Getting index counter on a data frame

Question

I have the following data frame:

import pandas as pd

data = {
    "id_1": [1, 1, 1, 2, 2, 2],
    "id_2": [1, 1, 1, 2, 2, 2],
    "foo": [0.1, 0.1, 0.1, 0.2, 0.2, 0.2],
}
df = pd.DataFrame(data)
df = df.set_index(["id_1", "id_2"])

which looks like this:

           foo
id_1 id_2
1    1     0.1
     1     0.1
     1     0.1
2    2     0.2
     2     0.2
     2     0.2

I want to have another column (index) that starts from 1 and goes up to the length of the index group, 3. The output should look like this:

           foo index
id_1 id_2
1    1     0.1 1
     1     0.1 2
     1     0.1 3
2    2     0.2 1
     2     0.2 2
     2     0.2 3

How do I go about this, please?

Asked By: alexjrlewis

||

Source

Answer 1

You can groupby the id_1 index whose level is 1 then transform on foo column to get length

df['index'] = (df.groupby(level=1)['foo']
               .transform(lambda col: range(1, len(col)+1)))

print(df)

           foo  index
id_1 id_2
1    1     0.1      1
     1     0.1      2
     1     0.1      3
2    2     0.2      1
     2     0.2      2
     2     0.2      3

Answered By: Ynjxsjmh

Answer 2

One approach could be as follows:

Use df.groupby with level=0, and apply Groupby.cumcount to get consecutive numbers for each item in each group.
Since the numbering will start at 0, we use Series.add to add 1 to the result.

df['index'] = df.groupby(level=0).cumcount().add(1)

df

           foo  index
id_1 id_2            
1    1     0.1      1
     1     0.1      2
     1     0.1      3
2    2     0.2      1
     2     0.2      2
     2     0.2      3

Answered By: ouroboros1

Getting index counter on a data frame

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