CODEWITHSUNDEEP
pythonpandasdataframenumpy
MTALY
MTALY

Reputation: 1772

Get percentage of each row grouped by a value

I have the following df:

df3 = pd.DataFrame(np.array([['Iza', 'Tuesday'],['Martin', 'Friday'],['John', 'Monday'],['Iza', 'Tuesday'],['Iza', 'Tuesday'],['Iza', 'Wednesday'],['Sara', 'Friday'], ['Sara', 'Friday'], ['Sara', 'Sunday'],['Silvia', 'Monday'],['Silvia', 'Wednesday'],['Paul', 'Monday'],['Paul', 'Tuesday'],['Paul', 'Wednesday']]),
                   columns=['Name', 'Day'])

df3:

    Name    Day
0   Iza     Tuesday
1   Martin  Friday
2   John    Monday
3   Iza     Tuesday
4   Iza     Tuesday
5   Iza     Wednesday
6   Sara    Friday
7   Sara    Friday
8   Sara    Sunday
9   Silvia  Monday
10  Silvia  Wednesday
11  Paul    Monday
12  Paul    Tuesday
13  Paul    Wednesday

I got the count of days for each user:

oo = df3.groupby(['Name','Day'])['Day'].size().reset_index(name='counts')

result:

    Name    Day      counts
0   Iza    Tuesday      3
1   Iza    Wednesday    1
2   John    Monday      1
3   Martin  Friday      1
4   Paul    Monday      1
5   Paul    Tuesday     1
6   Paul    Wednesday   1
7   Sara    Friday      2
8   Sara    Sunday      1
9   Silvia  Monday      1
10  Silvia  Wednesday   1

dropped unwanted users with only one day record;

uniq_us = oo[oo.duplicated(['Name'], keep=False)]

result:

    Name    Day       counts
0   Iza     Tuesday     3
1   Iza     Wednesday   1
4   Paul    Monday      1
5   Paul    Tuesday     1
6   Paul    Wednesday   1
7   Sara    Friday      2
8   Sara    Sunday      1
9   Silvia  Monday      1
10  Silvia  Wednesday   1

Now I want to get the percentage of counts in each grouped days by name:

uniq_us.groupby(['Name','Day'])['counts'].apply(lambda x: x.value_counts(normalize=True)) * 100

I got:

Name    Day         
Iza     Tuesday    3    100.0
        Wednesday  1    100.0
Paul    Monday     1    100.0
        Tuesday    1    100.0
        Wednesday  1    100.0
Sara    Friday     2    100.0
        Sunday     1    100.0
Silvia  Monday     1    100.0
        Wednesday  1    100.0
Name: counts, dtype: float64

I do not know how can I calculate it per grouped Name

Desired output:

     Name    Day         

    Iza     Tuesday    3    75.0
            Wednesday  1    25.0
    Paul    Monday     1    33.33
            Tuesday    1    33.33
            Wednesday  1    33.33
    Sara    Friday     2    66.66
            Sunday     1    33.34
    Silvia  Monday     1    50.0
            Wednesday  1    50.0
    Name: counts, dtype: float64

Upvotes: 2

Views: 57

Answers (3)

akuiper
akuiper

Reputation: 215057

Another option is to normalize the count at early stage:

(df3.groupby('Name')
    .Day
    .value_counts(normalize=True)
    .mul(100)
    .rename('Counts')
    .reset_index()
    .pipe(lambda x: x[x.duplicated(['Name'], keep=False)]))

#      Name        Day     Counts
#0      Iza    Tuesday  75.000000
#1      Iza  Wednesday  25.000000
#4     Paul     Monday  33.333333
#5     Paul    Tuesday  33.333333
#6     Paul  Wednesday  33.333333
#7     Sara     Friday  66.666667
#8     Sara     Sunday  33.333333
#9   Silvia     Monday  50.000000
#10  Silvia  Wednesday  50.000000

Upvotes: 1

Mustafa Aydın
Mustafa Aydın

Reputation: 18316

You can normalize the counts with their sum via transform:

uniq_us["pcnt"] = uniq_us.groupby("Name").counts.transform(lambda x: x / x.sum())

to get

>>> uniq_us

      Name        Day  counts      pcnt
0      Iza    Tuesday       3  0.750000
1      Iza  Wednesday       1  0.250000
4     Paul     Monday       1  0.333333
5     Paul    Tuesday       1  0.333333
6     Paul  Wednesday       1  0.333333
7     Sara     Friday       2  0.666667
8     Sara     Sunday       1  0.333333
9   Silvia     Monday       1  0.500000
10  Silvia  Wednesday       1  0.500000

You can put 100 * and round(2) in lambda and set the Name and Day as the index to match the output:

...transform(lambda x: (100 * x / x.sum()).round(2))

uniq_us = uniq_us.set_index(["Name", "Day"])

to get

                  counts   pcnt
Name   Day
Iza    Tuesday         3  75.00
       Wednesday       1  25.00
Paul   Monday          1  33.33
       Tuesday         1  33.33
       Wednesday       1  33.33
Sara   Friday          2  66.67
       Sunday          1  33.33
Silvia Monday          1  50.00
       Wednesday       1  50.00

Upvotes: 2

not_speshal
not_speshal

Reputation: 23146

You're almost there. Try:

>>> uniq_us.groupby(["Name", "Day"]).sum()/uniq_us.groupby("Name").sum()
                    counts
Name   Day                
Iza    Tuesday    0.750000
       Wednesday  0.250000
Paul   Monday     0.333333
       Tuesday    0.333333
       Wednesday  0.333333
Sara   Friday     0.666667
       Sunday     0.333333
Silvia Monday     0.500000
       Wednesday  0.500000

Upvotes: 1

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