Linear interpolation two array with target value in one array

Given two array

A=[483, 427, 306, 195]

B=[0, 0.25, 0.5, 0.75]

Given a target the value 241, which is between the values 306 and 195 in the Array A. I want to find the value in the array B which is proportional to the value 241 in A. For this case the wanted value is 0.65 because x=0.75- [(241-195)/(306-195)]*0.25

I have to do this for a lot of array. Is there a simple way to do it?

[Awais Saeed]

You are multiplying with 0.25 in the end. Should not it be 0.5? As it will cause array index errors if your target lies between A(1) and A(2).

[Yazan]

How do you define the target value 241 in your example? and explain better the derivation of x (why do you multiply by 0.25?)

[SYML2nd]

It is the step of the array B, 0.25=0.75-0.5 . 241 is a given target I decide it by myself

Answer 1

[Star Strider]

Interpolate using interp1:

A=[483, 427, 306, 195];
B=[0, 0.25, 0.5, 0.75];
target = 241;
Result = interp1(A, B, 241)

Result = 0.6464

Graphically:

figure
plot(A, B, '-b')
hold on
plot(target, Result, 'p', 'MarkerSize',10, 'MarkerFaceColor','g')
plot([1 1]*target, [min(ylim) Result], '--r')
plot([min(xlim) target], [1 1]*Result, '--r')
hold off
grid
xlabel('A')
ylabel('B')
text(target, Result, sprintf('  \\leftarrow (%d, %.2f)', target, Result), 'Horiz','left', 'Vert','middle')

.

Answer 2

[Awais Saeed]

I did not check it thoroughly but I think it will work

clc;clear all;close all
A=[483,427, 306, 195];
B=[0, 0.25, 0.5, 0.75];
target = 241;

for col = 1:1:size(A,2)
    if (target < A(col) && target > A(col+1))   % if target lies between the range
	    result = B(col+1)-((target-A(col+1))/(A(col)-A(col+1)))*(B(col+1)-B(col))   % desired operation
	    break
    end
end

Answer 3

[Yazan]

Try this

clc, clear

A = [483, 427, 306, 195];
B = [0, 0.25, 0.5, 0.75];

% assume that these are your target values
target = arrayfun(@(x, y) mean([x, y]), A(1:end-1), A(2:end));
% force last value to be 241 to check with your example
target(end) = 241;

step = B(2)-B(1);

% save values in c
c1 = nan(size(target));

% using for loop
for nv=1:length(c1)
    c1(nv) = B(nv+1) - step*(target(nv)-A(nv+1))/(A(nv)-A(nv+1));
end

% without foor
c2 = arrayfun(@(x0, x1, x2, y) y-step*(x0-x2)/(x1-x2), target, A(1:end-1), A(2:end), B(2:end));

disp(c1)

    0.1250    0.3750    0.6464

disp(c2)

    0.1250    0.3750    0.6464