3070

I want to change the column labels of a Pandas DataFrame from

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

to

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

33 Answers 33

4705

Rename Specific Columns

Use the df.rename() function and refer the columns to be renamed. Not all the columns have to be renamed:

df = df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'}) # Or rename the existing DataFrame (rather than creating a copy) df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'}, inplace=True) 

Minimal Code Example

df = pd.DataFrame('x', index=range(3), columns=list('abcde')) df a b c d e 0 x x x x x 1 x x x x x 2 x x x x x 

The following methods all work and produce the same output:

df2 = df.rename({'a': 'X', 'b': 'Y'}, axis=1) df2 = df.rename({'a': 'X', 'b': 'Y'}, axis='columns') df2 = df.rename(columns={'a': 'X', 'b': 'Y'}) df2 X Y c d e 0 x x x x x 1 x x x x x 2 x x x x x 

Remember to assign the result back, as the modification is not-inplace. Alternatively, specify inplace=True:

df.rename({'a': 'X', 'b': 'Y'}, axis=1, inplace=True) df X Y c d e 0 x x x x x 1 x x x x x 2 x x x x x 

You can specify errors='raise' to raise errors if an invalid column-to-rename is specified.


Reassign Column Headers

Use df.set_axis() with axis=1.

df2 = df.set_axis(['V', 'W', 'X', 'Y', 'Z'], axis=1) df2 V W X Y Z 0 x x x x x 1 x x x x x 2 x x x x x 

Headers can be assigned directly:

df.columns = ['V', 'W', 'X', 'Y', 'Z'] df V W X Y Z 0 x x x x x 1 x x x x x 2 x x x x x 
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Comments

2573

Just assign it to the .columns attribute:

>>> df = pd.DataFrame({'$a':[1,2], '$b': [10,20]}) >>> df $a $b 0 1 10 1 2 20 >>> df.columns = ['a', 'b'] >>> df a b 0 1 10 1 2 20 

Comments

524

The rename method can take a function, for example:

In [11]: df.columns Out[11]: Index([u'$a', u'$b', u'$c', u'$d', u'$e'], dtype=object) In [12]: df.rename(columns=lambda x: x[1:], inplace=True) In [13]: df.columns Out[13]: Index([u'a', u'b', u'c', u'd', u'e'], dtype=object) 

Comments

280

As documented in Working with text data:

df.columns = df.columns.str.replace('$', '') 

Comments

210

Pandas 0.21+ Answer

There have been some significant updates to column renaming in version 0.21.

  • The rename method has added the axis parameter which may be set to columns or 1. This update makes this method match the rest of the pandas API. It still has the index and columns parameters but you are no longer forced to use them.
  • The set_axis method with the inplace set to False enables you to rename all the index or column labels with a list.

Examples for Pandas 0.21+

Construct sample DataFrame:

df = pd.DataFrame({'$a':[1,2], '$b': [3,4], '$c':[5,6], '$d':[7,8], '$e':[9,10]}) $a $b $c $d $e 0 1 3 5 7 9 1 2 4 6 8 10 

Using rename with axis='columns' or axis=1

df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis='columns') 

or

df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis=1) 

Both result in the following:

 a b c d e 0 1 3 5 7 9 1 2 4 6 8 10 

It is still possible to use the old method signature:

df.rename(columns={'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}) 

The rename function also accepts functions that will be applied to each column name.

df.rename(lambda x: x[1:], axis='columns') 

or

df.rename(lambda x: x[1:], axis=1) 

Using set_axis with a list and inplace=False

You can supply a list to the set_axis method that is equal in length to the number of columns (or index). Currently, inplace defaults to True, but inplace will be defaulted to False in future releases.

df.set_axis(['a', 'b', 'c', 'd', 'e'], axis='columns', inplace=False) 

or

df.set_axis(['a', 'b', 'c', 'd', 'e'], axis=1, inplace=False) 

Why not use df.columns = ['a', 'b', 'c', 'd', 'e']?

There is nothing wrong with assigning columns directly like this. It is a perfectly good solution.

The advantage of using set_axis is that it can be used as part of a method chain and that it returns a new copy of the DataFrame. Without it, you would have to store your intermediate steps of the chain to another variable before reassigning the columns.

