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Linear interpolation two array with target value in one array

41 vues (au cours des 30 derniers jours)
SYML2nd
SYML2nd le 16 Août 2021
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?
  3 commentaires
Yazan
Yazan le 16 Août 2021
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
SYML2nd le 16 Août 2021
Modifié(e) : SYML2nd le 16 Août 2021
It is the step of the array B, 0.25=0.75-0.5 . 241 is a given target I decide it by myself

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Star Strider
Star Strider le 16 Août 2021
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')
.

Plus de réponses (2)

Awais Saeed
Awais Saeed le 16 Août 2021
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

Yazan
Yazan le 16 Août 2021
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

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