Apply function to pandas groupby

Posted on

Question :

Apply function to pandas groupby

I have a pandas dataframe with a column called my_labels which contains strings: 'A', 'B', 'C', 'D', 'E'. I would like to count the number of occurances of each of these strings then divide the number of counts by the sum of all the counts. I’m trying to do this in Pandas like this:

func = lambda x: x.size() / x.sum()
data = frame.groupby('my_labels').apply(func)

This code throws an error, ‘DataFrame object has no attribute ‘size’. How can I apply a function to calculate this in Pandas?

Asked By: turtle

||

Answer #1:

apply takes a function to apply to each value, not the series, and accepts kwargs.
So, the values do not have the .size() method.

Perhaps this would work:

from pandas import *

d = {"my_label": Series(['A','B','A','C','D','D','E'])}
df = DataFrame(d)


def as_perc(value, total):
    return value/float(total)

def get_count(values):
    return len(values)

grouped_count = df.groupby("my_label").my_label.agg(get_count)
data = grouped_count.apply(as_perc, total=df.my_label.count())

The .agg() method here takes a function that is applied to all values of the groupby object.

Answered By: monkut

Answer #2:

As of Pandas version 0.22, there exists also an alternative to apply: pipe, which can be considerably faster than using apply (you can also check this question for more differences between the two functionalities).

For your example:

df = pd.DataFrame({"my_label": ['A','B','A','C','D','D','E']})

  my_label
0        A
1        B
2        A
3        C
4        D
5        D
6        E

The apply version

df.groupby('my_label').apply(lambda grp: grp.count() / df.shape[0])

gives

          my_label
my_label          
A         0.285714
B         0.142857
C         0.142857
D         0.285714
E         0.142857

and the pipe version

df.groupby('my_label').pipe(lambda grp: grp.size() / grp.size().sum())

yields

my_label
A    0.285714
B    0.142857
C    0.142857
D    0.285714
E    0.142857

So the values are identical, however, the timings differ quite a lot (at least for this small dataframe):

%timeit df.groupby('my_label').apply(lambda grp: grp.count() / df.shape[0])
100 loops, best of 3: 5.52 ms per loop

and

%timeit df.groupby('my_label').pipe(lambda grp: grp.size() / grp.size().sum())
1000 loops, best of 3: 843 µs per loop

Wrapping it into a function is then also straightforward:

def get_perc(grp_obj):
    gr_size = grp_obj.size()
    return gr_size / gr_size.sum()

Now you can call

df.groupby('my_label').pipe(get_perc)

yielding

my_label
A    0.285714
B    0.142857
C    0.142857
D    0.285714
E    0.142857

However, for this particular case, you do not even need a groupby, but you can just use value_counts like this:

df['my_label'].value_counts(sort=False) / df.shape[0]

yielding

A    0.285714
C    0.142857
B    0.142857
E    0.142857
D    0.285714
Name: my_label, dtype: float64

For this small dataframe it is quite fast

%timeit df['my_label'].value_counts(sort=False) / df.shape[0]
1000 loops, best of 3: 770 µs per loop

As pointed out by @anmol, the last statement can also be simplified to

df['my_label'].value_counts(sort=False, normalize=True)
Answered By: Cleb

Answer #3:

Try:

g = pd.DataFrame(['A','B','A','C','D','D','E'])

# Group by the contents of column 0 
gg = g.groupby(0)  

# Create a DataFrame with the counts of each letter
histo = gg.apply(lambda x: x.count())

# Add a new column that is the count / total number of elements    
histo[1] = histo.astype(np.float)/len(g) 

print histo

Output:

   0         1
0             
A  2  0.285714
B  1  0.142857
C  1  0.142857
D  2  0.285714
E  1  0.142857
Answered By: Reservedegotist

Answer #4:

Regarding the issue with ‘size’, size is not a function on a dataframe, it is rather a property. So instead of using size(), plain size should work

Apart from that, a method like this should work

 def doCalculation(df):
    groupCount = df.size
    groupSum = df['my_labels'].notnull().sum()

    return groupCount / groupSum

dataFrame.groupby('my_labels').apply(doCalculation)
Answered By: Vaibhav

Answer #5:

I saw a nested function technique for computing a weighted average on S.O. one time, altering that technique can solve your issue.

def group_weight(overall_size):
    def inner(group):
        return len(group)/float(overall_size)
    inner.__name__ = 'weight'
    return inner

d = {"my_label": pd.Series(['A','B','A','C','D','D','E'])}
df = pd.DataFrame(d)
print df.groupby('my_label').apply(group_weight(len(df)))



my_label
A    0.285714
B    0.142857
C    0.142857
D    0.285714
E    0.142857
dtype: float64

Here is how to do a weighted average within groups

def wavg(val_col_name,wt_col_name):
    def inner(group):
        return (group[val_col_name] * group[wt_col_name]).sum() / group[wt_col_name].sum()
    inner.__name__ = 'wgt_avg'
    return inner



d = {"P": pd.Series(['A','B','A','C','D','D','E'])
     ,"Q": pd.Series([1,2,3,4,5,6,7])
    ,"R": pd.Series([0.1,0.2,0.3,0.4,0.5,0.6,0.7])
     }

df = pd.DataFrame(d)
print df.groupby('P').apply(wavg('Q','R'))

P
A    2.500000
B    2.000000
C    4.000000
D    5.545455
E    7.000000
dtype: float64
Answered By: Dickster

Leave a Reply

Your email address will not be published. Required fields are marked *