# new for pandas 0.21+ df.some_method1() .some_method2() .set_axis() .some_method3() # old way df1 = df.some_method1() .some_method2() df1.columns = columns df1.some_method3() 

Comments

162

Since you only want to remove the $ sign in all column names, you could just do:

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

OR

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

1 Comment

This one not only helps in OP's case but also in generic requirements. E.g.: to split a column name by a separator and use one part of it.
145

Renaming columns in Pandas is an easy task.

df.rename(columns={'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}, inplace=True) 

5 Comments

I will up this since It is naturally supported.
much better than approved solution
The columns arg here can also be a function. So if you want to remove the first char from each name you can do df.rename(columns=lambda name: name[1:], inplace=True) (ref)
It's very natural. You can do it for arbitrary columns. It should be an accepted answer.
also give a label to an unlabelled column using this method: df.rename(columns={0: "x", 1: "y", 2: "z"})
83

Use:

old_names = ['$a', '$b', '$c', '$d', '$e'] new_names = ['a', 'b', 'c', 'd', 'e'] df.rename(columns=dict(zip(old_names, new_names)), inplace=True) 

This way you can manually edit the new_names as you wish. It works great when you need to rename only a few columns to correct misspellings, accents, remove special characters, etc.

5 Comments

I like this approach, but I think df.columns = ['a', 'b', 'c', 'd', 'e'] is simpler.
I like this method of zipping old and new names. We can use df.columns.values to get the old names.
I display the tabular view and copy the columns to old_names. I copy the requirement array to new_names. Then use dict(zip(old_names, new_names)) Very elegant solution.
I often use subsets of lists from something like: myList = list(df) myList[10:20] , etc - so this is perfect.
Best to take the old names as @bkowshik suggested, then edit them and re-insert them, ie namez = df.columns.values followed by some edits, then df.columns = namez.
47

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 

Comments

45

Many of pandas functions have an inplace parameter. When setting it True, the transformation applies directly to the dataframe that you are calling it on. For example:

df = pd.DataFrame({'$a':[1,2], '$b': [3,4]}) df.rename(columns={'$a': 'a'}, inplace=True) df.columns >>> Index(['a', '$b'], dtype='object') 

Alternatively, there are cases where you want to preserve the original dataframe. I have often seen people fall into this case if creating the dataframe is an expensive task. For example, if creating the dataframe required querying a snowflake database. In this case, just make sure the the inplace parameter is set to False.

df = pd.DataFrame({'$a':[1,2], '$b': [3,4]}) df2 = df.rename(columns={'$a': 'a'}, inplace=False) df.columns >>> Index(['$a', '$b'], dtype='object') df2.columns >>> Index(['a', '$b'], dtype='object') 

If these types of transformations are something that you do often, you could also look into a number of different pandas GUI tools. I'm the creator of one called Mito. It’s a spreadsheet that automatically converts your edits to python code.

Comments

45

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']] 

Comments

33

Let's understand renaming by a small example...

  1. Renaming columns using mapping:

     df = pd.DataFrame({"A": [1, 2, 3], "B": [4, 5, 6]}) # Creating a df with column name A and B df.rename({"A": "new_a", "B": "new_b"}, axis='columns', inplace =True) # Renaming column A with 'new_a' and B with 'new_b' Output: new_a new_b 0 1 4 1 2 5 2 3 6 
  2. Renaming index/Row_Name using mapping:

     df.rename({0: "x", 1: "y", 2: "z"}, axis='index', inplace =True) # Row name are getting replaced by 'x', 'y', and 'z'. Output: new_a new_b x 1 4 y 2 5 z 3 6 

Comments

27

Let's say this is your dataframe.

enter image description here

You can rename the columns using two methods.

  1. Using dataframe.columns=[#list]

    df.columns=['a','b','c','d','e'] 

    enter image description here

    The limitation of this method is that if one column has to be changed, full column list has to be passed. Also, this method is not applicable on index labels. For example, if you passed this:

    df.columns = ['a','b','c','d'] 

    This will throw an error. Length mismatch: Expected axis has 5 elements, new values have 4 elements.

  2. Another method is the Pandas rename() method which is used to rename any index, column or row

    df = df.rename(columns={'$a':'a'}) 

    enter image description here

Similarly, you can change any rows or columns.

