我从0到3做了两个随机数。
a=0;
b=3;
A=round(a+(b-a)*rand(1,1000));
B=round(a+(b-a)*rand(1,1000));
然后我添加它们的每两位。然后我把它转换成二进制。
SUM = A + B;
binarySum = dec2bin(SUM);
因为我想计算过渡,我写这段代码:
s = 1;
for i = 1:1000
for j = 1:3
M(1,s) = binarySum(i,j);
s = s+1;
end
end
Tr = sum(diff(M)~=0);
现在我想分割M的每3个元素并用另一个元素对它们进行编码。例如000 By 000000,110 By 000001,001 By 00001,100 by 0001,101 By 001,010 By 01,011 By 1.
我使用过这种方法,但它不起作用。这有什么问题?
Lookup_In = [ 000 110 001 100 101 010 011 ] ;
Lookup_Out = {'000000','000001','00001','0001','101','01','1' } ;
StrOut = repmat({'Unknown'},size(M)) ;
[tf, idx] =ismember(M, Lookup_In) ;
StrOut(tf) = Lookup_Out(idx(tf))
答案 0 :(得分:2)
M
是一个可以使用Lookup_Out
以这种方式映射的字符串:
M2 = reshape(M, [3,1000] )';
Lookup_In = [ 000 110 001 100 101 010 011 ] ;
Lookup_Out = {'000000','000001','00001','0001','101','01','1' } ;
StrOut = repmat({''},[1,size(M,1)]);
for r=1:size(M2,1)
StrOut{r} = Lookup_Out{str2double(M2(r,:)) == Lookup_In};
end