# Problems with changing the order of a binary representation

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Jennifer Arellana 2021 年 4 月 27 日
コメント済み: Clayton Gotberg 2021 年 4 月 27 日
Hello everyone,
I have a problem to organize a binary representation that I calculate in Matlab, like this:
NJ=3
for i = 1:2^NJ-1
struct(i,:) = dec2bin(i,NJ);
end
This gives me this result:
ans:
struct =
7×3 char array
'001'
'010'
'011'
'100'
'101'
'110'
'111'
The problem is that I need this binary representation to have the following order:
100
010
110
001
101
011
111
I really need this to be done by the code because if you increase the number in NJ, it is no longer practical to do the manual change, as the amount of struct data increases exponentially.
I appreciate any help, Thank you.
##### 3 件のコメント表示非表示 2 件の古いコメント
Clayton Gotberg 2021 年 4 月 27 日
what is B_ampl? If it's your replacement for struct, the problem is probably with the formatting of the input. If you input a simple char array, the function assumes you're giving it one number. If you input a string array or a cell array containing char arrays, the function can see the numbers separately.

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### 採用された回答

Clayton Gotberg 2021 年 4 月 27 日

Use the fliplr function to swap the direction of arrays!
NJ=3
for i = 1:2^NJ-1
binary_save(i,:) = fliplr(dec2bin(i,NJ));
end
I also changed the name of struct to binary_save because struct is the name of a built-in MATLAB function, and it's best to avoid accidentally overwriting MATLAB functions. For example,
input = ones(5,4); % A 5x4 matrix
size = 3; % Setting the size for something later
for k = 1:size(input,1)
% Some operation
end
Will run from k = 1:3 instead of k = 1:5 (the number of rows in the input) and
input = magic(5);
size = 3; % Setting the size for something later
for k = 1:size(input,1)
% Some operation
end
Gives an error about how the index in position 1 doesn't make sense but doesn't even tell you in which line it's occured.
##### 3 件のコメント表示非表示 2 件の古いコメント
Paul Hoffrichter 2021 年 4 月 27 日
I will start hitting "follow" to get timely alerts. Not perfect since emails take time to arrive depending upon the email service.

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### その他の回答 (1 件)

Paul Hoffrichter 2021 年 4 月 27 日
NJ=4
for i = 1:2^NJ-1
struct(i,:) = fliplr(dec2bin(i,NJ));
end
struct =
15×4 char array
'1000'
'0100'
'1100'
'0010'
'1010'
'0110'
'1110'
'0001'
'1001'
'0101'
'1101'
'0011'
'1011'
'0111'
'1111'

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