If the interpreter is not smart enough and the vector is large, maybe
this is faster -- only one computation/element:
index = a<5;
b = zeros(a);
b(index) = a(index)^2;
b(~index) = a(~index)^0.5;
/Stefan
On 2013-04-21 21:49, Michael Dunn wrote:
--------------------
Say I want to do an array (vector) assignment, but the formula
requires a conditional. Is there a better way to do it than:
b = ((a^2).*(a<5)) + ((a^0.5).*(a>=5))
which returns a^2 for a<5, and sqrt(a) for a>=5.
And – just curious – if this is the only way, is the interpreter smart
enough to not waste its time computing the false expression?
Thanks,
Michael Dunn | Editor-in-Chief
Scope Junction | http://www.scopejunction.com/
(415) 947-6096 (USA)
(519) 744-9395 (Canada)
Skype: MichaelDunn_UBM
LinkedIn: http://www.linkedin.com/in/cantares
<http://www.tech.ubm.com/>
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