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Removing repeated in order elements from array?

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Nicholas Kavouris
Nicholas Kavouris 2022 年 3 月 30 日
コメント済み: Voss 2022 年 3 月 30 日
I have an array like [ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6]
Is there a method which will produce array [ 1 2 3 1 2 4 5 6] and remove all repeating orderly values but not remove all duplicates?
have tried code below with no luck
for k=length(array):-1:1
if array(k)==array(k+1)
array(k)=[]
end
end

回答 (2 件)

Rik
Rik 2022 年 3 月 30 日
No need for a loop (even if you should have replaced length by numel):
data=[ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6];
L=diff(data)==0; % note: this is 1 element shorter that data itself
data(L)=[]
data = 1×8
1 2 3 1 2 4 5 6

Voss
Voss 2022 年 3 月 30 日
Your approach will work; you just have to start at the second-to-last element instead of the last:
array = [ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6];
for k=length(array)-1:-1:1
if array(k)==array(k+1)
array(k)=[];
end
end
array
array = 1×8
1 2 3 1 2 4 5 6
Here's another approach:
array = [ 1 1 1 2 2 3 3 3 1 1 1 2 2 4 4 4 5 5 6];
array(diff(array) == 0) = [];
array
array = 1×8
1 2 3 1 2 4 5 6
  3 件のコメント
Torsten
Torsten 2022 年 3 月 30 日
編集済み: Torsten 2022 年 3 月 30 日
Why not ? I get
array = [2]
It's because diff(array) has one element less than array.
Voss
Voss 2022 年 3 月 30 日
As @Torsten says, it should still work (both ways work):
array = [2 2 2 2 2 2 2];
for k=length(array)-1:-1:1
if array(k)==array(k+1)
array(k)=[];
end
end
array
array = 2
array = [2 2 2 2 2 2 2];
array(diff(array) == 0) = [];
array
array = 2

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