How to use interp1 with different length?

3 次查看(过去 30 天)
How to use interp1 with different length?

采纳的回答

Walter Roberson
Walter Roberson 2012-1-10
The idea of using interp1 is obviously slowing you down, so use a different method.
MappedX = ((X - OldRangeMin) / (OldRangeMax - OldRangeMin) * (NewRangeMax - NewRangeMin) + NewrangeMin;
But to answer your question: you do not use different X and Y lengths: you use the same lengths, even if they represent different interval widths.
MappedX = interp1([OldRangeMin OldRangeMax], [NewRangeMin NewRangeMax], X);
  2 个评论
john john
john john 2012-1-10
great.. this is my problem now...my problem now....code below is for solving EL, which is he will read excel(contains the table) if the result same on the table he will give equivalent row value, if not he will interpolate the excel...so my problem how do i interpolate the table on excel...
a = get(handles.input1_gamma,'String'); %GAMMA
b = get(handles.input2_h,'String'); %H
c = get(handles.input3_q,'String'); %Q
d = get(handles.input4_power,'String'); %POWER
e = get(handles.input5_eout,'String');
out = str2num(b) + str2num(e);
set(handles.total_answer,'String',out);
data = xlsread('table.xlsx')
[m n] = size(data)
k = get(handles.total_answer,'String')
for i = 1 : m
A = data(i); %check xa 1st row then compare value
A = int2str(A);
if strcmp(A,k)
n = i;
score = data(n,2)%compare sa 2nd value
clc; %clear and cmd
set(handles.equalel,'String',score);
break;
else strcmp(A,k)
n = i;
score = data(n,2)
mHeight = (A)
mEout = (score);
outputArea = interp1(mHeight, mEout);
set(handles.equalel,'String',outputArea);
clc;
end
end
Walter Roberson
Walter Roberson 2012-1-17
Do you want the question addressed here or in your newer question http://www.mathworks.com/matlabcentral/answers/26268-can-you-help-me-how-to-interpolate-data-in-the-excel-table

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Annotations 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by