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', '$e']
to
['a', 'b', 'c', 'd', 'e'].
I have the edited column names stored it in a list, but I don't know how to replace the column names.
Use the df.rename() function and refer the columns to be renamed. Not all the columns have to be renamed:
df = df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'})
# Or rename the existing DataFrame (rather than creating a copy)
df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'}, inplace=True)
Minimal Code Example
df = pd.DataFrame('x', index=range(3), columns=list('abcde'))
df
a b c d e
0 x x x x x
1 x x x x x
2 x x x x x
The following methods all work and produce the same output:
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis=1) # new method
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis='columns')
df2 = df.rename(columns={'a': 'X', 'b': 'Y'}) # old method
df2
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
Remember to assign the result back, as the modification is not-inplace. Alternatively, specify inplace=True:
df.rename({'a': 'X', 'b': 'Y'}, axis=1, inplace=True)
df
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
From v0.25, you can also specify errors='raise' to raise errors if an invalid column-to-rename is specified. See v0.25 rename() docs.
Use df.set_axis() with axis=1 and inplace=False (to return a copy).
df2 = df.set_axis(['V', 'W', 'X', 'Y', 'Z'], axis=1, inplace=False)
df2
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x
This returns a copy, but you can modify the DataFrame in-place by setting inplace=True (this is the default behaviour for versions <=0.24 but is likely to change in the future).
You can also assign headers directly:
df.columns = ['V', 'W', 'X', 'Y', 'Z']
df
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x
The rename method can take a function, for example:
In [11]: df.columns
Out[11]: Index([u'$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)
There have been some significant updates to column renaming in version 0.21.
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. set_axis method with the inplace set to False enables you to rename all the index or column labels with a list.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 $e
0 1 3 5 7 9
1 2 4 6 8 10
rename with axis='columns' or axis=1df.rename({'$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)
set_axis with a list and inplace=FalseYou 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)
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()
Use:
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
df.rename(columns=dict(zip(old_names, new_names)), inplace=True)
This way you can manually edit the new_names as you wish. It works great when you need to rename only a few columns to correct misspellings, accents, remove special characters, etc.
I'll focus on two things:
OP clearly states
I have the edited column names stored it in a list, but I don't know how to replace the column names.
I do not want to solve the problem of how to replace '$' 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.
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 1pd.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 2pd.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
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.
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.
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'.
pd.DataFrame(df.one) will return
three
0 1
1 2
2 3
Because Pandas reuses the .name of the already defined Series.
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']]
Let's understand renaming by a small example...
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
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
Let's say this is your dataframe.
You can rename the columns using two methods.
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.
Another method is the Pandas rename() method which is used to rename any index, column or row
df = df.rename(columns={'$a':'a'})
Similarly, you can change any rows or columns.
df = pd.DataFrame({'$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] == '$' else col for col in df]
If you've got the dataframe, df.columns dumps everything into a list you can manipulate and then reassign into your dataframe as the names of columns...
columns = df.columns
columns = [row.replace("$", "") 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('$') for col in df.columns]
return df
print 'eumiro'
cProfile.run('eumiro(df, new_names)')
print 'lexual1'
cProfile.run('lexual1(df)')
print 'lexual2'
cProfile.run('lexual2(df, col_dict)')
print 'andy hayden'
cProfile.run('Panda_Master_Hayden(df)')
print 'paulo1'
cProfile.run('paulo1(df)')
print 'paulo2'
cProfile.run('paulo2(df)')
print 'migloo'
cProfile.run('migloo(df, old_names, new_names)')
print 'kadee'
cProfile.run('kadee(df)')
print 'awo'
cProfile.run('awo(df)')
print 'kaitlyn'
cProfile.run('kaitlyn(df)')
Another way we could replace the original column labels is by stripping the unwanted characters (here '$') 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('$') for col in df.columns]
(strip method in Python strips the given character from beginning and end of the string.)
My method is generic wherein you can add additional delimiters by comma separating delimiters= variable and future-proof it.
Working Code:
import pandas as pd
import re
df = pd.DataFrame({'$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 $e
0 1 3 5 7 9
1 2 4 6 8 10
>>> df
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10
Note that the approaches in previous answers do not work for a MultiIndex. For a MultiIndex, you need to do something like the following:
>>> df = pd.DataFrame({('$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','$y'):('b','y')}
>>> df.columns = pandas.MultiIndex.from_tuples([
rename.get(item, item) for item in df.columns.tolist()])
>>> df
a b e
x y f
0 1 3 5
1 2 4 6
If you have to deal with loads of columns named by the providing system out of your control, I came up with the following approach that is a combination of a general approach and specific replacements in one go.
First create a dictionary from the dataframe column names using regular expressions in order to throw away certain appendixes of column names and then add specific replacements to the dictionary to name core columns as expected later in the receiving database.
This is then applied to the dataframe in one go.
dict = dict(zip(df.columns, df.columns.str.replace('(:S$|: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)
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.
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')
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
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.
