Problem with interp1 function

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Sowmya MR
Sowmya MR 2016-9-29
回答： Walter Roberson 2016-9-29
I have some issues with interp1 function. I have a time vector
t1=1/fs:1/fs:1/fs*1462805 where fs=250;
Now i have the output of a classifier where each value corresponds to a 8s epoch
annot=[zeros(1,600) ones(1,570)];
Now when i do
annotnew=interp1(annot,t1),
I get NaN's. Can someone help me fix this?
  2 个评论
per isakson
per isakson 2016-9-29
What result do you expect?
Sowmya MR
Sowmya MR 2016-9-29
An output annotnew which is of length similar to t1 with real values (0's and 1's).

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回答（1 个）

Walter Roberson
Walter Roberson 2016-9-29
The syntax
annotnew=interp1(annot,t1)
is the same as
annotnew=interp1( 1 : length(annot), annot, t1)
and that implicitly indicates that the x values start from 1 and go upwards. You then ask to query at the positions in t1, some of which are probably less than 1, and some might be greater than length(annot) . Any position corresponding to a t1 whose value is less than 1 or greater than length(annot) is a request for extrapolation to before or after the data you provided, and the default behavior for extrapolation is to assign NaN there. See the 'extrap' keyword in interp1() if you want to change the extrapolation behavior.

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