I have the following data:
group cluster probabilityA probabilityB
0 a 0 0.28 0.153013
1 a 0 0.28 0.133686
2 a 0 0.28 0.058366
3 a 0 0.28 0.091937
4 a 1 0.50 0.040095
5 a 1 0.50 0.150359
6 a 2 0.32 0.043512
7 a 2 0.32 0.088408
8 a 2 0.32 0.005158
9 a 2 0.32 0.107054
10 a 2 0.32 0.029050
11 a 2 0.32 0.099361
12 b 0 0.40 0.057752
13 b 0 0.40 0.177103
14 b 1 0.60 0.218634
15 b 1 0.60 0.098535
16 b 1 0.60 0.065746
17 b 1 0.60 0.190805
18 b 1 0.60 0.191425
What I want to do, is to select top 5 (arbitrary number, can be N) of rows per each group based on probabilityB
AND on the share of the sizes of every cluster
. If we only look at group a
, there are 3 clusters: 0, 1 and 2. Their respective size shares are:
group cluster
a 0 0.333333
1 0.166667
2 0.500000
Name: probabilityA, dtype: float64
And here, if I want top 5 rows based on this shares I would take
(round
(df
.groupby(["group", "cluster"])["probabilityA"]
.count() /
df
.groupby(["group", "cluster"])["probabilityA"]
.count()
.sum(level = 0)
* 5)
group cluster
a 0 2.0
1 1.0
2 2.0
2 elements from cluster 0 and 2, and only 1 element from cluster 1 based on probabilityB
column. So, my result will look like this (index is irrelevant in the sample below):
group cluster probabilityA probabilityB
0 a 1 0.50 0.150359
1 a 2 0.32 0.107054
2 a 2 0.32 0.088408
3 a 0 0.28 0.153013
4 a 0 0.28 0.133686
5 b 0 0.40 0.177103
6 b 1 0.60 0.218634
7 b 1 0.60 0.191425
8 b 1 0.60 0.190805
9 b 1 0.60 0.098535
Is there a way I can achieve it?
thanks in advance!
I think, the most clear solution is to divide tke task into steps:
Generate counts for each top-level group:
c1 = df.groupby(["group"])["probabilityA"].count().rename('c1')
For your data, the result is:
group a 12 b 7 Name: c1, dtype: int64
Set the number of rows to take from each top-level group:
N = 5
Generate the counts of rows to take from each second-level group:
cnt = df.groupby(["group", "cluster"])["probabilityA"].count().rename('c2')\\ .reset_index(level=1).join(c1).set_index('cluster', append=True)\\ .apply(lambda row: N * row.c2 / row.c1, axis=1).round().astype(int)
For your data, the result is:
group cluster a 0 2 1 1 2 2 b 0 1 1 4 dtype: int32
Then define the function, retutning the respective number of "top" rows:
def takeFirst(grp): grpKey = tuple(grp.iloc[0, 0:2]) grpCnt = cnt.loc[grpKey] return grp.nlargest(grpCnt, 'probabilityB')
And the last step is to compute the result:
df.groupby(['group', 'cluster']).apply(takeFirst)
For your data, the result is:
group cluster probabilityA probabilityB group cluster a 0 0 a 0 0.28 0.153013 1 a 0 0.28 0.133686 1 5 a 1 0.50 0.150359 2 9 a 2 0.32 0.107054 11 a 2 0.32 0.099361 b 0 13 b 0 0.40 0.177103 1 14 b 1 0.60 0.218634 18 b 1 0.60 0.191425 17 b 1 0.60 0.190805 15 b 1 0.60 0.098535
I delibarately left group and cluster as index columns, to ease the identification from which group they were taken, but in the final version you can append .reset_index(level=[0,1], drop=True)
to drop them.
I think if you groupby ProbabilityA - you might be able to achieve this.
df.groupby(['group', 'cluster', 'probabilityA']).aggregate({
'group': 'first',
'cluster': 'first',
'probabilityA': lambda x: round(len(x)/(sum(x)*(len(x))*n),
'probabilityB': lambda x: sum(x)
})
The solution above was faulty because count().sum() is different on overall groupby and on probabilityA alone separately, which is why I did the following:
UPDATE - Full Solution:
df.sort_values(by=['group', 'cluster','probabilityB'], ascending=False)
cluster = pd.DataFrame(round(df.groupby(['group', 'cluster', 'probabilityA'])['probabilityA'].count()
/ df.groupby(['group', 'cluster', 'probabilityA'])['probabilityB'].count().sum(level=0)*5))
cluster.reset_index(level=['group', 'cluster', 'probabilityA'], inplace=True)
cluster = cluster.rename(columns={0: 'counts'})
cluster['counts'] = pd.to_numeric(cluster['counts'], downcast='integer')
output = pd.concat(cluster.apply(lambda x: df.loc[(df['group'] == x['group']) & (df['cluster'] == x['cluster'])].groupby(
['group', 'cluster']).head(x['counts']), axis=1).tolist())
Output: See Output DataFrame Here
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