How to use unique function to get data only when the number in one row increases

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Sharah
Sharah 2015 年 12 月 10 日
コメント済み: dpb 2015 年 12 月 11 日
Hello,
I am trying to figure out on how to extract data only when the number in one of the row increases; for example:
a = [1;1;1; 2;3;4;5;6;7;8;9;10;10;10;11;11];
b = rand(16,1);
c = [b,a]
1) i want to extract the b data only when the number in 'a' increases, which are 1, 2, 3, 4, 5, - 10 without taking the data when the number in a is the same.
2) The numbers in a will go back to 1 after it reaches 255, so i want the function to still take the data when the number goes back to 1. this means that only when the number in a are the same in consequtive order then only you remove the repetition
i appreciate your help in this, please let me know if my question is not clear.

採用された回答

dpb
dpb 2015 年 12 月 10 日
>> [a [true;diff(a)>0]]
ans =
1 1
1 0
1 0
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
10 0
10 0
11 1
11 0
>>
The "startover" case depends somewhat on how the series is defined; in the simplest case I think it would simply be
>> a=[a;a(1:5)]; % sample dataset w/ reset...
>> [a [true;diff(a)~=0]].'
ans =
Columns 1 through 15
1 1 1 2 3 4 5 6 7 8 9 10 10 10 11
1 0 0 1 1 1 1 1 1 1 1 1 0 0 1
Columns 16 through 21
11 1 1 1 2 3
0 1 0 0 1 1
>>
Or,
>> find([true;diff(a)~=0]).'
ans =
1 4 5 6 7 8 9 10 11 12 15 17 20 21
>>
  4 件のコメント
Sharah
Sharah 2015 年 12 月 10 日
thank you for your answer. i did not thought about using the derivatives though. thanks!
dpb
dpb 2015 年 12 月 11 日
It's surprising how often diff is the answer, indeed! :) The key is the fact you're looking for that difference from successive locations here. Ergo, the uninteresting positions are then zero when compared to each other. First finding (or recognizing) the underlying pattern is the key...

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