How to check if a column exists in Pandas

PythonPandasDataframe

Python Problem Overview


Is there a way to check if a column exists in a Pandas DataFrame?

Suppose that I have the following DataFrame:

>>> import pandas as pd
>>> from random import randint
>>> df = pd.DataFrame({'A': [randint(1, 9) for x in xrange(10)],
                       'B': [randint(1, 9)*10 for x in xrange(10)],
                       'C': [randint(1, 9)*100 for x in xrange(10)]})
>>> df
   A   B    C
0  3  40  100
1  6  30  200
2  7  70  800
3  3  50  200
4  7  50  400
5  4  10  400
6  3  70  500
7  8  30  200
8  3  40  800
9  6  60  200

and I want to calculate df['sum'] = df['A'] + df['C']

But first I want to check if df['A'] exists, and if not, I want to calculate df['sum'] = df['B'] + df['C'] instead.

Python Solutions


Solution 1 - Python

This will work:

if 'A' in df:

But for clarity, I'd probably write it as:

if 'A' in df.columns:

Solution 2 - Python

To check if one or more columns all exist, you can use set.issubset, as in:

if set(['A','C']).issubset(df.columns):
   df['sum'] = df['A'] + df['C']                

As @brianpck points out in a comment, set([]) can alternatively be constructed with curly braces,

if {'A', 'C'}.issubset(df.columns):

See this question for a discussion of the curly-braces syntax.

Or, you can use a generator comprehension, as in:

if all(item in df.columns for item in ['A','C']):

Solution 3 - Python

Just to suggest another way without using if statements, you can use the get() method for DataFrames. For performing the sum based on the question:

df['sum'] = df.get('A', df['B']) + df['C']

The DataFrame get method has similar behavior as python dictionaries.

Solution 4 - Python

You can use the set's method issuperset:

set(df).issuperset(['A', 'B'])
# set(df.columns).issuperset(['A', 'B'])

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionnpiresView Question on Stackoverflow
Solution 1 - PythonchrisbView Answer on Stackoverflow
Solution 2 - PythonC8H10N4O2View Answer on Stackoverflow
Solution 3 - PythonGergesView Answer on Stackoverflow
Solution 4 - PythonMykola ZotkoView Answer on Stackoverflow