将零填充到二进制数中的特定位置的命令？

Question

我需要将零填充到二进制数中的特定位置。循环二进制数的数组形式（例如dec2bin(43)）并添加零并调整大小听起来轮子的重新发明。

如何在 Matlab 中有效地将零填充到二进制数？

循环

positions=[1,3,6];
x=de2bi(43);
xx=flip(x);    

kk=1;
for ii=1:length(x)+length(positions)
    if isequal(positions(kk),xx(ii))
         %Transfer the tail from ii by one index ahead,
         %left out here because I think my method is bad!
         x(ii)=0;
         positions=positions+1;  %Increase positions by one now due to one padding
         kk=kk+1;
    end
end

这感觉太重创了：基本上如何在 Y 之前得到 0，比如从 xxxxYabcd 到 xxxx0Yabcd，其中 xxxxYabcd 是二进制数？

示例

输入

101011, (1,3,6)

输出

100100110

1x010x11x（在插入零的位置用 x 显示更清晰的零位置）

Answer 1

虽然没有太大的概念改进，但以下将自动重新索引并在预填充矩阵上分配旧值：

>> xx
xx =
     1     0     1     0     1     1

nPads = length(positions);
nPadsShifts = 1:nPads;           
y = ones(1, length(xx) + nPads); % re-indexing on the new array
y(positions + nPadsShifts) = 0;  % padding values
y(y==1) = xx;                    % set original bit values


>> y
y =
     1     0     0     1     0     0     1     1     0

Answer 2

您可以仅使用索引来完成。要在扩展向量中插入的零的位置是 insertzeros+(1:length(insertzeros))（在您的示例中，insertzeros 是 [1 3 6]）：

input = [1 0 1 0 1 1]; %// example data
insertzeros = [1 3 6]; %// example data

output = ones(1,length(input)+length(insertzeros));
output(insertzeros+(1:length(insertzeros))) = 0;
output(output==1) = input;

Answer 3

比赛开始！ Mendo 获胜，gevang 获得第二名，thewaywewalk 获得第三名，jaheruddin 获得第四名。 Shai 的bitshifting 是在 shai 函数中实现的，遗憾的是还没有让它运行得更快。

结果

gevang :2.2e-05
thewaywewalk :5.6975e-05
mendo :2.2102e-05
jaheruddin :0.0001693
shai (poor hhh-implementation) 5.3288e-04

答案的热身测试

function test_padding_zeros()

    function myOutput=gevang(xx,positions)
        nPads = length(positions);
        nPadsShifts = 1:nPads;
        myOutput = ones(1, length(xx) + nPads); % re-indexing on the new array
        myOutput(positions + nPadsShifts) = 0;  % padding values
        myOutput(myOutput==1) = xx;                    % set original bit values
    end

    function myOutput=thewaywewalk(x,positions)
        idx = numel(x):-1:1;
        myOutput = num2cell(x);
        myOutput(2,idx(positions)) = {0};
        myOutput = [myOutput{:}];
    end

    function myOutput=jaheruddin(myInput,positions) % myInput can be a row vector or a matrix!
        n = size(myInput,2)+numel(positions);
        myOutput = false(size(myInput,1),n);
        myOutput(:,setxor((1:length(positions)),1:n))=myInput;
    end

    function myOutput=mendo(myInput,positions)
        myOutput = ones(1,length(myInput)+length(positions));
        myOutput(positions+(1:length(positions))) = 0;
        myOutput(myOutput==1) = myInput;
    end

    function out = bizarreBitShift( bNum, fromBit, shiftAmount )
        % construct a mask
        msk = uint32( (2^( fromBit - 1 ) )-1 ); 
        shiftPart = bitand( uint32(bNum), bitcmp(msk) ); % bitcmp - complement of bits
        staticPart = bitand( uint32(bNum), msk );
        out = bitshift( shiftPart , shiftAmount );
        out = bitor( out, staticPart );
    end

    function myOutput=shai(myInput,positions)
        shiftAmount=1;
        myOutput=sprintf('%d',myInput);
        myOutput=bin2dec(myOutput);
        k=0;
        for ii=1:length(positions)
            fromBit=positions(ii)+k;
            myOutput=bizarreBitShift(myOutput, fromBit, shiftAmount);
            k=k+1;
        end

        myOutput=ismember(dec2bin(myOutput),'1');
    end




positions = [1 3 6]; %// example data
myInput = [1 0 1 0 1 1]; %// example data

ggevang=@() gevang(myInput,positions);
tthewaywewalk=@() thewaywewalk(myInput,positions);
mmendo=@() mendo(myInput,positions);
jjaheruddin=@() jaheruddin(myInput,positions);
sshai=@() shai(myInput,positions);

timeit(ggevang)
timeit(tthewaywewalk)
timeit(mmendo)
timeit(jjaheruddin)
timeit(sshai)

end

Answer 4

如果您正在寻找效率，那么尝试各种选择永远不会有坏处。这是另一种方法：

M = magic(5)>3; % Example matrix
positions = [2 3 6]; % Desired padding columns in resulting matrix

n = size(M,2)+numel(positions);
R = false(size(M,1),n);

R(:,setxor(positions,1:n))=M

Answer 5

Shai 为单个二进制数提供了here 位移解决方案。我在下面使用它来填充零，该解决方案可能会变得更快——目前是最慢的解决方案之一（一些不必要的 dec2bin）或与 Jaheruddin 的解决方案大致相同的速度。

function out = bizarreBitShift( bNum, fromBit, shiftAmount )
    % construct a mask
    msk = uint32( (2^( fromBit - 1 ) )-1 ); 
    shiftPart = bitand( uint32(bNum), bitcmp(msk) ); % bitcmp - complement of bits
    staticPart = bitand( uint32(bNum), msk );
    out = bitshift( shiftPart , shiftAmount );
    out = bitor( out, staticPart );
end

function myOutput=shai(myInput,positions)
    shiftAmount=1;
    myOutput=sprintf('%d',myInput);
    myOutput=bin2dec(myOutput);
    k=0;
    for ii=1:length(positions)
        fromBit=positions(ii)+k;
        myOutput=bizarreBitShift(myOutput, fromBit, shiftAmount);
        k=k+1;
    end

    myOutput=ismember(dec2bin(myOutput),'1');
end

positions = [1 3 6]; %// example data
myInput = [1 0 1 0 1 1]; %// example data

sshai=@() shai(myInput,positions);
timeit(sshai)

时机

ans =

   5.3288e-04