Question

我有以下问题，

使用 Matlab 我有一个非常大的向量包含让我们说的1和0，我想将它转换为十进制，事情是数字是大的所以变量不能握住它，所以我想把它制成可接受边界内的小块并将这些块转换成十进制数，然后存储在矩阵或向量中。我想知道你是否对如何实现这个或更好的方法有任何建议。

提前致谢。

Answer 1

<强>代码

%%// Given binary number
comp = '1010101011010110010010101001';
decnum_tobe_stored = bin2dec(comp) %%// To be used for verification

%%// Parameters
ndigits = 4; %%// Number of binary digits used per element for storage.

%%// Storing Process
app1 = ndigits - rem(numel(comp),ndigits);
comp = [repmat('0',[1 app1]) comp];

bin_array = reshape(comp,ndigits,[])' %%//' This is your binary storage medium
dec_array = bin2dec(bin_array) %%// This is your decimal storage medium

%%// Retrieving Process
pf = fliplr(ndigits.*(0:size(bin_array,1)-1));
v2 = power(2,pf)';%//'
ret_number = sum(dec_array.*v2); %%// Retrieved number

%%// Verfication
check_error = isequal(ret_number,decnum_tobe_stored) %%// org_number must be same as verify_decnum for this technique

<强>输出

decnum_tobe_stored =
   179135657

bin_array =
0000
1010
1010
1101
0110
0100
1010
1001

dec_array =
     0
    10
    10
    13
     6
     4
    10
     9

check_error =
     1

更大案例

我们假设您有一个二进制数 -

comp = '10101010110101100100101010011010101101010101010101101011001111'

您会知道bin2dec允许最大二进制数位为52的能力限制您，因此您可以使用ndigits = 52。运行代码后，您的十进制存储阵列将具有此分区数据 -

dec_array =
          683
   1.5686e+15

如果您启用了format longe，则可以轻松地显示整个数据。因此，

dec_array =

     6.830000000000000e+02
     1.568619453831887e+15

贪婪的存储空间：如果您真的想将大量数据存入非常小的存储阵列，这是适用的。

<强>代码

%%// Given binary number
comp = ['10101010110101100100101010011010101101010101010101101011001001' ...
    '01010011010101011010101010101101011001001010100110101010110101' ...
    '01010110110101011010110010010101001101010101101010101010110101' ...
    '10010010101001101010101101010100110010010101001101010101101010' ...
    '10101011010110010010101001101010101101010101100100101010011010' ...
    '10101101010101010110101100100101010011010101011010101010101010' ...
    '11010101010101101011001001010100110101010110101010101010110101' ...
    '10010010101001101010101101010101010110101100100101010011010101' ...
    '01101010101010101101011001001010100110101010110101010101011010' ...
    '11001001010100110101010110101010101011010110010010101001101010' ...
    '10110101010101011010110010010101001101010101101011001001010100' ...
    '11010101011010110010010101001101010101101011001001010100110101' ...
    '01011010110010010101001101010101101011001001010100110101010110' ...
    '10110010010101001101010101101011001001010100110101010110101100' ...
    '10010101001101010101101011001001010100110101010110101100100101' ...
    '01001101010101101011001001010100110101010110101100100101010011' ...
    '11010101010101101011001001010100110101010110101010101010110101' ...
    '10010010101001101010101101010101010110101100100101010011010101' ...
    '01101010101010101101011001001010100110101010110101010101011010' ...
    '11001001010100110101010110101010101011010110010010101001101010' ...
    '10110101010101011010110010010101001101010101101011001001010100' ...
    '11010101011010110010010101001101010101101011001001010100110101' ...
    '01011010110010010101001101010101101011001001010100110101010110' ...
    '10110010010101001101010101101011001001010100110101010110101100' ...
    '0101010110101100100101010011010101011010110010010101001'];

%%// Parameters
ndigits = 52; %%// Number of digits allowed

%%// Storing Process
app1 = ndigits - rem(numel(comp),ndigits);
comp = [repmat('0',[1 app1]) comp];

bin_array = reshape(comp,ndigits,[])'; %%//' This is your binary storage medium
dec_array = bin2dec(bin_array) %%// This is your decimal storage medium

%%// ******** Greedy storage approach ******************
%%// Maximum number of elements that can be packed together
num_ele_cumsum = find(isinf(cumsum(power(1E16,1:24))),1)-1;

dec_array = [zeros(num_ele_cumsum - rem(numel(dec_array),num_ele_cumsum),1) ;dec_array];
dec_array_mat = reshape(dec_array,num_ele_cumsum,[]);
exp_nums = repmat(power(1E16,num_ele_cumsum-1:-1:0)',[1 size(dec_array_mat,2)]);%%//'
cumsum_vals = cumsum(dec_array_mat.*exp_nums,1);
dec_array_compact = cumsum_vals(end,:)'

<强>输出

dec_array =
   2.2929e+10
   3.0024e+15
   1.5971e+15
   .......
   2.7211e+15
   3.0504e+15 (38 rows)

dec_array_compact =
  2.2929e+170
  1.5686e+303

另请注意，检索过程（此处未列出）与编码过程一样复杂。所以，在我认为极其有限的内存限制的情况下，这是有意义的。

巨大的二进制数转换

1 个答案: