getting the index of a row in a pandas apply function

Question

I am trying to access the index of a row in a function applied across an entire DataFrame in Pandas. I have something like this:

df = pandas.DataFrame([[1,2,3],[4,5,6]], columns=['a','b','c'])
>>> df
   a  b  c
0  1  2  3
1  4  5  6

and I’ll define a function that access elements with a given row

def rowFunc(row):
    return row['a'] + row['b'] * row['c']

I can apply it like so:

df['d'] = df.apply(rowFunc, axis=1)
>>> df
   a  b  c   d
0  1  2  3   7
1  4  5  6  34

Awesome! Now what if I want to incorporate the index into my function?
The index of any given row in this DataFrame before adding d would be Index([u'a', u'b', u'c', u'd'], dtype='object'), but I want the 0 and 1. So I can’t just access row.index.

I know I could create a temporary column in the table where I store the index, but I’m wondering if it is stored in the row object somewhere.

Asked By: Mike

||

Source

Answer 1

To access the index in this case you access the name attribute:

In [182]:

df = pd.DataFrame([[1,2,3],[4,5,6]], columns=['a','b','c'])
def rowFunc(row):
    return row['a'] + row['b'] * row['c']

def rowIndex(row):
    return row.name
df['d'] = df.apply(rowFunc, axis=1)
df['rowIndex'] = df.apply(rowIndex, axis=1)
df
Out[182]:
   a  b  c   d  rowIndex
0  1  2  3   7         0
1  4  5  6  34         1

Note that if this is really what you are trying to do that the following works and is much faster:

In [198]:

df['d'] = df['a'] + df['b'] * df['c']
df
Out[198]:
   a  b  c   d
0  1  2  3   7
1  4  5  6  34

In [199]:

%timeit df['a'] + df['b'] * df['c']
%timeit df.apply(rowIndex, axis=1)
10000 loops, best of 3: 163 µs per loop
1000 loops, best of 3: 286 µs per loop

EDIT

Looking at this question 3+ years later, you could just do:

In[15]:
df['d'],df['rowIndex'] = df['a'] + df['b'] * df['c'], df.index
df

Out[15]: 
   a  b  c   d  rowIndex
0  1  2  3   7         0
1  4  5  6  34         1

but assuming it isn’t as trivial as this, whatever your rowFunc is really doing, you should look to use the vectorised functions, and then use them against the df index:

In[16]:
df['newCol'] = df['a'] + df['b'] + df['c'] + df.index
df

Out[16]: 
   a  b  c   d  rowIndex  newCol
0  1  2  3   7         0       6
1  4  5  6  34         1      16

Answered By: EdChum

Answer 2

Either:

1. with `row.name` inside the `apply(..., axis=1)` call:

df = pandas.DataFrame([[1,2,3],[4,5,6]], columns=['a','b','c'], index=['x','y'])

   a  b  c
x  1  2  3
y  4  5  6

df.apply(lambda row: row.name, axis=1)

x    x
y    y

2. with `iterrows()` (slower)

DataFrame.iterrows() allows you to iterate over rows, and access their index:

for idx, row in df.iterrows():
    ...

Answered By: smci

Answer 3

To answer the original question: yes, you can access the index value of a row in apply(). It is available under the key name and requires that you specify axis=1 (because the lambda processes the columns of a row and not the rows of a column).

Working example (pandas 0.23.4):

>>> import pandas as pd
>>> df = pd.DataFrame([[1,2,3],[4,5,6]], columns=['a','b','c'])
>>> df.set_index('a', inplace=True)
>>> df
   b  c
a      
1  2  3
4  5  6
>>> df['index_x10'] = df.apply(lambda row: 10*row.name, axis=1)
>>> df
   b  c  index_x10
a                 
1  2  3         10
4  5  6         40

Answered By: Freek Wiekmeijer

getting the index of a row in a pandas apply function

Question:

Answers:

1. with `row.name` inside the `apply(..., axis=1)` call:

2. with `iterrows()` (slower)

getting the index of a row in a pandas apply function

Question:

Answers:

1. with row.name inside the apply(..., axis=1) call:

2. with iterrows() (slower)

1. with `row.name` inside the `apply(..., axis=1)` call:

2. with `iterrows()` (slower)