One-to-one correspondence anonymous function
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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
1 个评论
Walter Roberson
2013-2-11
are your x values always integer? if not there can be a problem.
采纳的回答
Azzi Abdelmalek
2013-2-10
编辑：Azzi Abdelmalek
2013-2-10
P=[10;100;105;700;700];
N=[5;8;4;11;20];
f=@(x) N(find(P==x));
f(700)
2 个评论
Walter Roberson
2013-2-11
The find() is not needed, P==x can be used directly.
更多回答（1 个）
Walter Roberson
2013-2-10
@(x) N(ismember(x, P))
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