Total length of elements, and subsets, in a pandas dataframe

Question

How I can count the total elements in a dataframe, including the subset, and put the result in the new column?

import pandas as pd
x = pd.Series([[1, (2,5,6)], [2, (3,4)], [3, 4], [(5,6), (7,8,9)]], \
              index=range(1, len(x)+1))
df = pd.DataFrame({'A': x})

I tried with the following code but it gives 2 in each of row:

df['Length'] = df['A'].apply(len)

print(df)

                         A  Length
    1       [1, (2, 5, 6)]       2
    2          [2, (3, 4)]       2
    3               [3, 4]       2
    4  [(5, 6), (7, 8, 9)]       2

However, what I want to get is as follow:

                         A  Length
    1       [1, (2, 5, 6)]       4
    2          [2, (3, 4)]       3
    3               [3, 4]       2
    4  [(5, 6), (7, 8, 9)]       5

thanks

Answer 1

Given:

import pandas as pd
x = pd.Series([[1, (2,5,6)], [2, (3,4)], [3, 4], [(5,6), (7,8,9)]])
df = pd.DataFrame({'A': x}) 

You can write a recursive generator that will yield 1 for each nested element that is not iterable. Something along these lines:

import collections 

def glen(LoS):
    def iselement(e):
        return not(isinstance(e, collections.Iterable) and not isinstance(e, str))
    for el in LoS:
        if iselement(el):
            yield 1
        else:
            for sub in glen(el): yield sub    

df['Length'] = df['A'].apply(lambda e: sum(glen(e)))

Yielding:

>>> df
                     A  Length
0       [1, (2, 5, 6)]       4
1          [2, (3, 4)]       3
2               [3, 4]       2
3  [(5, 6), (7, 8, 9)]       5

That will work in Python 2 or 3. With Python 3.3 or later, you can use yield from to replace the loop:

def glen(LoS):
    def iselement(e):
        return not(isinstance(e, collections.Iterable) and not isinstance(e, str))
    for el in LoS:
        if iselement(el):
            yield 1
        else:
            yield from glen(el)