Pandas - How to calculate for each group ,for each value in a column what percentage of value is equal and less than that

Question

Lets say I have a dataframe like this

x = pd.DataFrame({'person':['a','b']*5 , 'rating':[1,3,4,2,4,2,3,4,5,3]})

Now I want to calculate for each person ,each rating's 'preference score' . Now I define preference score for rating r as

freq of rating where rating <=r  -  freq of rating where rating ==r

For example a has the following rating

0   a   1
2   a   4
4   a   4
6   a   3
8   a   5

now for example rating =4 for person a

freq of rating where rating <=4  :  4/5 
freq of rating where rating ==4   : 2/5

so the preference score is 2/5

How do I do achieve the preference score for each record on that data frame . EDIT :Perhaps this makes it more clear

   person rating    pref_score
        a   1       0.0
        a   4       0.4
        a   4       0.4
        a   3       0.2
        a   5       0.8

Answer 1

so you need something like this ?

x.groupby('person').rating.apply(lambda x : (sum(x<=4)-sum(x==4))/len(x))
Out[7]: 
person
a    0.4
b    0.8
Name: rating, dtype: float64

Or transform ?

x.groupby('person').rating.transform(lambda x : (sum(x<=4)-sum(x==4))/len(x))
Out[8]: 
0    0.4
1    0.8
2    0.4
3    0.8
4    0.4
5    0.8
6    0.4
7    0.8
8    0.4
9    0.8
Name: rating, dtype: float64

EDIT :

x=x.sort_values('person')
x['ref']=x.groupby('person').rating.apply(lambda y : [(sum(y<=x)-sum(y==x))/len(y) for x in y]).apply(pd.Series).stack().values
x
Out[25]: 
  person  rating  ref
0      a       1  0.0
2      a       4  0.4
4      a       4  0.4
6      a       3  0.2
8      a       5  0.8
1      b       3  0.4
3      b       2  0.0
5      b       2  0.0
7      b       4  0.8
9      b       3  0.4

Since you are using python 2.7

x['map']=x.person.map(x.groupby('person').rating.apply(list))
x.apply(lambda x : sum(x['rating']<np.array(x['map']))/len(x['map']),1 )

Answer 2

You can do the following:

>> x.groupby("person").rating.apply(lambda x: x[x <= 4].count())
person
a    4
b    5

and

>> x.groupby("person").rating.apply(lambda x: x[x == 4].count())
person
a    2
b    1