Transform a matrix

Question

Write a program or function where a matrix A (maximum value of dimensions of the matrix is 100) is transformed to a new matrix B. Each element of the new matrix B is the sum of the positive neighbors of the corresponding element of the matrix A.

Only neighbors in the four orthogonal directions are considered, and the edges do not wrap around.

Example input:

 1  2  3
 4 -1 -2
-3 -4 10
 1  2 10

Output:

 6  4  2
 1  6 13
 5 12 10
 2 11 12

Rules:

• As input use integers.
• This is a code-golf, so the shortest code in bytes wins.

Answer 1

MATL, 9 bytes

t0>*1Y6Z+

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Explanation

The input matrix is multplied by an appropriate mask to make negative values equal to 0. Then a 2D convolution is applied to compute the sum of neighbours of each entry.

t     % Take input implicitly: 2D array. Duplicate
0>    % Is each entry positive? This gives a mask of positive values
*     % Multiply: set negative values of input array to zero
1Y6   % Predefined literal: [0 1 0; 1 0 1; 0 1 0]
Z+    % 2D convolution preserving size. Implicitly display

Answer 2

Octave, 46 44 40 bytes

Saved 2 bytes thanks to @flawr
@LuisMendo's kernel was 4 bytes shorter than @flawr's.

@(M)conv2(M.*(M>0),(x='aba')~=x','same')

Just like @LuisMendo's answer! Only less... golfy.

You can see it here on ideone.

Answer 3

JavaScript (ES6), 99 94 bytes

a=>a.map((b,i)=>b.map((_,j)=>(g=(c=j,k=j)=>c[k]>0&&c[k])(a[i-1])+g(a[i+1])+g(b,j-1)+g(b,j+1)))

Accepts and returns a two-dimensional array.

Edit: Completely rewritten when I discovered that default arguments work when you pass an explicitly undefined value, such as when you index off the end of an array.

Answer 4

JavaScript (ES6), 95 93 bytes

document.write("<pre>"+(

m=>m.map((a,r)=>a.map((_,c)=>(s=(x,y=0)=>(n=(m[r+y]||0)[c+x])>0&&n)(1)+s(-1)+s(0,1)+s(0,-1)))

)([[  1,  2,  3 ],
   [  4, -1, -2 ],
   [ -3, -4, 10 ],
   [  1,  2, 10 ]])

.join`\n`)

Answer 5

Python with SciPy, 127 bytes

from scipy.signal import*
lambda A,r=range(3):convolve2d([[x*(x>0)for x in y]for y in A],[[0,1,0],[1,0,1],[0,1,0]],mode='same')

This computes the result using Luis Mendo's method.

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Answer 6

Mathcad, bytes

Uses standard 2D convolution of the filtered array with a 3x3 kernel. Variants with negative element sums and diagonal kernel also added as part compensation for the program not being in the running for least bytes.

---

No byte count entered as Mathcad scoring has yet to be determined. However, using keyboard equivalence, it is the region of 28 bytes assuming that matrix input doesn't count towards the total.

Note that what you see in the image above is exactly how the solutionn is entered and displayed in Mathcad.

Answer 7

Jelly, 23 20 18 bytes

0;+ḊṖ
ZÇ€Z+Ç€
+AHÇ

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Algorithm

Let's say there is only one row: [1,2,3,4].

Let's say A is the result of prepending a zero, i.e. [0,1,2,3,4].

B is the result of removing the first item, i.e. [2,3,4].

Then the final result is simply A+B vectorized, then removing the last item.

Now, the algorithm is to apply this to every row as well as every column, then find their vectorized sum.

To each column?! I thought Jelly does not support this...

You're right. Therefore, I transposed it, applied to each row, then transposed it again.

Algorithm for removing negative numbers

Here, you just add to each number their absolute. It effectively eliminates negative numbers while doubling each positive number. Then, just halve the whole matrix.

Answer 8

Pyth, 36 bytes

Lm+Vt+d0+0dbJmm/+.akk2dQ.b+VNYyJCyCJ

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Direct translation of my answer in Jelly.