Comments

24

If you already have a list for the new column names, you can try this:

new_cols = ['a', 'b', 'c', 'd', 'e'] new_names_map = {df.columns[i]:new_cols[i] for i in range(len(new_cols))} df.rename(new_names_map, axis=1, inplace=True) 

Comments

22
df = pd.DataFrame({'$a': [1], '$b': [1], '$c': [1], '$d': [1], '$e': [1]}) 

If your new list of columns is in the same order as the existing columns, the assignment is simple:

new_cols = ['a', 'b', 'c', 'd', 'e'] df.columns = new_cols >>> df a b c d e 0 1 1 1 1 1 

If you had a dictionary keyed on old column names to new column names, you could do the following:

d = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'} df.columns = df.columns.map(lambda col: d[col]) # Or `.map(d.get)` as pointed out by @PiRSquared. >>> df a b c d e 0 1 1 1 1 1 

If you don't have a list or dictionary mapping, you could strip the leading $ symbol via a list comprehension:

df.columns = [col[1:] if col[0] == '$' else col for col in df] 

1 Comment

Instead of lambda col: d[col] you could pass d.get... so it would look like df.columns.map(d.get)
22

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? I don't know. 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 and execution time. @kadee, @kaitlyn, and @eumiro had the functions with the fastest execution times - though these functions are so fast we're comparing the rounding of 0.000 and 0.001 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)') 

Comments

22
df.rename(index=str, columns={'A':'a', 'B':'b'}) 

pandas.DataFrame.rename

1 Comment

An explanation would be in order.
21

Another way we could replace the original column labels is by stripping the unwanted characters (here '$') from the original column labels.

This could have been done by running a for loop over df.columns and appending the stripped columns to df.columns.

Instead, we can do this neatly in a single statement by using list comprehension like below:

df.columns = [col.strip('$') for col in df.columns] 

(strip method in Python strips the given character from beginning and end of the string.)

1 Comment

Can you explain how/why this works? That will make the answer more valuable for future readers.
13

You could use str.slice for that:

df.columns = df.columns.str.slice(1) 

1 Comment

PS: This is a more verbose equivalent to df.columns.str[1:]... probably better to use that, it's shorter and more obvious.
13

Another option is to rename using a regular expression:

import pandas as pd import re df = pd.DataFrame({'$a':[1,2], '$b':[3,4], '$c':[5,6]}) df = df.rename(columns=lambda x: re.sub('\$','',x)) >>> df a b c 0 1 3 5 1 2 4 6 

Comments

12

My method is generic wherein you can add additional delimiters by comma separating delimiters= variable and future-proof it.

Working Code:

import pandas as pd import re df = pd.DataFrame({'$a':[1,2], '$b': [3,4],'$c':[5,6], '$d': [7,8], '$e': [9,10]}) delimiters = '$' matchPattern = '|'.join(map(re.escape, delimiters)) df.columns = [re.split(matchPattern, i)[1] for i in df.columns ] 

Output:

>>> df $a $b $c $d $e 0 1 3 5 7 9 1 2 4 6 8 10 >>> df a b c d e 0 1 3 5 7 9 1 2 4 6 8 10 

Comments

11

Note that the approaches in previous answers do not work for a MultiIndex. For a MultiIndex, you need to do something like the following:

>>> df = pd.DataFrame({('$a','$x'):[1,2], ('$b','$y'): [3,4], ('e','f'):[5,6]}) >>> df $a $b e $x $y f 0 1 3 5 1 2 4 6 >>> rename = {('$a','$x'):('a','x'), ('$b','$y'):('b','y')} >>> df.columns = pandas.MultiIndex.from_tuples([ rename.get(item, item) for item in df.columns.tolist()]) >>> df a b e x y f 0 1 3 5 1 2 4 6 

Comments

10

If you have to deal with loads of columns named by the providing system out of your control, I came up with the following approach that is a combination of a general approach and specific replacements in one go.

First create a dictionary from the dataframe column names using regular expressions in order to throw away certain appendixes of column names and then add specific replacements to the dictionary to name core columns as expected later in the receiving database.

This is then applied to the dataframe in one go.

dict = dict(zip(df.columns, df.columns.str.replace('(:S$|:C1$|:L$|:D$|\.Serial:L$)', ''))) dict['brand_timeseries:C1'] = 'BTS' dict['respid:L'] = 'RespID' dict['country:C1'] = 'CountryID' dict['pim1:D'] = 'pim_actual' df.rename(columns=dict, inplace=True) 

Comments

10

If you just want to remove the '$' sign then use the below code

df.columns = pd.Series(df.columns.str.replace("$", "")) 

Comments

10

Rename column at a specific position

One use-case not mentioned on this page is how to rename a column by index, i.e. rename a column name at a specific position. If the column names are unique, then rename() would work. For example, if we want to rename the second column, the following would work.

df = pd.DataFrame({'$A': [1, 2], '$B': ['a', 'b']}) df.rename(columns={df.columns[1]: 'new'}, inplace=True) # ^^^^^^^^^^^^^ <--- second column is renamed 

result1

However, if the column labels are non-unique (which is a common reason to rename it by index in the first place), the above would change all duplicate column names. However, pd.DataFrame().columns is an immutable pandas Index object that is built on a (mutable) numpy ndarray which can be accessed as a view using .values/.to_numpy(). Modifying the underlying array by index does the job.

