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I want to shift vector values one by one to the left

17 Ansichten (letzte 30 Tage)
Muhammad
Muhammad am 30 Aug. 2022
Beantwortet: Bruno Luong am 30 Aug. 2022
Hello everyone,
I have a binary vector with five 0 and three 1.
num=[1 1 1 0 0 0 0 0]
and I want to shift each 1 left, shift one value as
num=[1 1 0 1 0 0 0 0]
untill I get a complete shift of the vector values and printing of each vector shift
num=[0 0 0 0 0 1 1 1]
any helpfull code of the above program with nested for loop will be highly appreciated
Thanks
  2 Kommentare
Bruno Luong
Bruno Luong am 30 Aug. 2022
=> direction is on the right to my book.
John D'Errico
John D'Errico am 30 Aug. 2022
PLEASE STOP POSTING MULTIPLE TIMES. You have posted the exact same question now three times. One I have now closed.

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Antworten (4)

Chunru
Chunru am 30 Aug. 2022
num=[1 1 1 0 0 0 0 0]
num = 1×8
1 1 1 0 0 0 0 0
for i=1:3
num = circshift(num, -1)
end
num = 1×8
1 1 0 0 0 0 0 1
num = 1×8
1 0 0 0 0 0 1 1
num = 1×8
0 0 0 0 0 1 1 1

Abderrahim. B
Abderrahim. B am 30 Aug. 2022
What about this:
num = [1 1 1 1 1 0 0 0 0 0] ;
for ii = 1: nnz(num)
num = circshift(num, -1)
end
num = 1×10
1 1 1 1 0 0 0 0 0 1
num = 1×10
1 1 1 0 0 0 0 0 1 1
num = 1×10
1 1 0 0 0 0 0 1 1 1
num = 1×10
1 0 0 0 0 0 1 1 1 1
num = 1×10
0 0 0 0 0 1 1 1 1 1
  4 Kommentare
Muhammad
Muhammad am 30 Aug. 2022
I already had an idea of shifting but it does not fullfill the requirement that i want.
Abderrahim. B
Abderrahim. B am 30 Aug. 2022
if you only need the last vector,then use sort function.
sort(yourVectorHere)

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Bruno Luong
Bruno Luong am 30 Aug. 2022
Bearbeitet: Bruno Luong am 30 Aug. 2022
Is it what you want?
num = [1 1 0 1 0 1 0 1 0 0];
j = find(num);
m = length(j);
l = (-m+1:0)+(length(num))-j;
q = sum(l)+1;
J = zeros(q,m);
i = 1;
J(i,:) = j;
for k=m:-1:1
for n=1:l(k)
i = i+1;
j(k) = j(k)+1;
J(i,:) = j;
end
end
I = repmat((1:q)',1,m);
B = accumarray([I(:) J(:)],1)
B = 20×10
1 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 0 1 1 1 0 1 0 0 1 0 0 1 1 1 0 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 1 1 1

Bruno Luong
Bruno Luong am 30 Aug. 2022
May be this?
num = [1 1 0 1 0 1 0 1 0 0];
j = find(num);
m = length(j);
J=repmat(j,m,1);
for i=m:-1:1
J(m-i+1:end,i) = length(num)+i-m;
end
J = [j; J];
I = repmat((1:m+1)',1,m);
B = accumarray([I(:) J(:)],1)
B = 6×10
1 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1

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