Sort one set of data to correspond to another.

1 Ansicht (letzte 30 Tage)
Ronan
Ronan am 16 Nov. 2015
Bearbeitet: Stephen23 am 17 Nov. 2015
Say I have an ordered set of data.
a = [100,200,300,400,500];
And say I have another set of data,
b = [300,200,500,400,100]
I m trying to find the index where a is sorted to b. I could use a nested for loop of course but is there a better way to get the index shown below?
c = [5,2,1,4,3]

Akzeptierte Antwort

Thorsten
Thorsten am 16 Nov. 2015
Bearbeitet: Thorsten am 16 Nov. 2015
If b is an unsorted version of a, i.e., all elements in b occur once and only once in a, you can use
[~, idx] = sort(b);
In the more general case where b can have some elements of a, and elements can occur more than once, use
for i=1:numel(b), idx(i) = find(ismember(a, b(i))); end
Instead of the for loop, you can also use
idx = arrayfun(@(x) find(ismember(a,x)), b);
  4 Kommentare
Ronan
Ronan am 16 Nov. 2015
Thank you very much.
Guillaume
Guillaume am 16 Nov. 2015
Bearbeitet: Guillaume am 16 Nov. 2015
z = arrayfun(@(i) find(ismember(k,i)), x)
Is actually not going to work in the general case, because find may return none or several indices, which would then require a 'UniformOutput', false to arrayfun.
If you assume that find is always going to return one and only one value. Then the second output of ismember is a much more efficient way (no loop) to obtain the same result.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Stephen23
Stephen23 am 16 Nov. 2015
Bearbeitet: Stephen23 am 17 Nov. 2015
>> a = [100,200,300,400,500];
>> b = [300,200,500,400,100];
>> [~,Xa] = sort(a);
>> [~,Xb] = sort(b);
>> Xa(Xb)
ans =
5 2 1 4 3

Guillaume
Guillaume am 16 Nov. 2015
Following your comment to Thorsten's answer, use the 2nd return value of ismember:
x = [350,420,245,100]
k = [420,100,350,245]
z = ismember(x, k)

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by