, df.dtypes)))

   x098  y765  z432
0     1     3     5
1     2     4     6

---

Solution 5
Use a lambda in pd.DataFrame.rename that cycles through each element of new
In this solution, we pass a lambda that takes x but then ignores it. It also takes a y but doesn't expect it. Instead, an iterator is given as a default value and I can then use that to cycle through one at a time without regard to what the value of x is.

df.rename(columns=lambda x, y=iter(new): next(y))

   x098  y765  z432
0     1     3     5
1     2     4     6

And as pointed out to me by the folks in sopython chat, if I add a * in between x and y, I can protect my y variable. Though, in this context I don't believe it needs protecting. It is still worth mentioning.

df.rename(columns=lambda x, *, y=iter(new): next(y))

   x098  y765  z432
0     1     3     5
1     2     4     6

Answer 3

I think this method is useful:

df.rename(columns={"old_column_name1":"new_column_name1", "old_column_name2":"new_column_name2"})

This method allows you to change column names individually.

Answer 4

DataFrame -- df.rename() will work.

df.rename(columns = {'Old Name':'New Name'})

df is the DataFrame you have, and the Old Name is the column name you want to change, then the New Name is the new name you change to. This DataFrame built-in method makes things very easier.

Answer 5

In case you don't want the row names df.columns = ['a', 'b',index=False]

Answer 6

If you've got the dataframe, df.columns dumps everything into a list you can manipulate and then reassign into your dataframe as the names of columns...

columns = df.columns
columns = [row.replace("$","") for row in columns]
df.rename(columns=dict(zip(columns, things)), inplace=True)
df.head() #to validate the output

Best way? IDK. A way - yes.

A better way of evaluating all the main techniques put forward in the answers to the question is below using cProfile to gage memory & execution time. @kadee, @kaitlyn, & @eumiro had the functions with the fastest execution times - though these functions are so fast we're comparing the rounding of .000 and .001 seconds for all the answers. Moral: my answer above likely isn't the 'Best' way.

import pandas as pd
import cProfile, pstats, re

old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
col_dict = {'$a': 'a', '$b': 'b','$c':'c','$d':'d','$e':'e'}

df = pd.DataFrame({'$a':[1,2], '$b': [10,20],'$c':['bleep','blorp'],'$d':[1,2],'$e':['texa$','']})

df.head()

def eumiro(df,nn):
    df.columns = nn
    #This direct renaming approach is duplicated in methodology in several other answers: 
    return df

def lexual1(df):
    return df.rename(columns=col_dict)

def lexual2(df,col_dict):
    return df.rename(columns=col_dict, inplace=True)

def Panda_Master_Hayden(df):
    return df.rename(columns=lambda x: x[1:], inplace=True)

def paulo1(df):
    return df.rename(columns=lambda x: x.replace('$', ''))

def paulo2(df):
    return df.rename(columns=lambda x: x.replace('$', ''), inplace=True)

def migloo(df,on,nn):
    return df.rename(columns=dict(zip(on, nn)), inplace=True)

def kadee(df):
    return df.columns.str.replace('$','')

def awo(df):
    columns = df.columns
    columns = [row.replace("$","") for row in columns]
    return df.rename(columns=dict(zip(columns, '')), inplace=True)

def kaitlyn(df):
    df.columns = [col.strip('$') for col in df.columns]
    return df

print 'eumiro'
cProfile.run('eumiro(df,new_names)')
print 'lexual1'
cProfile.run('lexual1(df)')
print 'lexual2'
cProfile.run('lexual2(df,col_dict)')
print 'andy hayden'
cProfile.run('Panda_Master_Hayden(df)')
print 'paulo1'
cProfile.run('paulo1(df)')
print 'paulo2'
cProfile.run('paulo2(df)')
print 'migloo'
cProfile.run('migloo(df,old_names,new_names)')
print 'kadee'
cProfile.run('kadee(df)')
print 'awo'
cProfile.run('awo(df)')
print 'kaitlyn'
cProfile.run('kaitlyn(df)')