self made sorting numbers function
6 ビュー (過去 30 日間)
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I am trying to write my own sort function.. it gave me an error idk how to fix.
for j = 1:length(x)
for i = 1:length(x)-1
if x(i) < x(i+1)
minval = x(i);
val = x(i);
x(i) = [];
end
end
sortval(j) = val;
end
fprintf(sortval);
2 件のコメント
Rik
2017 年 3 月 6 日
If you want to do this, you should at the very least loop backwards through your vector. You are removing elements, so at some point x will no longer have the length that your for-loops expect.
Cameron Stark
2017 年 9 月 16 日
function b=downsort(b)
i=.5;
n = length(b);
while i<1
for x =1:n
if x ==n
i = 2;
break
end
if b(x) < b(x+1)
a = b(x);
b(x) = b(x+1);
b(x+1) = a;
break
end
end
採用された回答
Walter Roberson
2017 年 3 月 6 日
What is the point of assigning the same value to both minval and val ?
What is the point of assigning to minval when you never use minval again?
Note:
You are deleting elements from an array using a forward-indexing "for" loop. Suppose you have 4 elements and you delete the 2nd one in the loop:
i = 2, x = [A B C D], x(i) = []
now x = [A C D] because x(2) was deleted.
Now proceed to i = 3 in your for loop, and look at x(i). That is looking at D -- what happened to looking at C ?
When you delete something from vector, all the later elements "fall down" to fill the hole, getting renumbered along the way. If you increase the loop counter then you skip over the entry that fell down into the place you just looked.
You can deal with this in one of a few ways:
- avoid deleting elements at all (for example, copy elements into a different array instead of deleting them out of the existing array); or
- keep track of what is to be deleted but do not deleted it while you are looping; instead, at the end delete all of the ones you marked for deletion; or
- trick: run the loop backwards, from the end towards the beginning.
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