Forming loop to simplify codes

2019 12 10
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回答已采纳
更新时间：2019 12 11
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0 个投票

Can you help me use loop to simplify the following code and produce the same graph?
c=5
dista=log([12:-1:8 6 4:-1:1]+retint)
distb=log([14:-1:9 6 3:-1:1]+retint)
distc=log([16:-1:10 6 2:-1:1]+retint)
etaa = exp(-c*abs(dista-dista(:)));
discrima = 1./sum(etaa,1)
etab = exp(-c*abs(distb-distb(:)));
discrimb = 1./sum(etab,1)
etac = exp(-c*abs(distc-distc(:)));
discrimc = 1./sum(etac,1)
figure (11)
hold on
plot(discrima,'-mo','Displayname','gap = 12(8,6,4)1')
plot(discrimb,'-ro','Displayname','gap = 14(9,6,3)1')
plot(discrimc,'-bo','Displayname','gap = 16(10,6,2)1')
axis([0 length(dista) 0 1])
xlabel('Serial Position','fontsize',14)
ylabel('Discriminability','fontsize',14)
title({'SIMPLE Memory Model'...
'\fontsize{9}\color{blue}Length = 10; Various Gaps'})
lgd = legend ('Location','northwest');
lgd.NumColumns = 1;
lgd.FontSize = 9;
set(gcf, 'color', 'white');
hold off

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 采纳的回答

per isakson
per isakson 2019-12-11

0 个投票

"produce the same graph" The code below produces the same graph.
However
  • it may be disputed whether it's simpler
  • less intermediate results are saved
%%
c = 5;
retint = 0;
%%
cac = { '-go','Displayname','gap = 12(8,6,4)1'
'-ro','Displayname','gap = 14(9,6,3)1'
'-bo','Displayname','gap = 16(10,6,2)1' };
num = [ [12:-1:8,6,4:-1:1]+retint
[14:-1:9 6 3:-1:1]+retint
[16:-1:10 6 2:-1:1]+retint ];
%%
figure(11);
hold on
%%
for jj = 1 : 3
dist = log(num(jj,:));
eta = exp(-c*abs(dist-reshape(dist,[],1)));
discrim = 1./sum(eta,1);
plot( discrim, cac{jj,:} )
end
%%
axis([0,length(dist(1,:)),0,1.02*max(max(discrim))])
xlabel('Serial Position','fontsize',14)
ylabel('Discriminability','fontsize',14)
title({'SIMPLE Memory Model'...
'\fontsize{9}\color{blue}Length = 10; Various Gaps'})
lgd = legend ('Location','northwest');
lgd.NumColumns = 1;
lgd.FontSize = 9;
set(gcf, 'color', 'white');
hold off

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