# matlab table, find y values from x values

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Serhat Unal on 4 Nov 2020
Commented: Mathieu NOE on 7 Nov 2020 I have a few questions which are:
1: I have plotted 13 x-values and 13 y-values, but I want this curve to be smooth and want
more values in between each point to make it smoother.
2: From this picture above, we want to find the y values from known x-values, but in this case different x-values.

Mathieu NOE on 4 Nov 2020
hi
for 1 /
• create a higher number of points using interp1 (see help)
• smooting can be done by sliding averaging use movingmean - or my code below
fo 2/
• from the refined / smoothed data points, use again interp1 at the requested x points to get the corresponding y values.
Code for sliding averaging (curve smoothing) :
function out = myslidingavg(in, N)
% OUTPUT_ARRAY = MYSLIDINGAVG(INPUT_ARRAY, N)
%
% The function 'slidingavg' implements a one-dimensional filtering, applying a sliding window to a sequence. Such filtering replaces the center value in
% the window with the average value of all the points within the window. When the sliding window is exceeding the lower or upper boundaries of the input
% vector INPUT_ARRAY, the average is computed among the available points. Indicating with nx the length of the the input sequence, we note that for values
% of N larger or equal to 2*(nx - 1), each value of the output data array are identical and equal to mean(in).
%
% * The input argument INPUT_ARRAY is the numerical data array to be processed.
% * The input argument N is the number of neighboring data points to average over for each point of IN.
%
% * The output argument OUTPUT_ARRAY is the output data array.
if (isempty(in)) | (N<=0) % If the input array is empty or N is non-positive,
disp(sprintf('SlidingAvg: (Error) empty input data or N null.')); % an error is reported to the standard output and the
return; % execution of the routine is stopped.
end % if
if (N==1) % If the number of neighbouring points over which the sliding
out = in; % average will be performed is '1', then no average actually occur and
return; % OUTPUT_ARRAY will be the copy of INPUT_ARRAY and the execution of the routine
end % if % is stopped.
nx = length(in); % The length of the input data structure is acquired to later evaluate the 'mean' over the appropriate boundaries.
if (N>=(2*(nx-1))) % If the number of neighbouring points over which the sliding
out = mean(in)*ones(size(in)); % average will be performed is large enough, then the average actually covers all the points
return; % of INPUT_ARRAY, for each index of OUTPUT_ARRAY and some CPU time can be gained by such an approach.
end % if % The execution of the routine is stopped.
out = zeros(size(in)); % In all the other situations, the initialization of the output data structure is performed.
if rem(N,2)~=1 % When N is even, then we proceed in taking the half of it:
m = N/2; % m = N / 2.
else % Otherwise (N >= 3, N odd), N-1 is even ( N-1 >= 2) and we proceed taking the half of it:
m = (N-1)/2; % m = (N-1) / 2.
end % if
for i=1:nx, % For each element (i-th) contained in the input numerical array, a check must be performed:
dist2start = i-1; % index distance from current index to start index (1)
dist2end = nx-i; % index distance from current index to end index (nx)
if dist2start<nx | dist2end<nx % if we are close to start / end of data, reduce the mean calculation on centered data vector reduced to available samples
dd = min(dist2start,dist2end); % min of the two distance (start or end)
out(i) = mean(in(i-dd:i+dd)); % mean of centered data , reduced to available samples
end % if
end % for i
Mathieu NOE on 7 Nov 2020
hi
just figured out there was a bug in my function
here is the correct code :
function out = myslidingavg(in, N)
% OUTPUT_ARRAY = MYSLIDINGAVG(INPUT_ARRAY, N)
%
% The function 'slidingavg' implements a one-dimensional filtering, applying a sliding window to a sequence. Such filtering replaces the center value in
% the window with the average value of all the points within the window. When the sliding window is exceeding the lower or upper boundaries of the input
% vector INPUT_ARRAY, the average is computed among the available points. Indicating with nx the length of the the input sequence, we note that for values
% of N larger or equal to 2*(nx - 1), each value of the output data array are identical and equal to mean(in).
%
% * The input argument INPUT_ARRAY is the numerical data array to be processed.
% * The input argument N is the number of neighboring data points to average over for each point of IN.
%
% * The output argument OUTPUT_ARRAY is the output data array.
if (isempty(in)) | (N<=0) % If the input array is empty or N is non-positive,
disp(sprintf('SlidingAvg: (Error) empty input data or N null.')); % an error is reported to the standard output and the
return; % execution of the routine is stopped.
end % if
if (N==1) % If the number of neighbouring points over which the sliding
out = in; % average will be performed is '1', then no average actually occur and
return; % OUTPUT_ARRAY will be the copy of INPUT_ARRAY and the execution of the routine
end % if % is stopped.
nx = length(in); % The length of the input data structure is acquired to later evaluate the 'mean' over the appropriate boundaries.
if (N>=(2*(nx-1))) % If the number of neighbouring points over which the sliding
out = mean(in)*ones(size(in)); % average will be performed is large enough, then the average actually covers all the points
return; % of INPUT_ARRAY, for each index of OUTPUT_ARRAY and some CPU time can be gained by such an approach.
end % if % The execution of the routine is stopped.
out = zeros(size(in)); % In all the other situations, the initialization of the output data structure is performed.
if rem(N,2)~=1 % When N is even, then we proceed in taking the half of it:
m = N/2; % m = N / 2.
else % Otherwise (N >= 3, N odd), N-1 is even ( N-1 >= 2) and we proceed taking the half of it:
m = (N-1)/2; % m = (N-1) / 2.
end % if
for i=1:nx, % For each element (i-th) contained in the input numerical array, a check must be performed:
dist2start = i-1; % index distance from current index to start index (1)
dist2end = nx-i; % index distance from current index to end index (nx)
if dist2start<N || dist2end<N % if we are close to start / end of data, reduce the mean calculation on centered data vector reduced to available samples
dd = min(dist2start,dist2end); % min of the two distance (start or end)
else
dd = m;
end % if
out(i) = mean(in(i-dd:i+dd)); % mean of centered data , reduced to available samples at both ends of the data vector
end % for i

