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19 views (last 30 days)

Stelios Fanourakis
on 20 Apr 2019

Commented: Stelios Fanourakis
on 24 Apr 2019

Hi

I have two datasets (X,Y). For every value of Y that corresponds to a value of X.

I am thinking of applying a polyfit command to get the best curve fitting for the data, and as a result I will come up with a function (slope+intercept).

How do I find the difference between those two functions? What command do I apply to subtract Function A from Function B?

Walter Roberson
on 21 Apr 2019

P1 = polyfit((1:length(X)).', X(:), 6);

Likewise for Y producing P2

Pd = polyval(P1, 1:length(X)) - polyval(P2, 1:length(X))

Stelios Fanourakis
on 21 Apr 2019

@Walter.

I guess this is the correct answer but I need a few clarifications.

The two datasets are comprised a series of X data that corresponds to a series of Y data.

I guess the polyfit command would be like

P1 = polyfit((1:length(X1)).', Y1(:), 6);

P2 = polyfit((1:length(X2).',Y2(:),6);

Pd = polyval(P1, 1:length(X)) - polyval(P2, 1:length(X2))

Am I doing something wrong? I need to estimate numerically the difference between two curves.

Can you also remind me what the ' means in programming?

Walter Roberson
on 21 Apr 2019

Supposing that X and Y have two columns, with X(K,1) corresponding to Y(K,1) and X(K,2) corresponding to Y(K,2) then

P1 = polyfit(X(:,1), Y(:,1), 6);

P2 = polyfit(X(:,2), Y(:,2), 6);

allX = union(X(:,1), X(:,2));

Pd = polyval(P1, allX) - polyval(P2, allX);

plot(allX, Pd);

In MATLAB, ' is complex conjugate transpose, formal name ctranspose() . But I did not use it. I used .' which is regular transpose without complex conjugate, formal name transpose() .

polyfit() can accept X and Y that are both rows, or X and Y that are both columns, but it does not like a mix such as a row X = 1:20 and a column Y = rand(20,1) . I do not know the orientation of your dataset so I cannot assume that your Y is a row, so it is not safe for me to program polyfit(1:length(Y), Y, 6) because Y might be a column. Therefore I need to force both X and Y to be rows, or force both to be columns. Using (:) forces a column so Y(:) to force a column for Y is easy, but the two ways to force Y to be a row are Y(:).' or else reshape(Y,1,[]) which are a bit ugly. So it is easiest to force X to be a column by using the .' operator on what is known to be a row, 1:length(X), and use Y(:) . I could also have written

P1 = polyfit(1:length(X), Y(:).', 6)

Now, if someone happened to be writing this as part of a sequence of statements and had reason to know that Y was a row already, then they could just use

P1 = polyfit(1:length(X), Y, 6)

but since I do not know because the person asking the problem did not specify the shape, then I have to put in the protections to make the code work either way.

Stelios Fanourakis
on 21 Apr 2019

You can say Y are rows and X columns. For every Y is a X and the same applies at both data sets

Walter Roberson
on 21 Apr 2019

%"Y are rows and X columns"

P1 = polyfit(X(:,1), Y(1,:).', 6);

P2 = polyfit(X(:,2), Y(2,:).', 6);

allX = unique(X);

Pd = polyval(P1, allX) - polyval(P2, allX);

plot(allX, Pd)

Stelios Fanourakis
on 21 Apr 2019

@Walter

I use this code

clc;

clear all;

x1 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','M2:M384');

y1 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','A2:A376');

x2 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','N2:N867');

y2 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','B2:B868');

P1 = polyfit(x1(:,1), y1(1,:).', 6);

P2 = polyfit(x2(:,2), y2(2,:).', 6);

allX = unique(X);

Pd = polyval(P1, allX) - polyval(P2, allX);

plot(allX, Pd)

And I get the error

Error using .' (line 191)

Undefined function 'transpose' for input arguments of type 'table'.

Error in curvefitdif (line 10)

P1 = polyfit(x1(:,1), y1(1,:).', 6);

Stelios Fanourakis
on 21 Apr 2019

I also get the error

Error using polyfit (line 44)

X and Y vectors must be the same size.

Error in curvefitdif (line 10)

P1 = polyfit(x1(:,1), y1(1,:), 6);

Why?

Walter Roberson
on 21 Apr 2019

You seem to be reading 384-2+1 = 383 values for x1

You seem to be reading 376-2+1 = 375 values for y1

You seem to be reading 867-2+1 = 866 values for x2

You seem to be reading 868-2+1 = 867 values for y2.

I do not know what it means to fit 383 x values against 375 y values.