# modify the second column name df = pd.DataFrame([[1, 'a', 1.2], [2, 'b', 3.4]], columns=['$A', '$B', '$B']) df.columns[1] = 'new' # <---- TypeError df.columns.values[1] = 'new' # <---- OK df.columns.to_numpy()[1] = 'new' # <---- OK 

To perform the same in a chained method or create a new copy of a dataframe would require changing the entire columns object and assign using set_axis():

# change the second column name df = df.set_axis([*df.columns[:1], 'new', *df.columns[2:]], axis=1) 

result2

str methods

pd.DataFrame().columns also defines a .str accessor, which enables one to call specific string methods. For the use case in the question, one could use removeprefix() to remove leading '$'s.

df = pd.DataFrame({'$A': [1, 2], '$B': ['a', 'b']}) df.columns = df.columns.str.removeprefix('$') 

result3

Comments

9

My one line answer is

df.columns = df_new_cols

It is the best one with 1/3rd the processing time.

timeit comparison:

df has seven columns. I am trying to change a few of the names.

%timeit df.rename(columns={old_col:new_col for (old_col,new_col) in zip(df_old_cols,df_new_cols)},inplace=True) 214 µs ± 10.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each) %timeit df.rename(columns=dict(zip(df_old_cols,df_new_cols)),inplace=True) 212 µs ± 7.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each) %timeit df.columns = df_new_cols 72.9 µs ± 17.2 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each) 

Comments

8

In addition to the solution already provided, you can replace all the columns while you are reading the file. We can use names and header=0 to do that.

First, we create a list of the names that we like to use as our column names:

import pandas as pd ufo_cols = ['city', 'color reported', 'shape reported', 'state', 'time'] ufo.columns = ufo_cols ufo = pd.read_csv('link to the file you are using', names = ufo_cols, header = 0) 

In this case, all the column names will be replaced with the names you have in your list.

Comments

8

Assuming you can use a regular expression, this solution removes the need of manual encoding using a regular expression:

import pandas as pd import re srch = re.compile(r"\w+") data = pd.read_csv("CSV_FILE.csv") cols = data.columns new_cols = list(map(lambda v:v.group(), (list(map(srch.search, cols))))) data.columns = new_cols 

Comments

7

Here's a nifty little function I like to use to cut down on typing:

def rename(data, oldnames, newname): if type(oldnames) == str: # Input can be a string or list of strings oldnames = [oldnames] # When renaming multiple columns newname = [newname] # Make sure you pass the corresponding list of new names i = 0 for name in oldnames: oldvar = [c for c in data.columns if name in c] if len(oldvar) == 0: raise ValueError("Sorry, couldn't find that column in the dataset") if len(oldvar) > 1: # Doesn't have to be an exact match print("Found multiple columns that matched " + str(name) + ": ") for c in oldvar: print(str(oldvar.index(c)) + ": " + str(c)) ind = input('Please enter the index of the column you would like to rename: ') oldvar = oldvar[int(ind)] if len(oldvar) == 1: oldvar = oldvar[0] data = data.rename(columns = {oldvar : newname[i]}) i += 1 return data 

Here is an example of how it works:

In [2]: df = pd.DataFrame(np.random.randint(0, 10, size=(10, 4)), columns = ['col1', 'col2', 'omg', 'idk']) # First list = existing variables # Second list = new names for those variables In [3]: df = rename(df, ['col', 'omg'],['first', 'ohmy']) Found multiple columns that matched col: 0: col1 1: col2 Please enter the index of the column you would like to rename: 0 In [4]: df.columns Out[5]: Index(['first', 'col2', 'ohmy', 'idk'], dtype='object') 

2 Comments

The use case for a function like this is extremely rare. In most cases, I know what I'm looking for and what I want to rename it to, I'd just assign/modify it myself.
@cs95 I tend to work with large national or international surveys where variables will have coded variable names that begin with prefixes depending on answer options, likert scales, and branching (such as EDU_2913.443, EDU_2913.421,...). This function has been very useful for me in working with those types of sets, I understand if its not for you though :)
7

I needed to rename features for XGBoost, and it didn't like any of these:

import re regex = r"[!\"#$%&'()*+,\-.\/:;<=>?@[\\\]^_`{|}~ ]+" X_trn.columns = X_trn.columns.str.replace(regex, '_', regex=True) X_tst.columns = X_tst.columns.str.replace(regex, '_', regex=True) 

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