matlab table, find y values from x values

11 Ansichten (letzte 30 Tage)
Serhat Unal
Serhat Unal am 4 Nov. 2020
Kommentiert: Mathieu NOE am 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.
  2 Kommentare
Mathieu NOE
Mathieu NOE am 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
Mathieu NOE am 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

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Image Analyst
Image Analyst am 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)

Kategorien

Mehr zu Spline Postprocessing finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by