When you use readtable(), you get back a table() object as the result, not a numeric array. You will need to use {} indexing to get to the numeric results, such as

x1 = x1{:,}; %replace the table object with numeric object it contains

Walter Roberson
on 21 Apr 2019

x1r = readtable('ValidationTest.xls', 'Sheet',5, 'Range','M2:M384');

y1r = readtable('ValidationTest.xls', 'Sheet',5, 'Range','A2:A376');

x2r = readtable('ValidationTest.xls', 'Sheet',5, 'Range','N2:N867');

y2r = readtable('ValidationTest.xls', 'Sheet',5, 'Range','B2:B868');

h1 = min([size(x1r,1), size(y1r,1)]);

h2 = min([size(x2r,1), size(y2r,1)]);

x1 = x1r{1:h1, 1};

y1 = y1r{1:h1, 1};

x2 = x2r{1:h2, 1};

y2 = y2r{1:h2, 1};

P1 = polyfit(x1(:,1), y1(1,:), 6);

P2 = polyfit(x2(:,2), y2(2,:), 6);

allX = unique(X);

Pd = polyval(P1, allX) - polyval(P2, allX);

plot(allX, Pd)

Stelios Fanourakis
on 21 Apr 2019

@walter

I adjusted a bit the code

x1 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','M2:M384');

y1 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','A2:A384');

x2 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','N2:N868');

y2 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','B2:B868');

P1 = polyfit(x1{:,1}, y1{1,:}, 6);

P2 = polyfit(x2{:,2}, y2{2,:}, 6);

allX = unique(X);

Pd = polyval(P1, allX) - polyval(P2, allX);

plot(allX, Pd)

But still I get the error

Error using polyfit (line 44)

X and Y vectors must be the same size.

Error in curvefitdif (line 11)

P1 = polyfit(x1{:,1}, y1{1,:}, 6);

Why are they not the same size now??

Since we are talking about x1, y1.

x2,y2 are different dataset

Stelios Fanourakis
on 21 Apr 2019

@Walter

I don't know what to do.

Even if I assign the same number of cells to all variables x1,y1,x2,y2 still I get the error

Error using polyfit (line 44)

X and Y vectors must be the same size.

Error in curvefitdif (line 19)

P1 = polyfit(x1(:,1), y1(1,:), 6);

And it especially stucks at the specific line of code, where it shouldn't since I have made both x1 and y1 same size.

Is it something wrong with my Matlab?

Walter Roberson
on 22 Apr 2019

x1 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','M2:M384');

y1 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','A2:A384');

x2 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','N2:N868');

y2 = readtable('ValidationTest.xls', 'Sheet',5, 'Range','B2:B868');

P1 = polyfit(x1{:,:}, y1{:,:}, 6);

P2 = polyfit(x2{:,:}, y2{:,:}, 6);

allX = unique([x1{:,}; x2{:,:}]);

Pd = polyval(P1, allX) - polyval(P2, allX);

plot(allX, Pd)

Stelios Fanourakis
on 22 Apr 2019

I get the error

Error using unique (line 117)

Invalid input. Valid flags are 'rows', 'first', 'last', 'stable', 'sorted', 'legacy'.

Error in curvefitdif (line 14)

allX = unique(x1{:,:}, x2 {:,:});

Stelios Fanourakis
on 22 Apr 2019

@Walter

I get NaN for P2. Why?

P1 =

-0.0000 0.0000 -0.0019 0.0344 -0.2979 1.1208 4.9503

P2 =

NaN NaN NaN NaN NaN NaN NaN

Stelios Fanourakis
on 22 Apr 2019

I found the error of Nan but now I get the new error

Error using horzcat

Dimensions of arrays being concatenated are not consistent.

Error in curvefitdif (line 16)

allX = unique([x1{:,:}, x2{:,:}]);

Walter Roberson
on 22 Apr 2019

You did not copy the assignment to allX from my code. My code clearly had

allX = unique([x1{:,:}; x2{:,:}]);

with [] in place and using semi-colon between the two arrays. You used

allX = unique(x1{:,:}, x2{:,:});

with no [] and using comma instead of semi-colon. Then you put back the [] but you used

allX = unique([x1{:,:}, x2{:,:}]);

with comma instead of semi-colon.

My guess about the P2 and NaN is that when you said to use N2:N868 and B2:B868 that you should have said N2:N867 and B2:B867 . The code I offered you with h1 and h2 and min() of size() would have taken care of that problem for you, but you decided that your version was more correct for your needs and you got burned because you did not understand what it was you were discarding.

Stelios Fanourakis
on 22 Apr 2019

Yes thank you so much Walter. Now I see a result. I used to come up with various errors of brackets and columns that's why I experimented a bit but now it is sorted.

