【求助】想实现非线性数据的插值
x=y=
假如想得到在y=25处的两个x值，应该怎么做
interp1是只能实现线性插值，高手指点 找到一个解决办法
table=.'
xi=mminterp(table, 2, 25)
函数定义如下
function y=mminterp(tab, col, val)
%MMINTERP 1-D Table Search by Linear Interpolation.
%Y=MMINTERP(TAB,COL,VAL) linearly interpolates the table
%TAB searching for the scalar value VAL in the column COL.
%All crossings are found and TAB(:,COL) need not be monotonic.
%Each crossing is returned as a separate row in Y and Y has as
%many columns as TAB.Naturally,the column COL of Y contains
%the value VAL. If VAL is not found in the table,Y=.
%Copyright (c) 1996 by Prentice-Hall,Inc.
=size(tab);
if length(val) > 1,error(' VAL must be a scalar. '),end
if col>ct|col < 1,error(' Chosen column outside table width. '),end
if rt < 2,error(' Table too small or not oriented in columns. '),end
above=tab(: , col) > val;%True where > VAL
below=tab(: , col) < val;%True where < VAL
equal=tab(: , col) == val;%True where = VAL
if all(above == 0) | all(below == 0),%handle simplest case
y=tab(find(equal), : );return
end
pslope=find(below(1:rt-1)&above(2:rt));%indices where slope is +
nslope=find(below(2:rt)&above(1:rt-1));%indices where slope is -
ib=sort();%put indices below in order
ia=sort();%put indices above in order
ie=find(equal);%indices where equal to val
=sort( );%find where equals fit in result
ieq=ix > length(ib);%True where equals values fit
ry=length(tmp);%# of rows in result y
y=zeros(ry, ct);%poke data into a zero matrix
alpha=(val-tab(ib,col))./(tab(ia,col)-tab(ib,col));
alpha=alpha(: , ones(1, ct));%duplicate for all columns
y(~ieq, : )=alpha.*tab(ia, : )+(1-alpha).*tab(ib, : );%interpolated values
y(ieq, : )=tab(ie, : );%equal values
y( : , col)=val*ones(ry, 1);%remove roundoff error 强，试一下。 有这么复杂吗？
xx=linspace(1,5,150);
yy=spline(x,y,xx);
f=xx(yy>=24.9&yy<=25.001)
result return:
f =
2.3423 3.6577
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