Get every first value above a treshold out of an array

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Mike Mierlo van
Mike Mierlo van 2019 年 12 月 8 日
コメント済み: Mike Mierlo van 2019 年 12 月 9 日
Hi guys, Lets say I have an array with elements with
A = [1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 7 8 1 2 1 2 1 2 3 4 5 6]
Now I want to have only the first values that are above the treshold 4.5 when reading from left to right. So reading from left to right when an element is lower than 4.5 it is a 0, when it is higher than 4.5, only the first element is a 1. When the next element is also higher than 4.5 it should be a zero. Only when the value of an element drops below 4.5 the next element above 4.5 should again be a one.
The result should be:
B = [0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0]
Please help me with the magic between A and B.
  2 件のコメント
Mike Mierlo van
Mike Mierlo van 2019 年 12 月 9 日
Hi. It is not homework. It is a simplified array. The actual array contains measurements of degradation of a component. This fails at a certain treshold. After that a new component is build in. The acutal array is way to big to share here, so I asked the question with a simplified code. I am going to try to fix it with the hints. If not, I'll be back.

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採用された回答

dpb
dpb 2019 年 12 月 8 日
編集済み: dpb 2019 年 12 月 9 日
B=([0 diff((A>trsh)-1)]==1);
For comparison,
>> [A;B]
ans =
Columns 1 through 27
1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 7 8 1 2 1 2 1 2 3 4 5 6
0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0
>>

その他の回答 (1 件)

Image Analyst
Image Analyst 2019 年 12 月 8 日
Looks like homework so we're only giving hints. Hint:
da = [0, diff(A > 4.5)]
and then look up strfind() and zeros() to help you get the answer in the final 3 lines of code.
  1 件のコメント
dpb
dpb 2019 年 12 月 8 日
"...so we're only giving hints."
Oh, I didn't see this in time...sorry! :)

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