How to delete all occurrences of a value from a matrix?

Question

Let h be my matrix

h=[[0,346,34,6,7,0,0,34634,6,0],[2352,205230,523,50,5023,502,350,0,0]]

and I want to delete all the zeros from h.

So the result should be:

h=[[346,34,6,7,34634,6],[2352,205230,523,50,5023,502,350]]

I tried several different things from HERE and other questions but nothing seems to work for a matrix. Is there any nice solution to that?

score 2 · Answer 1 · answered Mar 02 '17 at 17:17

2

You can simply use list comprehension:

h = [[element for element in row if element] for row in h]

So here we iterate over every row in h. For each row we construct a new list [element for element in row if element]. This means that for every element in that row, we check if it is equal to 0 (with if element). If it is not, bool(element) is True and thus we add element to that list. Otherwise we omit element.

answered Mar 02 '17 at 17:17

Willem Van Onsem

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`chain.from_iterable(h.tolist())` in place of `h` if it's a `numpy` matrix. – Nickil Maveli Mar 02 '17 at 18:08
@NickilMaveli: why do you need to `chain`? – Willem Van Onsem Mar 02 '17 at 18:14
It's because `tolist()` creates an additional outer `list` which you'd need to flatten. – Nickil Maveli Mar 02 '17 at 18:16
@NickilMaveli: I've tested it and you get a 2d list back which is what we want. – Willem Van Onsem Mar 02 '17 at 18:17

martineau · Accepted Answer · 2017-03-02T18:31:09.173

2

If you want to change the matrix in-place, you could do it with a function like the following. The trick is to go through the elements of each row backwards so that the indices of successive elements aren't changed because of ones being removed earlier.

def remove_zeros(matrix):
    for row in matrix:
        length = len(row)-1
        for i, elem in enumerate(reversed(row)):
            if not elem:  # zero value?
                row.pop(length-i)

h = [[0, 346, 34, 6, 7, 0, 0, 34634, 6, 0], [2352, 205230, 523, 50, 5023, 502, 350, 0, 0]]

remove_zeros(h)

print(h)  # -> [[346, 34, 6, 7, 34634, 6], [2352, 205230, 523, 50, 5023, 502, 350]]

edited Mar 02 '17 at 18:31

answered Mar 02 '17 at 17:40

martineau

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I would use `reversed(row)` instead of `row[::-1]` – Felipe Nardi Batista Mar 02 '17 at 17:46
@FelipeNardiBatista: Sure, that would work, too, and is more readable. However most "Pythonistas" would immediately recognize the `[::=1]` since it was what was commonly used before the `reversed()` built-in was added to the language. – martineau Mar 02 '17 at 18:09
yes, but as of right now, it is more readable and more efficient as it does not copy the row, it just iterate over it – Felipe Nardi Batista Mar 02 '17 at 18:19
@FelipeNardiBatista: Ok, ok... `;-)` – martineau Mar 02 '17 at 18:32

How to delete all occurrences of a value from a matrix?

2 Answers2