考虑一个整数数组:
X = [1, 3, 3, 3,2,3 ,1,1,2,3]
我想将每个数字转换为4位二进制等价物。我做了以下,但结果不正确。似乎MSB和LSB是相反的。我尝试使用MATLAB&decimal_to_bin =dec2bin(X)
的函数,但即使这个输出也是错误的。
我该如何解决这个问题?
X = [1, 3, 3, 3,2,3 ,1,1,2,3];
b = 4;
lookup_table = generate_lookupTable(b);
BinaryX = lookup_table(X,:);
function result = generate_lookupTable(b)
% generate binary code
k = 2^b;
result = zeros(k,b);
for i=1:k
for j=1:b
result(i,j) = bitand(uint8(2^(j-1)),uint8(i-1))/uint8(2^(j-1));
end
end
end
我得到的输出是BinaryX
0 0 0 0
0 1 0 0
0 1 0 0
0 1 0 0
1 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0
1 0 0 0
0 1 0 0
答案 0 :(得分:4)
我认为烧杯意味着以下几点:
% Your vector
X = [1, 3, 3, 3,2,3 ,1,1,2,3];
% number of bits to use
numbits = 4;
% Create lookup table
little_endian = true;
LUT = dec2bin(0:15, numbits) == '1';
if ~little_endian
LUT = fliplr(LUT); end
% The conversion
bits = LUT(X(:) + 1, :)
或者,您可以执行dec2bin
所做的事情,但不必大惊小怪:
bits = (rem(floor(X(:)*pow2(1-numbits:0)),2) == 1);