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?
Solution 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.
Solution 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)
Solution 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
Solution 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)
Solution 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