finding elements in a vector from another vector

I'd like to create a vector y with the positions of the elements of x in m.
i.e.
x=0 (first element of x) has and index of 1 in m. So y(1)=1
x=241 (second element of x) has index of 242 in x. So y(2)=242
m and x are in attached

 Accepted Answer

Try this:
M = load('m.mat');
X = load('x.mat');
m = M.m;
x = X.x;
[~,y] = ismembertol(x, m, 1E-4)
producing:
y =
1
242
937
2001
3306
4695
6001
7065
7759
8001
I checked that separately for a few values using find, and it appears to produce the correct result.
.

8 Comments

can't make it work since I think m and x are in my script, I don't need to lload them.
I needed to load them. If they are already in your workspace as ‘x’ and ’m’, just run:
[~,y] = ismembertol(x, m, 1E-4)
You should get the same result that I did.
.
I did but I get this.
maybe it's because I have r2014b?
Undefined function 'ismembertol' for input arguments of type 'double'.
The ismembertol function was introduced in the next release, R2015a.
The only work-around I can offer is this loop:
for k = 1:numel(x)
y(k) = find(m <= x(k), 1, 'last');
end
It produces almost the same result. They are identical, except for ‘y(3)’ that using ismembertol is 937 and using find is 936.
EDIT — (18 Jul 2020 at 20:52)
Unfortunately, ismember only works for 4 of the members of ‘x’. However that can be remedied by multiplying them by :
[~,y] = ismember(fix(x*1E+4), fix(m*1E+4))
producing essentially the same result as ismembertol, and the same results as the find loop.
this worrks better:
[~,y] = ismember(fix(x*1E+4), fix(m*1E+4))
thanks.
Great!
I did not initially pick up on your posting the Release information as R2014b (so my apologies for that oversight if you posted it originally, since I only checked later).
.
I think it's best we don't edit comment so people can read different options and figure it out based on their version

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