Turn logical matrix into string vector

Hi. I have the logical matrix idx and the string array vec.

idx = logical([0 1 0; 0 0 1; 1 0 0; 1 0 0; 0 1 0])
vec = ["A","B","C"]

Each row of idx indicates which letter (from vec) should be assigned to that row.

The desired result is:

result = ["B";"C";"A";"A";"B"]

Is there a way to 'apply' matrix idx to a string matrix, or something like the following:

temp = repmat(vec, [size(idx,1), 1])
temp(idx) %This returns something but not as expected

Answer 1

[Walter Roberson]

idx = logical([0 1 0; 0 0 1; 1 0 0; 1 0 0; 0 1 0])

idx = 5×3 logical array
   0   1   0
   0   0   1
   1   0   0
   1   0   0
   0   1   0

vec = ["A","B","C"]

vec = 1×3 string array
    "A"    "B"    "C"

vec(idx(:,1)*1 + idx(:,2)*2 + idx(:,3) * 3).'

ans = 5×1 string array
    "B"
    "C"
    "A"
    "A"
    "B"

vec(idx*(1:size(idx,2)).').'

ans = 5×1 string array
    "B"
    "C"
    "A"
    "A"
    "B"

[Haris K.]

Great, thank you!

Answer 2

[Haris K.]

Btw, here is one solution:

arrayfun(@(i) vec(idx(i,:)),1:size(idx,1))'

But I was wondering if there is something that does not use any for-loop directly or indirectly.

Answer 3

[Bruno Luong]

Assuming idx has one 1 per row

idx = logical([0 1 0; 0 0 1; 1 0 0; 1 0 0; 0 1 0]);
vec = ["A","B","C"];

[r,c]=find(idx);
result(r)=vec(c)

result = 1×5 string array
    "B"    "C"    "A"    "A"    "B"

[Haris K.]

Thanks all for your effort!

Answer 4

[Yvan Lengwiler]

(I wrote my answer as a comment above. Sorry. Here it is as a formal answer.)

Of course, even if a solution does not appear to use a loop as seen in the Matlab program code, there will be a loop in the ultimapte machine code anyway. Not having a code in the Matlab program can improve legibility of the text for humans, but it is not obvious that it improves the speed of execusion.

Anyway, here is a solution:

idx = logical([0 1 0; 0 0 1; 1 0 0; 1 0 0; 0 1 0]);
vec = ["A","B","C"];
tempvec = repmat(vec', [size(idx',2), 1]);
tempidx = idx'; tempidx = tempidx(:);
result = tempvec(tempidx)'

[Walter Roberson]

Seems over-complicated compared to

vec(idx*(1:size(idx,2)).').'

[Yvan Lengwiler]

True :-)