Shifting number to end of an array

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Katelin Cherry
Katelin Cherry 2016-8-31
评论： Sim 2020-9-28
Hi, I'm trying to write a function that takes an array (v) and a value(a) as its input then outputs an array(w). The output array takes any instance of the value out of its current position and moves it to the end of the array. This is a hw problem for school. My function works perfectly when the value a is nonzero but it doesn't work for zero values. I have included my function below. I'm afraid it's something with the for loop and iterating through it. I'm pretty new to Matlab and can't figure this out. Thanks!
function v=move_me(v,a)
if nargin<2
a=0;
end
ii=0;
for ii=1:length(v)
if ii==a
v(ii)=[]
v(end+1)=a;
w=v;
end
end
if true
% code
end

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采纳的回答

Star Strider
Star Strider 2016-8-31
See if changing your if test to:
if v(ii)==a
helps.
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Guillaume
Guillaume 2016-8-31
编辑：Guillaume 2016-8-31
You cannot delete elements of a vector while iterating over the vector index. Your index will get out-of-sync.
The above code will fail for example on:
v = [1 5 5 1 5 5 1]
a = 5
You will either have to restart the loop from scratch every time you delete an element, or, better, store the indices of elements to move in the loop, and move them all at once after the loop.
And of course, the whole thing can be achieved without a loop:
v = [v(v ~= a), v(v == a)];
Sim
Sim 2020-9-28
@Guillaume,
What if i want to move 2 or more elements to the end of the array?
v = [1 5 3 1 5 5 1]
a = [3 5];
I did it with a "for loop", but maybe there is a way without a loop...
for i = 1 : length(a)
v = [v(v ~= a(i)), v(v == a(i))]
end

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更多回答（2 个）

ledinh lam
ledinh lam 2016-11-30
as they said above . you can use this code.
function v = move_me(v,a)
if nargin <2
a = 0;
end
v = [v(v ~= a), v(v == a)];
end
Maybe, it will help you !
John BG
John BG 2016-8-31
When you say 'to the end' of the array, what happens to the values shifted beyond the size of the array, do they show up at the beginning of the array?
or do you pad zeros at the beginning, losing data?
In any case you may want to use circshift?
A=[1:10]
circshift(A,3,2)
=
Columns 1 through 5
8.00 9.00 10.00 1.00 2.00
Columns 6 through 10
3.00 4.00 5.00 6.00 7.00
A=[1:10]'
circshift(A,3)
=
8.00
9.00
10.00
1.00
2.00
3.00
4.00
5.00
6.00
7.00
or
circshift(A,[-1 -1])
=
1.00 0 0 1.00
0 0 0 0
0 0 0 0
1.00 0 0 1.00
circshift(A,[1 1])
=
0 0 0 0
0 1.00 1.00 0
0 1.00 1.00 0
0 0 0 0
may be you mean shifting linearly, in such case you may want to combine circshift with reshape
A=magic(4)
=
16.00 2.00 3.00 13.00
5.00 11.00 10.00 8.00
9.00 7.00 6.00 12.00
4.00 14.00 15.00 1.00
[sz1 sz2]=size(A)
B=reshape(A,[sz1*sz2,1]);
B2=circshift(B,3,1); % the shift factor is 3
C=reshape(B2,size(A))
=
8.00 5.00 11.00 10.00
12.00 9.00 7.00 6.00
1.00 4.00 14.00 15.00
16.00 2.00 3.00 13.00
Regarding the what dimension to shift along, or whether its circular shift, please clarify so a satisfactory answer can be supplied.
If you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG
<mailto:jgb2012@sky.com jgb2012@sky.com>
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John BG
John BG 2016-9-11
and what do you think that circshift does?
Guillaume
Guillaume 2016-9-12
编辑：Guillaume 2016-9-12
I'll repeat (not that it matter anymore as this thread is out of date):
"They are being asked to develop an algorithm using a loop that does:"
v = [v(v~=a), v(v==a)];
What do you think circshift does? Certainly not the above. Given
v = [1 2 1 2 3 1 2 3 4]
a = 2
you should end up with
v = [1 1 3 1 3 4 2 2 2]
Your whole business about circshifting matrices misses the point, the problem is only concerned with vectors. Your question about what happens to values shifted beyond the size of the array is also irrelevant.

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