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One-to-one correspondence anonymous function

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Arash
Arash am 10 Feb. 2013
I was wondering if can make an injective (one-to-one correspondence) anonymous function to do this;
if the input was one member from P then function convert it to correspondence member in N, for example if input is x=10 out put will be y=5, x=105-->y=4 and ...
P=[10;100;105;700;943]
N=[5;8;4;11;20]
Thanks in advance, Arash

Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 10 Feb. 2013
Bearbeitet: Azzi Abdelmalek am 10 Feb. 2013
P=[10;100;105;700;700];
N=[5;8;4;11;20];
f=@(x) N(find(P==x));
f(700)
  2 Kommentare
Walter Roberson
Walter Roberson am 11 Feb. 2013
The find() is not needed, P==x can be used directly.
Arash
Arash am 11 Feb. 2013
Bearbeitet: Arash am 11 Feb. 2013
Thank you Azzi, you helped a lot.

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Walter Roberson
Walter Roberson am 10 Feb. 2013
@(x) N(ismember(x, P))

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