How do you search for a number in a cell array containing different length arrays?

Is there a way to search for an element inside a cell array? I have the following cell array:

1×10 cell array

Columns 1 through 5

{[2 5 8 9]} {[1 2 4 5 7 10]} {[3 5 6 7 8 9 10]} {[1 2 3 5 8 9 10]} {[1 2 4 6 8 9]}

Columns 6 through 10

{[1 3 4 5 6 8]} {[1 2 4]} {[4 5 7 9]} {[1 2 5 7 10]} {[1 2 3 4 6 8]}

I'm trying to determine the index of the array that contains a number, such as 1 for example. So for this example 1 is in array elements 2, 4, 5, 6, 7,9 and 10.

Answer 1

[Dyuman Joshi]

in = {[2 5 8 9] [1 2 4 5 7 10] [3 5 6 7 8 9 10] [1 2 3 5 8 9 10] [1 2 4 6 8 9] [1 3 4 5 6 8] [1 2 4] [4 5 7 9] [1 2 5 7 10] [1 2 3 4 6 8]}

in = 1×10 cell array
    {[2 5 8 9]}    {[1 2 4 5 7 10]}    {[3 5 6 7 8 9 10]}    {[1 2 3 5 8 9 10]}    {[1 2 4 6 8 9]}    {[1 3 4 5 6 8]}    {[1 2 4]}    {[4 5 7 9]}    {[1 2 5 7 10]}    {[1 2 3 4 6 8]}

idx = find(cellfun(@(x) ismember(1, x), in))

idx = 1×7

2 4 5 6 7 9 10

[Dyuman Joshi]

Though, the fastest approach would be to use a for loop.

in = {[2 5 8 9] [1 2 4 5 7 10] [3 5 6 7 8 9 10] [1 2 3 5 8 9 10] [1 2 4 6 8 9] [1 3 4 5 6 8] [1 2 4] [4 5 7 9] [1 2 5 7 10] [1 2 3 4 6 8]};

in = 1×10 cell array
    {[2 5 8 9]}    {[1 2 4 5 7 10]}    {[3 5 6 7 8 9 10]}    {[1 2 3 5 8 9 10]}    {[1 2 4 6 8 9]}    {[1 3 4 5 6 8]}    {[1 2 4]}    {[4 5 7 9]}    {[1 2 5 7 10]}    {[1 2 3 4 6 8]}

vec = 1:numel(in);
%Lazy preallocation
z = vec~=0;

for k=vec
    z(k) = any(in{k}==1);
end
z

z = 1×10 logical array
   0   1   0   1   1   1   1   0   1   1

out = vec(z)

out = 1×7

2 4 5 6 7 9 10

[Dyuman Joshi]

Some timings -

%Generate random data
vec = randi([25 50], 1, 50000);
s = sum(vec)

s = 1873943

arr = randi([1 100], 1, s);
in = mat2cell(arr, 1, vec);


f1 = @(x) funloop(x);
f2 = @(x) funcellany(x);
f3 = @(x) funcellismember(x);

F1 = @() f1(in);
F2 = @() f2(in);
F3 = @() f3(in);


%Check for equality first
isequal(F1(), F2(), F3())

ans = logical
   1

fprintf('Time taken by for loop = %f seconds', timeit(F1))

Time taken by for loop = 0.008287 seconds

fprintf('Time taken by cellfun and any() = %f seconds', timeit(F2))

Time taken by cellfun and any() = 0.090079 seconds

fprintf('Time taken by cellfun and ismember() = %f seconds', timeit(F3))

Time taken by cellfun and ismember() = 0.128586 seconds

As can be observed, for loop is atleast 10x faster than both of the cellfun() methods.

function out = funloop(in)
    vec = 1:numel(in);
    %Lazy preallocation
    z = vec~=0;
    for idx=vec
        z(idx) = any(in{idx}==1);
    end
    out = vec(z);
end


function out = funcellany(in)
    f=@(x) any(x==1);
    out = find(cellfun(f,in));
end

function out = funcellismember(in)
    out = find(cellfun(@(x) ismember(1, x), in));
end

Answer 2

[Voss]

C = {[2 5 8 9], [1 2 4 5 7 10], [3 5 6 7 8 9 10], [1 2 3 5 8 9 10], [1 2 4 6 8 9], [1 3 4 5 6 8], [1 2 4], [4 5 7 9], [1 2 5 7 10], [1 2 3 4 6 8]}

C = 1×10 cell array
    {[2 5 8 9]}    {[1 2 4 5 7 10]}    {[3 5 6 7 8 9 10]}    {[1 2 3 5 8 9 10]}    {[1 2 4 6 8 9]}    {[1 3 4 5 6 8]}    {[1 2 4]}    {[4 5 7 9]}    {[1 2 5 7 10]}    {[1 2 3 4 6 8]}

val = 1; % value to check for

One way is to write a for loop to check each element of C for whether its contents contain a 1 (or whatever value):

nC = numel(C);
has_val = false(1,nC);
for ii = 1:nC
    has_val(ii) = any(C{ii} == val);
end
has_val

has_val = 1×10 logical array
   0   1   0   1   1   1   1   0   1   1

result = find(has_val)

result = 1×7

2 4 5 6 7 9 10

Another way is to let cellfun do the loop for you:

has_val = cellfun(@(v)any(v == val),C)

has_val = 1×10 logical array
   0   1   0   1   1   1   1   0   1   1

result = find(has_val)

result = 1×7

2 4 5 6 7 9 10

Answer 3

[Fangjun Jiang]

a=[{[2 5 8 9]} {[1 2 4 5 7 10]} {[3 5 6 7 8 9 10]} {[1 2 3 5 8 9 10]} {[1 2 4 6 8 9]},...
    {[1 3 4 5 6 8]} {[1 2 4]} {[4 5 7 9]} {[1 2 5 7 10]} {[1 2 3 4 6 8]}]

a = 1×10 cell array
    {[2 5 8 9]}    {[1 2 4 5 7 10]}    {[3 5 6 7 8 9 10]}    {[1 2 3 5 8 9 10]}    {[1 2 4 6 8 9]}    {[1 3 4 5 6 8]}    {[1 2 4]}    {[4 5 7 9]}    {[1 2 5 7 10]}    {[1 2 3 4 6 8]}

f=@(x) any(x==1);
find(cellfun(f,a))

ans = 1×7

2 4 5 6 7 9 10