问题
I have a vector that could look like this:
v = [1 1 2 2 2 3 3 3 3 2 2 1 1 1];
that is, the number of equal elements can vary, but they always increase and decrease stepwise by 1.
What I want is an easy way to be left with a new vector looking like this:
v2 = [ 1 2 3 2 1];
holding all the different elements (in the right order as they appear in v), but only one of each. Preferably without looping, since generally my vectors are about 10 000 elements long, and already inside a loop that's taking for ever to run.
Thank you so much for any answers!
回答1:
You can use diff for this. All you're really asking for is: Delete any element that's equal to the one in front of it.
diff return the difference between all adjacent elements in a vector. If there is no difference, it will return 0. v(ind~=0) will give you all elements that have a value different than zero. The 1 in the beginning is to make sure the first element is counted. As diff returns the difference between elements, numel(diff(v)) = numel(v)-1.
v = [1 1 2 2 2 3 3 3 3 2 2 1 1 1];
ind = [1 diff(v)];
v(ind~=0)
ans =
1 2 3 2 1
This can of course be done in a single line if you want:
v([1, diff(v)]~=0)
回答2:
You could try using diff which, for a vector X, returns [X(2)-X(1) X(3)-X(2) ... X(n)-X(n-1)] (type help diff for details on this function). Since the elements in your vector always increase or decrease by 1, then
diff(v)
will be a vector (of size one less than v) with zeros and ones where a one indicates a step up or down. We can ignore all the zeros as they imply repeated numbers. We can convert this to a logical array as
logical(diff(v))
so that we can index into v and access its elements as
v(logical(diff(v)))
which returns
1 2 3 2
This is almost what you want, just without the final number which can be added as
[v(logical(diff(v))) v(end)]
Try the above and see what happens!
来源：https://stackoverflow.com/questions/24327771/remove-duplicates-appearing-next-to-each-other-but-keep-it-if-it-appears-again