How do I count the values from a pandas column which is a list of strings?

Question

I have a dataframe column which is a list of strings:

df[\'colors\']

0              [\'blue\',\'green\',\'brown\']
1              []
2              [\'green\         


        

Answer 1

You can use Counter from the collections module:

import pandas as pd
from collections import Counter
from  itertools import chain

df = pd.DataFrame({'colors':[['blue','green','brown'],
                             [],
                             ['green','red','blue'],
                             ['purple'],
                             ['brown']]})

df = pd.Series(Counter(chain(*df.colors)))

print (df)

Output:

blue      2
green     2
brown     2
red       1
purple    1
dtype: int64

Answer 2

I would use .apply with pd.Series to accomplish this:

# 1. Expand columns and count them
df_temp = df["colors"].apply(pd.Series.value_counts)

    blue    brown   green   purple  red
0   1.0 1.0 1.0 NaN NaN
1   NaN NaN NaN NaN NaN
2   1.0 NaN 1.0 NaN 1.0
3   NaN NaN NaN 1.0 NaN
4   NaN 1.0 NaN NaN NaN

# 2. Get the value counts from this:
df_temp.sum()

blue      2.0
brown     2.0
green     2.0
purple    1.0
red       1.0

# Alternatively, convert to a dict
df_temp.sum().to_dict()
# {'blue': 2.0, 'brown': 2.0, 'green': 2.0, 'purple': 1.0, 'red': 1.0}

Answer 3

Use a Counter + chain, which is meant to do exactly this. Then construct the Series from the Counter object.

import pandas as pd
from collections import Counter
from itertools import chain

s = pd.Series([['blue','green','brown'], [], ['green','red','blue']])

pd.Series(Counter(chain.from_iterable(s)))
#blue     2
#green    2
#brown    1
#red      1
#dtype: int64

---

While explode + value_counts are the pandas way to do things, they're slower for shorter lists.

import perfplot
import pandas as pd
import numpy as np

from collections import Counter
from itertools import chain

def counter(s):
    return pd.Series(Counter(chain.from_iterable(s)))

def explode(s):
    return s.explode().value_counts()

perfplot.show(
    setup=lambda n: pd.Series([['blue','green','brown'], [], ['green','red','blue']]*n), 
    kernels=[
        lambda s: counter(s),
        lambda s: explode(s),
    ],
    labels=['counter', 'explode'],
    n_range=[2 ** k for k in range(17)],
    equality_check=np.allclose,  
    xlabel='~len(s)'
)

Answer 4

A quick and dirty solution would be something like this I imagine.

You'd still have to add a condition to get the empty list, though.

colors = df.colors.tolist()
d = {}
for l in colors:
    for c in l:
        if c not in d.keys():
            d.update({c: 1})
        else:
            current_val = d.get(c)
            d.update({c: current_val+1})

this produces a dictionary looking like this:

{'blue': 2, 'green': 2, 'brown': 2, 'red': 1, 'purple': 1}

Answer 5

Solution

Best option: df.colors.explode().dropna().value_counts().

However, if you also want to have counts for empty lists ([]), use Method-1.B/C similar to what was suggested by Quang Hoang in the comments.

You can use any of the following two methods.

• Method-1: Use pandas methods alone ⭐⭐⭐

  explode --> dropna --> value_counts

• Method-2: Use list.extend --> pd.Series.value_counts

## Method-1
# A. If you don't want counts for empty []
df.colors.explode().dropna().value_counts() 

# B. If you want counts for empty [] (classified as NaN)
df.colors.explode().value_counts(dropna=False) # returns [] as Nan

# C. If you want counts for empty [] (classified as [])
df.colors.explode().fillna('[]').value_counts() # returns [] as []

## Method-2
colors = []
_ = [colors.extend(e) for e in df.colors if len(e)>0]
pd.Series(colors).value_counts()

Output:

green     2
blue      2
brown     2
red       1
purple    1
# NaN     1  ## For Method-1.B
# []      1  ## For Method-1.C
dtype: int64

Dummy Data

import pandas as pd

df = pd.DataFrame({'colors':[['blue','green','brown'],
                             [],
                             ['green','red','blue'],
                             ['purple'],
                             ['brown']]})