Pandas count occurence of value in dictionary

You can do an apply:

df['count'] = df.B.apply(pd.Series).eq('Closed').sum(1)

Output:

   A                                                  B  count
0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}      2
1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}      1
2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}      0

You can try converting the series of dictionary into a dataframe and then stack , then take sum of Closed values on level=0 to get Count per row:

df['Count_closed'] = pd.DataFrame(df['B'].tolist()).stack().eq("Closed").sum(level=0)

   A                                                  B  Count_closed
0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}           2.0
1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}           1.0
2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}           0.0

I will do

df.B.astype(str).str.count('Closed')
Out[30]: 
0    2
1    1
2    0
Name: B, dtype: int64

Or

df['Cnt']=pd.DataFrame(df.B.tolist()).eq('Closed').sum(1).values
Out[35]: 
0    2
1    1
2    0
dtype: int64

Straightforward .apply() solution:

df['Count'] = df.B.apply(lambda x: sum('Closed' in v for v in x.values()))
print(df)

Prints:

   A                                                  B  Count
0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}      2
1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}      1
2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}      0

Benchmark:

import perfplot
import pandas as pd


def f1(df):
    df['Count'] = df.B.apply(lambda x: sum('Closed' in v for v in x.values()))
    return df

def f2(df):
    df['count'] = df.B.astype(str).str.count('Closed')
    return df

# Commented out because of timed-out:
# def f3(df):
#     df['count'] = df.B.apply(pd.Series).eq('Closed').sum(1)
#     return df

def f4(df):
    df['count'] = pd.DataFrame(df['B'].tolist()).stack().eq("Closed").sum(level=0)
    return df

def setup(n):
    A = [*range(n)]
    B = [{'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'} for _ in range(n)]
    df = pd.DataFrame({'A': A,
                       'B': B})
    return df

perfplot.show(
    setup=setup,
    kernels=[f1, f2, f4],
    labels=['apply(sum)', 'str.count()', 'stack.eq()'],
    n_range=[10**i for i in range(1, 7)],
    xlabel='N (* len(df))',
    equality_check=None,
    logx=True,
    logy=True)

Result:

So it seems that straightforward apply() with sum() is fastest.

Please don't put dictionaries into dataframe columns. You are losing all the speed of vectorized operations and make values hard to access.

Clean your df:

>>> df = pd.concat([df['A'], df['B'].apply(pd.Series)], axis=1)
>>> df 
   A     Mon   Tue     Wed
0  1  Closed  Open  Closed
1  2    Open  Open  Closed
2  3    Open  Open    Open

Now counting 'Closed' is easy.

>>> df['count'] = df.eq('Closed').sum(1)
>>> df
   A     Mon   Tue     Wed  count
0  1  Closed  Open  Closed      2
1  2    Open  Open  Closed      1
2  3    Open  Open    Open      0

Use an auxiliary function:

def aux_func(x):

    week_days = x.keys()
    count=0
    for day in week_days:
        if x[day]=='Closed':
            count+=1

    return count

counts = [aux_func(c) for c in df.loc[:,'B'] ]

df['counts'] = counts

You can use a Counter in a simple list comprehension.

from collections import Counter

df['count'] = [Counter(x.values())['Closed'] for x in df.B]

#   A                                                  B  Count
#0  1  {'Mon': 'Closed', 'Tue': 'Open', 'Wed': 'Closed'}      2
#1  2    {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Closed'}      1
#2  3      {'Mon': 'Open', 'Tue': 'Open', 'Wed': 'Open'}      0