Problem with Interp1

조회 수: 3 (최근 30일)
Jason
Jason 2018년 1월 26일
댓글: Star Strider 2018년 1월 26일
Hi, I cannot get interp1 to give me the correct answer. I have a simple, data set (see below) and all I want to do if use interpolation to give me the x value at y=thresh
thresh=225.5
My code is:
vq1 = interp1(x,y,thresh,'pchip')
This results in vq1=12207908.76
x=
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
15.00
16.00
17.00
18.00
19.00
20.00
21.00
22.00
23.00
24.00
25.00
26.00
27.00
28.00
29.00
30.00
31.00
32.00
33.00
34.00
35.00
36.00
37.00
38.00
39.00
40.00
41.00
42.00
y=
430.00
422.00
445.00
438.00
423.00
435.00
432.00
422.00
419.00
440.00
403.00
427.00
444.00
437.00
429.00
417.00
440.00
441.00
432.00
425.00
413.00
370.00
271.00
139.00
63.00
28.00
36.00
27.00
25.00
27.00
15.00
27.00
22.00
29.00
20.00
19.00
27.00
17.00
17.00
15.00
22.00
17.00

채택된 답변

Star Strider
Star Strider 2018년 1월 26일
If you want to find the ‘x’-values where ‘y’ equals ‘thresh’, you need to reverse the order of ‘x’ and ‘y’ in the interp1 call. However with your data, you first have to find the approximate index of that value, since you cannot simply reverse the order of the entire vector.
Try this:
yval = find(y >= thresh, 1, 'last'); % Define Region To Interpolate
vq1 = interp1(y(yval-1:yval+1), x(yval-1:yval+1), thresh,'pchip') % Interpolate In Region
vq1 =
23.3818
  댓글 수: 2
Jason
Jason 2018년 1월 26일
Thankyou (and to Stephen)
Star Strider
Star Strider 2018년 1월 26일
As always, my pleasure.

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