我已经如下所示:
感谢您的帮助!
到目前为止,我有:
data = 14
out = dec2bin(data, 4)
这将会得到:
out = 1110
但我希望以这种格式获得二进制数:
out = [1 1 1 0]
感谢您的帮助!
de2bi
和 bi2de
函数。de2bi = @(x) dec2bin(x)>48;
bi2de = @(x) x*2.^(size(x,2)-1:-1:0)';
测试:
dec = 1:10
bin = de2bi(dec)
dec = bi2de(bin)
输出:
dec =
1 2 3 4 5 6 7 8 9 10
bin =
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
dec =
1
2
3
4
5
6
7
8
9
10
附言:如果出于某种原因,您不想使用dec2bin
,您可以定义de2bi
函数如下:
在Matlab/Octave中:
de2bi = @(x) 2.^[(floor(log2(max(x(:)))):-1:1),0];
de2bi = @(x) rem(x(:),2*de2bi(x))>(de2bi(x)-1);
de2bi = @(dec,bit=[(1+floor(log2(max(dec(:))))):-1:2, 1]) rem(dec(:),2.^bit)>(2.^(bit-1)-1);
%Call examples:
x=0:10;
n=3:-1:1;
de2bi(x)
de2bi(x,n) % returns only the three least significant bits
P.S. 这里提供了更常见的答案:{{link1:dec2base with independent bits/digits calculation
}}
data = 14
out = bitget (data, 4:-1:1)
out =
1 1 1 0