How to get indices from a list/ndarray?

Question

I have a list which looks like:

[[0,1,2], [1,2,3], [2,3,4], [3,4,5]]

I can make it to an array like:

array([[0,1,2],
       [1,2,3],
       [2,3,4],
       [3,4,5]])

So all together I have 4 rows and each row has 3 columns. Now I want to find the indices of all the elements which are greater than 2, so for the whole matrix, the indices should be:

((1,2),(2,1),(2,2),(3,1),(3,2),(3,3))

Then for each row, I will randomly picked out a col index which indicates a value greater than 2. Now my code is like:

a = np.array([[0,1,2],[1,2,3],[2,3,4],[3,4,5]]
out = np.ones(4)*-1
cur_row = 0
col_list = []
for r,c in np.nonzero(a>2):
    if r == cur_row:
        col_list.append(c)
    else:
        cur_row = r
        shuffled_list = shuffle(col_list)
        out[r-1] = shuffled_list[0]
        col_list = []
        col_list.append(c)

I hope to get a out which looks like:

array([-1, 2, 1, 2])

However, now when I run my code, it shows

ValueError: too many values to unpack

Anyone knows how I fix this problem? Or how should I do to achieve my goal? I just want to run the code as fast as possible, so any other good ideas is also more than welcome.

Loop over zip(*np.nonzero(a>2)).

Ashwini Chaudhary
– Ashwini Chaudhary

2015-05-11 13:50:04 +00:00
Commented May 11, 2015 at 13:50 — Ashwini Chaudhary
– Ashwini Chaudhary, Commented May 11, 2015 at 13:50

Eli Korvigo · Accepted Answer · 2015-05-11 14:02:14Z

3

Try this.

import numpy as np

arr = np.array([[0,1,2],
               [1,2,3],
               [2,3,4],
               [3,4,5]])
indices = np.where(arr>2)

for r, c in zip(*indices):
    print(r, c)

Prints

So, it should work. You can use itertools.izip as well, it would even be a better choice in this case.

A pure numpy solution (thanks to @AshwiniChaudhary for the proposition):

for r, c in np.vstack(np.where(arr>2)).T:
    ...

though I'm not sure this will be faster than using izip or zip.

edited May 11, 2015 at 14:02

answered May 11, 2015 at 13:45

Eli Korvigo

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12 Comments

Malik Brahimi Over a year ago

Of course, it is much simpler with numpy but it can easily be done without it.

canyon289 Over a year ago

He's already loaded numpy, I feel like its even easier, and more pythonic, to use the built in methods

Sean Summers Over a year ago

I still get the ValueError, unpacking zip(*arr) into r,c

Ashwini Chaudhary Over a year ago

You should use np.vstack(np.where(arr>2)).T rather than zip.

Eli Korvigo Over a year ago

@gladys0313 you're welcome. Btw, you've got two answers that do the job, you might consider accepting one of them.

|

canyon289 · Accepted Answer · 2015-05-11 13:49:57Z

0

You could just compare the array to your value and use where.

a = np.array([[0,1,2],[1,2,3],[2,3,4],[3,4,5]])
np.where(a>2)

(array([1, 2, 2, 3, 3, 3], dtype=int64), array([2, 1, 2, 0, 1, 2], dtype=int64))

To get your tuples

list(zip(*np.where(a>2)))

[(1, 2), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)]

answered May 11, 2015 at 13:49

canyon289

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Comments

gladys0313 · Accepted Answer · 2015-05-11 14:13:26Z

0

I have made it out, the code should be:

a = np.array([[0,1,2],[1,2,3],[2,3,4],[3,4,5]])
out = np.ones(4)*-1
cur_row = 0
col_list = []
for r,c in zip(*(np.nonzero(a>2))):
    if  r == cur_row:
        col_list.append(c)
    else:
        cur_row = r
        shuffle(col_list)
        if len(col_list) == 0:
            out[r-1] = -1
        else:
            out[r-1] = col_list[0]
        col_list = []
        col_list.append(c)

shuffle(col_list)
if len(col_list) == 0:
    out[len(out)-1] = -1
else:
    out[len(out)-1] = col_list[0]

The part in the end but outside the forloop is to make sure that the last row will be taken care of.

It works in my case.

answered May 11, 2015 at 14:13

gladys0313

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