I am trying to plot everything now in one graph. How to plot P1, P2 and their difference together in one plot?

Actually, when I try to use those lines

Pd = polyval(P1, allX) - polyval(P2, allX)

P11 = polyval(P1, allX)

P22 = polyval(P2, allX)

hold on

plot(allX, Pd)

plot(allX, P11)

plot(allX, P22)

I get three plots. But one of the datasets is not correct

Stelios Fanourakis
on 22 Apr 2019

Stelios Fanourakis
on 22 Apr 2019

@Walter.

I just changed excel sheet to read a different set of datasets with the same row number and now I get again NaN as result for both P1 and P2. Why is this?

If I apply the same code for Sheet 5 and different columns but same number of rows I get numerical values. Why am I not getting them for sheet 6?

x1 = readtable('ValidationTest.xls', 'Sheet',6, 'Range','N2:N384 ');

y1 = readtable('ValidationTest.xls', 'Sheet',6, 'Range','A2:A384 ');

x2 = readtable('ValidationTest.xls', 'Sheet',6, 'Range','O2:O867');

y2 = readtable('ValidationTest.xls', 'Sheet',6, 'Range','B2:B867');

P1 = polyfit(x1{:,:}, y1{:,:}, 6 )

P2 = polyfit(x2{:,:}, y2{:,:}, 6)

allX = unique([x1{:,:}; x2{:,:}]);

Pd = polyval(P1, allX) - polyval(P2, allX)

P11 = polyval(P1, allX)

P22 = polyval(P2, allX)

hold on

plot(allX, Pd, 'b')

plot(allX, P11, 'g')

plot(allX, P22, 'r')

Walter Roberson
on 22 Apr 2019

Dang, I posted complete code for this, but somehow it did not get saved!

Walter Roberson
on 22 Apr 2019

%after the readtable

mask = isnan(x1{:,:}) | isnan(y1{:,:});

x1(mask,:) = [];

y1(mask,:) = [];

mask = isnan(x2{:,:}) | isnan(y2{:,:});

x2(mask,:) = [];

y2(mask,:) = [];

Stelios Fanourakis
on 23 Apr 2019

I get the below error

P1 =

0 0 0 0 0 0 0

Warning: Polynomial is not unique; degree >= number of data points.

> In polyfit (line 74)

In curvefitdif (line 19)

P2 =

0 0 0 0 0 0 0

Pd =

0×1 empty double column vector

P11 =

0×1 empty double column vector

P22 =

0×1 empty double column vector

Stelios Fanourakis
on 24 Apr 2019

I also get the error

Error using isnan

Too many input arguments.

Error in see (line 38)

mask = isnan(x1{:,:}) | isnan(y1{:,:});

Why?

John D'Errico
on 21 Apr 2019

John D'Errico
on 21 Apr 2019

You claim to have two polynomials, each of degree 6. You want the difference between them. SUBTRACT THE COEFFICIENTS.

x = rand(1,50);

y1 = rand(1,50);

y2 = rand(1,50);

Fit.

p1 = polyfit(x,y1,6);

p2 = polyfit(x,y2,6);

Subtract.

pdiff = p1 - p2;

pdiff is a polynomial of order 6, that represents the difference between the two. You can evaluate it.

fplot(@(x) polyval(p1,x),[0 1])

hold on

fplot(@(x) polyval(p2,x),[0 1],'g')

fplot(@(x) polyval(pdiff,x),[0 1],'r')

axis([0 1 -1,1])

Is there a reason why you refuse to believe me? You can evaluate that difference function at any point.

John D'Errico
on 22 Apr 2019

So effectively, your question has nothing to do with your real problem?

Problem 1: Read the data in. xlsread should suffice.

Problem 2: Extract the data from a table. There is no need to create a table. See problem 1.

Problem 3: Apply polyfit. Um, use polyfit? Read the help for polyfit. Or read my comment.

Problem 4: Subtracting the polynomials. I told you how to do that.

You are making this wildly overly complex.

Stelios Fanourakis
on 22 Apr 2019

It has to do with P2. I don;t know why it gives NaN. I am trying to find out at the moment

Stelios Fanourakis
on 22 Apr 2019

John D'Errico
on 23 Apr 2019

If pdiff gives a result of NaN, then one or both of the polynomials was already garbage. LOOK at the cofficients that you got from those two fits.

NaN can only ever result as a subtraction by things like inf - inf, or anything plus a NaN. So the problem lies in your original fits.

Stelios Fanourakis
on 23 Apr 2019

Actually, the problem lies where I use readtable and import the columns from excel.

They are imported as NaNs

Stelios Fanourakis
on 21 Apr 2019

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