Is it possible to use an array in an if statement?

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Rebecca Ward
Rebecca Ward 2011-11-8
Hi, I need to compare each member of an array against a fixed value and then modify that memeber according to whether it is smaller or larger than the fixed value. I can loop through the array to do this, but I'm trying to speed the code up and was wondering whether you could simply operate on the array directly? I've tried it and I get the same operation on each member irrespective of whether it is smaller or larger than the fixed value.
The code I've used is as follows:
if Gr<10^5
Nu_G=0.5.*Gr.^0.25;
else
Nu_G=0.13.*Gr.^0.33;
end
Can anyone advise whether this is possible or whether there may be any other way to achieve the same effect i.e. avoid the loop and/or speed up the code
Thanks,
Rebecca

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采纳的回答

Daniel Shub
Daniel Shub 2011-11-8
Nu_G = zeros(size(Gr));
Nu_G(Gr<10^5) = 0.5.*Gr(Gr<10^5).^0.25;
Nu_G(Gr>=10^5) = 0.13.*Gr(Gr>=10^5).^0.33;
This is potentially slower than the loop method. In your if/else you only need to test Gr<10^5 once for every element, in the vectorized version you need to test twice.

更多回答（1 个）

Jakob Sievers
Jakob Sievers 2011-11-8
Assuming your GR vector is LGR long, then perhaps something like:
Nu_G=zeros(LGR,1);
Nu_g(Gr<10^5)=0.5.*Gr(Gr<10^5).^25;
Nu_g(Gr>=10^5)=0.13.*Gr(Gr>=10^5).^33;
There's probably even faster ways but this would definitely be faster than doing loops.
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Daniel Shub
Daniel Shub 2011-11-8
On my machine:
Gr = randn(1e8, 1);
tic
LGR = length(Gr);
Nu_G=zeros(LGR,1);
Nu_g(Gr<10^5)=0.5.*Gr(Gr<10^5).^25;
Nu_g(Gr>=10^5)=0.13.*Gr(Gr>=10^5).^33;
toc
tic
LGR = length(Gr);
Nu_G=zeros(LGR,1);
for iGr = 1:LGR
if Gr(iGr)<10^5
Nu_g(iGr)=0.5.*Gr(iGr).^25;
else
Nu_g(iGr)=0.13.*Gr(iGr).^33;
end
end
toc
Elapsed time is 104.696162 seconds.
Elapsed time is 41.326568 seconds.
For smaller N, the vectorization is faster.
Rebecca Ward
Rebecca Ward 2011-11-8
I found another way of doing this:
comp_Gr=Gr>10^5
Nu_G=(1-comp_Gr).*(0.5.*Gr.^0.25)+comp_Gr.*(0.13.*Gr.^0.33)
Don't know whether this is faster or slower?
Thanks for your help,
Rebecca

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