Truncation of a vactor

1 回表示 (過去 30 日間)
Mohammad Sayeed
Mohammad Sayeed 2014 年 4 月 8 日
コメント済み: Mohammad Sayeed 2014 年 4 月 8 日
Hello
I have a vector like this x=[4 7 17 4 6 9 12]
I need to get rid of any values which are higher than 10. I tried this:
for i=1:length(x);
if x(i)>10 x(i)=[];
end
end
This works but with this error message:
'Attempted to access x(7); index out of bounds because numel(x)=5.'
How can I do it without any error message because I need to compute further with x.
Can anyone help?
Kind regards
Sayeed

採用された回答

Walter Roberson
Walter Roberson 2014 年 4 月 8 日
After you delete x(3) then what was at x(4) "falls down" to occupy x(3), what was at x(5) "falls down" to x(4), and so on, leaving x one element shorter. You then go to the next "i" element, 4, examining what is now in x(4) which was in x(5)... and you have skipped examining what was in x(4) and is now in x(3).
Eventually you get to the point where you have deleted the last entry in x, making x two elements shorter, but your "for" loop over length(x) is going to be based on the original length of x, so you are going to end up attempting to access after the last element of what x has become.
Lesson: if you delete elements of a vector using a forward-going loop, you need to take into account that the remainder of the vector moves down to fill up the hole. You can code that in, or you can recode to avoid using a loop to delete elements, or you can try your hand at using a loop that does not go forward.
  1 件のコメント
Mohammad Sayeed
Mohammad Sayeed 2014 年 4 月 8 日
Thanks for your valuable insights. That helps.

サインインしてコメントする。

その他の回答 (1 件)

ES
ES 2014 年 4 月 8 日
Or in other words, you might do
x(x>10)
  1 件のコメント
Mohammad Sayeed
Mohammad Sayeed 2014 年 4 月 8 日
Yes that works, x(x>10)=[]; No need to make any loop
Thank you very much.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeGain Scheduling についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by