Select specific elements from an array

Question

I have an array:

  X = [[5*,  0,     0,      0,      0,      0,      0,      0],
       [9*,  6,     0,      0,      0,      0,      0,      0],
       [4,   6*,    8,      0,      0,      0,      0,      0],
       [0,   7*,    1,      5,      0,      0,      0,      0],
       [9,   3,     3*,     4,      4,      0,      0,      0],
       [4,   5,     5*,     6,      7,      5,      0,      0],
       [4,   5,     6,      8*,     7,      7,      8,      0], 
       [4,   7,     8,      9*,     7,      3,      9,      6]]

I want to select and append all the values that are marked by *. The approach is basically to select 0th element from 0th and 1st row...1th element from 2nd and 3rd row..and so on.

The resulting set should be:

  Result = ((X[0][0], (X[1][0]), (X[2][1], X[3][1]), (X[4][2], X[5][2]), (X[6][3], X[7][3]))

Which can be written as:

  Result = ((X[n+0][n], (X[n+1][n]), (X[n+2][n+1], X[n+3][n+1]), (X[n+4][n+2], X[n+5][n+2]), (X[n+6][n+3], X[n+7][n+3]))

  Where n = 0

How do I do that? I have applied this but its not working:

 Result = []

 for a in X:
     Result.append([[[ a[i][j] ] for i in range(0,8)] for j in range(0,8)])

But no results. Any guesses?

You should probably move to numpy ...

wim
– wim

2013-04-16 04:05:30 +00:00
Commented Apr 16, 2013 at 4:05 — wim
– wim, Commented Apr 16, 2013 at 4:05

Bi Rico · Accepted Answer · 2013-04-16 18:46:22Z

3

Because of the numpy tag I thought I would add this:

import numpy as np
X = np.array([[5 ,  0,     0,      0,      0,      0,      0,      0],
              [9 ,  6,     0,      0,      0,      0,      0,      0],
              [4,   6 ,    8,      0,      0,      0,      0,      0],
              [0,   7 ,    1,      5,      0,      0,      0,      0],
              [9,   3,     3 ,     4,      4,      0,      0,      0],
              [4,   5,     5 ,     6,      7,      5,      0,      0],
              [4,   5,     6,      8 ,     7,      7,      8,      0], 
              [4,   7,     8,      9 ,     7,      3,      9,      6]])
i = np.array([0, 1, 2, 3, 4, 5, 6, 7])
j = np.array([0, 0, 1, 1, 2, 2, 3, 3])

result = X[i, j]
print result
# [5 9 6 7 3 5 8 9]

To generate i and j in the general case you can do something like:

n = 8
i = np.arange(n)
j = np.arange(n) // 2

edited Apr 16, 2013 at 18:46

answered Apr 16, 2013 at 4:16

Bi Rico

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2 Comments

Blair Over a year ago

Are you sure that last line is correct? On my version of NumPy (1.5.1), repeat needs two arguments. Should that also be arange?

Bi Rico Over a year ago

@Blair, indeed it should have been arange, I've fixed it.

Volatility · Accepted Answer · 2013-04-16 04:13:58Z

2

This will work if X has an even number of lists in it:

>>> [(X[2*i][i], X[2*i+1][i]) for i in range(len(X)//2)]
[(5, 9), (6, 7), (3, 5), (8, 9)]

If you don't mind flattened lists, then then will work for X of any length:

>>> [lst[idx//2] for idx, lst in enumerate(X)]
[5, 9, 6, 7, 3, 5, 8, 9]

edited Apr 16, 2013 at 4:13

answered Apr 16, 2013 at 4:03

Volatility

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Comments

thkang · Accepted Answer · 2013-04-16 04:02:52Z

1

try this:

from itertools import chain, count, tee

lst = [row[i] for row, i in zip(array, chain.from_iterable(zip(*tee(count(), 2))))]

answered Apr 16, 2013 at 4:02

thkang

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dawg · Accepted Answer · 2013-04-16 15:18:58Z

0

In Numpy:

import numpy as np
x = [['5*','0 ','0 ','0 ','0 ','0 ','0 ','0 '],
     ['9*','6 ','0 ','0 ','0 ','0 ','0 ','0 '],
     ['4 ','6*','8 ','0 ','0 ','0 ','0 ','0 '],
     ['0 ','7*','1 ','5 ','0 ','0 ','0 ','0 '],
     ['9 ','3 ','3*','4 ','4 ','0 ','0 ','0 '],
     ['4 ','5 ','5*','6 ','7 ','5 ','0 ','0 '],
     ['4 ','5 ','6 ','8*','7 ','7 ','8 ','0 '],
     ['4 ','7 ','8 ','9*','7 ','3 ','9 ','6 ']]

a=np.array(x)

Then do a list comprehension and/or Numpy slice to get the items:

[a[i:,j][:2].tolist() for i,j in zip(range(0,7,2),range(0,7,1))]

or

[a[i*2:,i][:2].tolist() for i in range(len(a)//2)]

or

a[range(len(a)),np.repeat(range(len(a)//2),2)].reshape(4,2).tolist()

Any case, output is:

[['5*', '9*'], ['6*', '7*'], ['3*', '5*'], ['8*', '9*']]

edited Apr 16, 2013 at 15:18

answered Apr 16, 2013 at 5:48

dawg

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L Tyrone · Accepted Answer · 2024-07-13 08:04:41Z

0

I think the simpliest way in one line:

Result = [[X[index-1][int(index/2-0.5)],X[index][int(index/2-0.5)]] for index in range(1,len(X),2)]

edited Jul 13, 2024 at 8:04

L Tyrone

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answered Jul 12, 2024 at 14:44

Dr Youssef Zied Elhechmi

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1 Comment

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