check ismember for each element individually

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Xingda Chen
Xingda Chen 2021-7-22
编辑： Jan 2021-7-23
Hello all,
I wonder if there is any smart way checking the indices of every individual elements in an array, in another array.
For example,given an input
a=[ 1 2 3 3 5 6 6];
b=[1 2 3 5 6];
I want to check where is each element of b is located in a
for example: for b(1), the location in a) would be:
[1 0 0 0 0 0 0]
%or
a(1)
for b(3), it then would be:
[0 0 1 1 0 0 0]
%or
a(3,4)
of course I can do it using For loop, for speed reason I need to avoid it.

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采纳的回答

Jan
Jan 2021-7-23
编辑：Jan 2021-7-23
Ran in:
You want to get a logical vector for each element, which is TRUE for the matching elements. Then this is efficient:
a = [1 2 3 3 5 6 6];
b = [1 2 3 5 6];
match = (b.' == a)
match = 5×7 logical array
1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1
This works with auto-expanding since R2016b. For older Matlab versions:
match = bsxfun(@eq, b.', a)
If you want the indices instead, a loop is nice and efficient (I assume, you have this already, but maybe another user might be interested):
indices = cell(size(b)); % Pre-allocate
for k = 1:numel(b)
indices{k} = find(b(k) == a);
% This would waste time to create a temporary cell:
% indices(k) = {find(b(k) == a)};
end

更多回答（2 个）

Chunru
Chunru 2021-7-23
Ran in:
Doc ismember to see if it meets your requirement:
a=[ 1 2 3 3 5 6 6];
b=[1 2 3 5 6];
[Lib, Loca] = ismember(b, a)
Lib = 1×5 logical array
1 1 1 1 1
Loca = 1×5
1 2 3 5 6
a(Loca) % Loca is the the location of the first appearance of b in a
ans = 1×5
1 2 3 5 6
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Chunru
Chunru 2021-7-23
编辑：Chunru 2021-7-23
Ran in:
%% arrayfun
a=[ 1 2 3 3 5 6 6];
b=[1 2 3 5 6];
a = randi(100, 1e6, 1);
b = randi(100, 1e3, 1);
tic
cac = arrayfun( @(ii) find(a==ii), b, 'uni',false );
toc
Elapsed time is 2.493500 seconds.
% Elapsed time is 1.798082 seconds.
%% for loop
tic
cac1 = cell(length(b), 1);
for i=1:length(b)
cac1{i} = find(a==b(i));
end
toc
Elapsed time is 2.050463 seconds.
% Elapsed time is 1.571855 seconds.
%% ismember
% This is fastest.
% it use sort (faster) and stop when 1st appearance is found
tic
[Lib, Loca] = ismember(b, a);
toc
Elapsed time is 0.108364 seconds.
%% array expansion, Jan's answer below
tic
match = (b.' == a)';
toc
Elapsed time is 1.722124 seconds.
Jan
Jan 2021-7-23
编辑：Jan 2021-7-23
Thanks for the useful speed comparison. arrayfun is 10% slower than a loop. That ismember is much faster with its binary search is interesting, but it replies a different result.
In your code for the array expansion method, 60% are spent in the final transposition, which can be omitted:
a = randi(100, 1e6, 1);
b = randi(100, 1e3, 1);
tic
match = (b.' == a).';
toc
% Elapsed time is 1.836326 seconds. (i7, Matlab R2018b)
tic
match = (b.' == a);
toc
% Elapsed time is 0.642080 seconds.
tic
match = (b == a.');
toc
% Elapsed time is 0.553570 seconds.
And if you store the rows of the output in a cell:
tic
cac1 = cell(length(b), 1);
for i=1:length(b)
cac1{i} = (a==b(i)); % Without FIND
end
toc
% Elapsed time is 0.528209 seconds.
By the way: Matlab R2009a:
% Elapsed time is 3.368870 seconds.
% Elapsed time is 1.358478 seconds.
% Elapsed time is 1.219993 seconds.

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per isakson
per isakson 2021-7-23
Ran in:
The function arrayfun() does the trick (i.e. hides the foor-loop)
a=[ 1 2 3 3 5 6 6];
b=[1 2 3 5 6];
cac = arrayfun( @(ii) find(a==ii), b, 'uni',false )
cac = 1×5 cell array
{[1]} {[2]} {[3 4]} {[5]} {[6 7]}
cac{3}
ans = 1×2
3 4
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Xingda Chen
Xingda Chen 2021-7-23
i guess i will for loop then.
My process is quite large. at big picture i am dealing with ~500M iterations averaging 100 times, most of the temp data are not stored so memeory isn't an issue but i am trying the avoid for loop as much as i can
Jan
Jan 2021-7-23
@Chunru: "Loop in newer matlab is no longer the enemy of speed." I can confirm this. In addition for loops tend to be easier to write, read and maintain. By the way, "newer" means the Matlab version, which has introduced the JIT acceleration for loops, and this was Matlab R6.5, so the rumor of slow loops is outdated for over 20 years now.

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