CODEWITHSUNDEEP
pythonpandas
jxn
jxn

Reputation: 8025

Pandas count occurence of value in dictionary

Given a DF:

pd.DataFrame({"A":[1,2,3],
              "B": [{"Mon":"Closed", "Tue":"Open", "Wed":"Closed"},
                    {"Mon":"Open", "Tue":"Open", "Wed":"Closed"},
                    {"Mon":"Open", "Tue":"Open", "Wed":"Open"}]
              })

How do i get a count of number of times "Closed" appears in the dict?

A  B    count
1 {..}  2
2 {..}  1 
3 {..}  0

I really don't know how to start on this to try

Upvotes: 5

Views: 80

Answers (7)

ALollz
ALollz

Reputation: 59549

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

Upvotes: 0

Andrej Kesely
Andrej Kesely

Reputation: 195438

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:

enter image description here

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

Upvotes: 0

mangelfdz
mangelfdz

Reputation: 306

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

Upvotes: 0

timgeb
timgeb

Reputation: 78700

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

Upvotes: 0

BENY
BENY

Reputation: 323276

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

Upvotes: 2

Quang Hoang
Quang Hoang

Reputation: 150745

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

Upvotes: 4

anky
anky

Reputation: 75080

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

Upvotes: 3

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