Indexing with intercept for repeating values

a = [1;1;1;2;3;3;4;5;6;7];
b = [1;1;1;1;1;2;3;4;4;4;5;6;7;7];

% I should get the following logical index as answer. Find position in b with equivalent in a.
idx = [1;1;1;0;0;1;1;1;0;0;1;1;1;0];

I tried intersect and ismember but cant seem to find an elegant way without looping. Thanks in advance!

[madhan ravi]

Are a and b always sorted?

[hal9k]

yes. In ascending order.

[Matt J]

Couldn't this be a solution,

idx = [0;1;1;1;0;1;1;1;0;0;1;1;1;0];

[hal9k]

Sorry no! The idx should have first 3 values set as 1,1,1 and 4th value is 0 (since there is no other 1's left in a).

Answer 1

[Matt J]

a = [1;1;1;2;3;3;4;5;6;7];
b = [1;1;1;1;1;2;3;4;4;4;5;6;7;7];

c = splitapply(@(x){(1:numel(x)).'<=sum(a==x(1))} , b , findgroups(b));

idx=cell2mat(c)

idx = 14x1 logical array
   1
   1
   1
   0
   0
   1
   1
   1
   0
   0

[madhan ravi]

Could it be

c = splitapply(@(x){(1:numel(x)).'<=sum(a==x(1))} , b , b); %? since b is already sorted?

[Matt J]

No, for example,

a = [1;1;1;2;3;3;4;5;6;7]+1;
b = [1;1;1;1;1;2;3;4;4;4;5;6;7;7]+1;

c = splitapply(@(x){(1:numel(x)).'<=sum(a==x(1))} , b , b);

Error using splitapply (line 111)
For N groups, every integer between 1 and N must occur at least once in the vector of group numbers.

[madhan ravi]

Thanks Matt for the clarification.

[hal9k]

Elegant solution but it took really long time for large arrays which I am working with.

[Matt J]

Well, you didn't say it had to be fast. You just asked for a method that avoided loops.

Answer 2

[the cyclist]

Here is one way:

a = [1;1;1;2;3;3;4;5;6;7];
b = [1;1;1;1;1;2;3;4;4;4;5;6;7;7];

idx = zeros(numel(b),1);

ii = 1;
while any(b) && any(a)
    if a(1)==b(1)
        idx(ii) = 1;
        a(1) = [];
        b(1) = [];
        ii = ii+1;
    elseif b(1) < a(1)
        b(1) = [];
        ii = ii+1;
    else
        a(1) = [];
    end
end

idx

idx = 14×1

1 1 1 0 0 1 1 1 0 0

This method "cannibalizes" a and b, so you should make copies of those first if you need them.

[hal9k]

This works but I wanted to avoid loops.

Answer 3

[the cyclist]

Here is one way:

a = [1;1;1;2;3;3;4;5;6;7];
b = [1;1;1;1;1;2;3;4;4;4;5;6;7;7];

ha = histcounts(a,1:max(a)+1);
hb = histcounts(b,1:max(b)+1);

d = max(0,min(hb,ha));

idx = zeros(numel(b),1);

loc = cumsum([1,hb]);

for ii = 1:numel(d)
    idx(loc(ii):loc(ii)+d(ii)-1) = 1;
end

idx

idx = 14×1

1 1 1 0 0 1 1 1 0 0

[hal9k]

I tested this but it would fail for this:

a = [1;1;1;2;3;3;4;5;6;7];

b = [1;1;1;1;1;2;3;4;4;4;5;6;7;7;8;8]; (a might not have all the elements(

Correct result is idx = [1;1;1;0;0;1;1;1;0;0;1;1;1;0;0;0];

But I got error with the code. So the above code would only work if a and b has same number of unique elements.

[the cyclist]

This had a very simple fix.

a = [1;1;1;2;3;3;4;5;6;7];
b = [1;1;1;1;1;2;3;4;4;4;5;6;7;7;8;8];

maxval = max([a;b]);

ha = histcounts(a,1:maxval+1);
hb = histcounts(b,1:maxval+1);

d = max(0,min(hb,ha));

idx = zeros(numel(b),1);

loc = cumsum([1,hb]);

for ii = 1:numel(d)
    idx(loc(ii):loc(ii)+d(ii)-1) = 1;
end

idx

idx = 16×1

1 1 1 0 0 1 1 1 0 0