Creating time series vectors of unequal duration

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Sowmya MR
Sowmya MR 2017-5-15
评论： dpb 2017-5-16
Hi,
I have two vectors: a1=[ 2 4 6 8 10...20] with an increment of 2 and a2=[2 7 12 17....50] with an increment of 5. How do i create a 3rd vector a3=[2 4 6 8....50] with an increment of 2? Bacically a3 should have a start point of a1(1) and an end point of a2(end) with an increment of 2.

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dpb
dpb 2017-5-16
One possible interpretation...
>> inear=interp1(a2,1:length(a2),a1(end),'nearest');
>> a3=[a2(1:inear-1) a2(inear):2:a2(end)]
a3 =
2 7 12 17 22 24 26 28 30 32 34 36 38 40 42 44 46
>>
NB: You can't create your a2 above as 2:5:50 with 50 as the last element because 2+n*5=50 --> n=48/5 --> noninteger. Hence the above ending at 46.
You're second description in original posting said you wanted "a 3rd vector a3=[2 4 6 8....50]"_ which is what the first solution above gives.
As noted, the request is ambiguous at best, contradictory in its details as given.

更多回答（1 个）

dpb
dpb 2017-5-15
"... a3 should have a start point of a1(1) and an end point of a2(end) with an increment of 2."
a3=[a1(1):2:a2(end)]; % done! :)
  2 个评论
Sowmya MR
Sowmya MR 2017-5-16
Thanks. But i want to know the closest point in a2 where a1 ends and then append remaining values to the length of a2.
dpb
dpb 2017-5-16
Say wha'!!!??? I've no klew what that means nor how relates to the above problem description?
Show what you would expect the result to be from the two input vectors and explain how you deduced that result.

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