Find elements from A in B and get the index of found element in B
60 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Nathan Kennedy
il 19 Dic 2017
Commentato: Jos (10584)
il 20 Dic 2017
Hi
I have two arrays, A and B, A has 12,000 elements and B has 260,000 elements.
I need to find every instance when an element in A equals B and get the index of B when this occurs.
So for example at A(5) I need to search all elements of B for the value in A(5) and return the index location of B where it is found.
var_found = []
for i : length(A)
B( find(A(i)) ) = var_found(i);
end
Please can someone advise
0 Commenti
Risposta accettata
Jos (10584)
il 19 Dic 2017
Modificato: Jos (10584)
il 19 Dic 2017
You want to find all the indices of all elements of A in B? So, some elements of A might occur once, other multiple times, and some not at all. Therefore you need to resort to cell arrays. With arrayfun you can loop over all elements of A:
A = [1 2 3 4]
B = [2 2 1 4 2 1]
C = arrayfun(@(x) find(B==x), A, 'un', 0)
% C{k} now holds the indices into B, where B equals A(k)
6 Commenti
Jos (10584)
il 19 Dic 2017
You write: For every value contained in array A, find the equivalent value in array B but also get the index position of that value in B.
- The equivalent value in B will be the same value of A, by definition.
- The index in B will be the second (or third) output of intersect.
Jos (10584)
il 19 Dic 2017
btw, did you take floating point problems into account? Two floating points may look the same but may in fact differ minutely, so they are treated as not being equal:
a = 0.3 % looks like 0.3000
b = 0.1 + 0.2 % looks like 0.3000
a==b % false ...
fprintf('%.20f\n',[a b]) % ... as they do differ!
I do begin to suspect your problems are arising from this ...
Più risposte (4)
Harish Ramachandran
il 19 Dic 2017
You can use the 'intersect' command in order to extract the common values along with the indices from base arrays A and B.
[output_array,index_a,index_b] = intersect(A,B,'stable');
2 Commenti
Harish Ramachandran
il 19 Dic 2017
Modificato: Harish Ramachandran
il 19 Dic 2017
Intersect returns all data elements common to A and B. There is no need to loop. If output_array has only one element, it implies that only one common element exists between A and B. However, this method neglects indices in data containing repetitive elements and is suitable for a unique set.
>> A = [ 1.01 2.02 3.03 4.04 5.05 6.06];
>> B = [ 1 2 3.03 4 5.05 6 ];
>> [output_array,index_a,index_b] = intersect(A,B,'stable');
>> output_array
output_array =
3.0300 5.0500
>> index_a
index_a =
3
5
>> index_b
index_b =
3
5
Nathan Kennedy
il 19 Dic 2017
Modificato: Nathan Kennedy
il 19 Dic 2017
2 Commenti
Jos (10584)
il 19 Dic 2017
I'm 99.99999999...% sure that floating points is one of the issues here when using intersect.
Also did you plot waveform en power_in. waveform contains negative values, whereas power_in does not!
Stephen23
il 20 Dic 2017
"I am interested in what Jo says, but I don't think that's the issue?"
Actually the floating point error is the exact cause of what you are experiencing. Why do you think it isn't?
Jos (10584)
il 19 Dic 2017
I think you'll find my function NEARESTPOINT pretty helpful in this respect. You can download it from the File Exchange:
Based on your code above:
IDX = nearestpoint(waveform, power_in)
will return an index into B for each value of A, so that the difference between A(k) and B(IDX(k)) is minimal. Nearestpoint is quite fast :)
3 Commenti
Stephen23
il 20 Dic 2017
"Is there another way around this without using nearest point?"
Of course! Did you read Jan Simon's answer? You could also use bsxfun and abs, then compare against a tolerance. You are not limited using nearestpoint, although it does provide a very simple calling syntax.
Jos (10584)
il 20 Dic 2017
Indeed, of course you can! There are many ways to Rome ...
However, nearestpoint is quite optimised in both speed and memory efficiency. Note that bsxfun creates an intermediate matrix of M-by-N for two vectors with M and N elements.
Vedere anche
Categorie
Scopri di più su Matrix Indexing in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!