重命名 Pandas 中的列名

Question

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

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

to

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

Answer 1

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)  # new method
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis='columns')
df2 = df.rename(columns={'a': 'X', 'b': 'Y'})  # old method  

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
 

From v0.25, you can also specify errors='raise' to raise errors if an invalid column-to-rename is specified. See v0.25 rename() docs .

---

REASSIGN COLUMN HEADERS

Use df.set_axis() with axis=1 and inplace=False (to return a copy).

df2 = df.set_axis(['V', 'W', 'X', 'Y', 'Z'], axis=1, inplace=False)
df2

   V  W  X  Y  Z
0  x  x  x  x  x
1  x  x  x  x  x
2  x  x  x  x  x

This returns a copy, but you can modify the DataFrame in-place by setting inplace=True (this is the default behaviour for versions <=0.24 but is likely to change in the future).

You can also assign headers 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

Answer 2

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

Answer 3

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)

Answer 4

如使用文本数据中所述：

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

Answer 5

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()

Answer 6

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)

Answer 7

在 Pandas 中重命名列是一项简单的任务。

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

Answer 8

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

它将按照您提供的顺序将现有名称替换为您提供的名称。

Answer 9

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.

Answer 10

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.array s 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.

Artefacts 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 11

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 12

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

Answer 13

Suppose your dataset name is df, and df has.

df = ['$a', '$b', '$c', '$d', '$e']`

So, to rename these, we would simply do.

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

Answer 14

Let's say this is your dataframe.

You can rename the columns using two methods.

1. Using dataframe.columns=[#list]

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

   

   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'})

   

Similarly, you can change any rows or columns.

Answer 15

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 . Its a spreadsheet that automatically converts your edits to python code.

Answer 16

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

pandas.DataFrame.rename

Answer 17

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)')

Answer 18

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]

Answer 19

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.)

Answer 20

It is real simple. Just use:

df.columns = ['Name1', 'Name2', 'Name3'...]

And it will assign the column names by the order you put them in.

Answer 21

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)

Answer 22

# This way it will work
import pandas as pd

# Define a dictionary 
rankings = {'test': ['a'],
        'odi': ['E'],
        't20': ['P']}

# Convert the dictionary into DataFrame
rankings_pd = pd.DataFrame(rankings)

# Before renaming the columns
print(rankings_pd)

rankings_pd.rename(columns = {'test':'TEST'}, inplace = True)

Answer 23

你可以使用str.slice ：

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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)

Answer 28

如果您只想删除“$”符号，请使用以下代码

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

Answer 29

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.

Answer 30

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')

Answer 31

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

Answer 32

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)

Answer 33

You can use lstrip or strip methods with index:

df.columns = df.columns.str.lstrip('$')

or

cols = ['$a', '$b', '$c', '$d', '$e']
pd.Series(cols).str.lstrip('$').tolist()

Output:

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

Answer 34

My one line answer is df.columns = df_new_cols is the best one with 1/3rd processing time.

timeit Comparison: df has 7 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)