Image Analyst on 4 Nov 2020
1. Try spline(). See attached example.
% Demo to show spline interpolation.
% Clean up / initialize
clc;
close all;
clear all;
workspace; % Display workspace panel.
% Create the original knot points.
lengthX = 10;
x = 1:lengthX;
y = rand (lengthX,1);
% Plot it and show how the line has sharp bends.
plot(x, y, '-sr', 'LineWidth', 2);
set(gcf, 'Position', get(0,'Screensize')); % Maximize figure.
% Use splines to interpolate a smoother curve,
% with 10 times as many points,
% that goes exactly through the same data points.
samplingRateIncrease = 10;
newXSamplePoints = linspace(1, max(x), lengthX * samplingRateIncrease);
smoothedY = spline(x, y, newXSamplePoints);
% Plot smoothedY and show how the line is
% smooth, and has no sharp bends.
hold on; % Don't destroy the first curve we plotted.
plot(newXSamplePoints, smoothedY, '-ob');
title('Spline Interpolation Demo', 'FontSize', 20);
legend('Original Points', 'Spline Points');
% Mathworks Demo code from their Help
% x = 0:10;
% y = sin(x);
% xx = 0:.25:10;
% yy = spline(x,y,xx);
% plot(x,y,'o',xx,yy)
slopes = [0, diff(smoothedY)];
plot(newXSamplePoints, slopes, 'k-', 'LineWidth', 3);
% Draw x axis
line(xlim, [0,0], 'Color', 'k', 'LineWidth', 2);
grid on;
legend('Original Points', 'Spline Points', 'Slope');
1. Did you try interp1?
xEstimated = interp1(x, y, xSpecified)

#### 1 Comment

Image Analyst on 7 Nov 2020
Serhat - an update please. Did this work for you?