Grouping column indices of an array based on max value of another list in python

Question

Suppose I have an array

k= array([[1, 2, 3, 4, 5],
          [5, 6, 7, 8, 9],
          [2, 5, 4, 7, 3],
          [4, 7, 6, 8, 2],
          [1, 2, 4, 3, 6],
          [7, 8, 9, 5, 4]])

Assuming that after computing for each columns in an array I got array([0.6,0.4,0.75,0.2,0.75]) respectively, such that:

computation on column1 i.e. computation on array([1,5,2,4,1,7]) results in 0.6, computation on column2 i.e. computation on array([2,6,5,7,2,8]) results in 0.4, computation on column3 i.e. computation on array([3,7,4,6,4,9]) results in 0.75, and so on.

Let the computed list be m. such that

m=array([0.6,0.4,0.75,0.2,0.75])

So far I have computed for single columns in array k. Now I would like to group the elements in list m based on the largest floating point element in the list m and compute again on k. For example:

m[2]=m[4]=0.75 (largest number in the array), that would mean index 2 and index 4 of column in array k is largest. So keeping that index number common I would like to group k[:,2] with k[:,0],k[:,2] with k[:,1],k[:,2] with k[:,3] and similarly k[:,4] with k[:,0],k[:,4] with k[:,1],k[:,4] with k[:,3] and compute again on k,

such that grouping k[:,2] with k[:,0] means :

k0_2=array([[1,3],          k1_2=array([[2,3],          k3_2=array([[4,3],
            [5,7],                      [6,7],                      [8,7],
            [2,4],                      [5,4],                      [7,4],
            [4,6],                      [7,6],                      [8,6],
            [1,4],                      [2,4],                      [3,4],  
            [7,9]])                     [8,9]])                     [5,9]])

k0_4=array([[1,5],          k1_4=array([[2,5],          k3_4=array([[4,5],
            [5,9],                      [6,9],                      [8,9],
            [2,9],                      [5,3],                      [7,3],
            [4,2],                      [7,2],                      [8,2],
            [1,6],                      [2,6],                      [3,6],  
            [7,4]])                     [8,4]])                     [5,4]])

Can anyone please give me any clue regarding grouping column indices of array k based on max value of list m as shown above .

I don't quite understand your paragraph with the sentence "So keeping that index number common...". What do you mean by "compute again on k"? And how do you get the groups you've described in that sentence? Could you make this a bit clearer, for example describe to us in more details the steps we would take to get these groups? — YXD
– YXD, Commented Sep 12, 2015 at 11:34
By saying i would like to keep the index number common i meant that all I want to keep the index number of greater element in the list M common and use that index, to group with non max index as shown above. — Sayantan Mukherjee
– Sayantan Mukherjee, Commented Sep 12, 2015 at 11:47
I am trying to understand what the paragraph I mentioned means and how you work out this grouping. If that means you also need to explain how you got the array m then sure. — YXD
– YXD, Commented Sep 12, 2015 at 11:50
Based on lower approximation i got m. You will get good enough literature on that topic. — Sayantan Mukherjee
– Sayantan Mukherjee, Commented Sep 12, 2015 at 12:15
Your question has a lot of irrelevant detail but doesn't specify what you actually want. What does the required answer look like? Do you want a list of 6-by-2 arrays? If so, what order should they come in? If not, what is the form of the answer? — saulspatz
– saulspatz, Commented Sep 12, 2015 at 15:39

saulspatz · Accepted Answer · 2015-09-12 16:08:48Z

1

Does this help?

import numpy as np

k= np.array(
    [[1, 2, 3, 4, 5],
     [5, 6, 7, 8, 9],
     [2, 5, 4, 7, 3],
     [4, 7, 6, 8, 2],
     [1, 2, 4, 3, 6],
     [7, 8, 9, 5, 4]])

cols=[2,4]
others = [c for c in range(k.shape[1]) if c not in cols]
groups = [k[:,[o, c]] for c in cols for o in others]
for g in groups:
    print(g)
    print('')

This gives me

[[1 3]
 [5 7]
 [2 4]
 [4 6]
 [1 4]
 [7 9]]

[[2 3]
 [6 7]
 [5 4]
 [7 6]
 [2 4]
 [8 9]]

[[4 3]
 [8 7]
 [7 4]
 [8 6]
 [3 4]
 [5 9]]

[[1 5]
 [5 9]
 [2 3]
 [4 2]
 [1 6]
 [7 4]]

[[2 5]
 [6 9]
 [5 3]
 [7 2]
 [2 6]
 [8 4]]

[[4 5]
 [8 9]
 [7 3]
 [8 2]
 [3 6]
 [5 4]]

I think there's a typographical error in your k_04 array. I think k_04[2,10 should be 3, not 9, isn't that so?

answered Sep 12, 2015 at 16:08

saulspatz

5,3117 gold badges43 silver badges49 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Sayantan Mukherjee Over a year ago

Thanks a lot :) and yeah my mistake. It's an error anyway thanks bro

Sayantan Mukherjee · Accepted Answer · 2015-09-12 18:35:15Z

0

Here is another solution which I along with one of my friend figured it out :

import numpy as np

k= np.array(
    [[1, 2, 3, 4, 5],
     [5, 6, 7, 8, 9],
     [2, 5, 4, 7, 3],
     [4, 7, 6, 8, 2],
     [1, 2, 4, 3, 6],
     [7, 8, 9, 5, 4]])

cols=[2,4]
kl = np.delete(k, np.s_[cols],1)

 for i in range(len(cols)):
     for j in range(len(kl[0])):
          print np.column_stack((k[:,cols[i]],kl[:,j]))

answered Sep 12, 2015 at 18:35

Sayantan Mukherjee

2051 gold badge5 silver badges13 bronze badges

Collectives™ on Stack Overflow

Grouping column indices of an array based on max value of another list in python

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related