CODEWITHSUNDEEP
pandasnumpypandas-groupby
Dmitriy Grankin
Dmitriy Grankin

Reputation: 608

groupby list of lists of indexes

I have a list of np. arrays, representing indexes of pandas dataframe.

I need to groupby index to get each group for each array

let's say, that is the df:

index values
0     2
1     3
2     2
3     2
4     4
5     4
6     1
7     4
8     4
9     4

and that is the list of np.arrays:

[array([0, 1, 2, 3]), array([6, 7, 8])]

from this data I expect to get 2 groups without loop opertaions as a single groupby object:

group1:

index values
0     2
1     3
2     2
3     2

group2:

index values
6     1
7     4
8     4

I would stress again that finally I need to get a single groupby object.

Thank you!

Upvotes: 0

Views: 3396

Answers (3)

BENY
BENY

Reputation: 323226

I still using for-loop to create the groupby key dict

l=[np.array([0, 1, 2, 3]), np.array([6, 7, 8])]
df=pd.DataFrame([2, 3, 2, 2, 4, 4, 1, 4, 4, 4],columns=['values'])

from collections import ChainMap
L=dict(ChainMap(*[dict.fromkeys(y,x) for x, y in enumerate(l)]))
list(df.groupby(L))
Out[33]: 
[(0.0,        values
  index        
  0           2
  1           3
  2           2
  3           2), (1.0,        values
  index        
  6           1
  7           4
  8           4)]

Upvotes: 3

Scott Boston
Scott Boston

Reputation: 153460

This seems like an X-Y problem:

l = [np.array([0,1,2,3]), np.array([6,7,8])]
df_indx = pd.DataFrame(l).stack().reset_index()
df_new = df.assign(foo=df['index'].map(df_indx.set_index(0)['level_0']))
for n,g in df_new.groupby('foo'):
    print(g)

Output:

   index  values  foo
0      0       2  0.0
1      1       3  0.0
2      2       2  0.0
3      3       2  0.0
   index  values  foo
6      6       1  1.0
7      7       4  1.0
8      8       4  1.0

Upvotes: 0

Andrea Mannari
Andrea Mannari

Reputation: 1012

df=pd.DataFrame([2,3,2,2,4,4,1,4,4,4],columns=['values'])
df.index.name ='index'
l=[np.array([0, 1, 2, 3]), np.array([6, 7, 8])]

group1= df.loc[pd.Series(l[0])]
group2= df.loc[pd.Series(l[1])]

Upvotes: 2

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