Function "diff" and the loss of an element in the array

I have an issue with MATLAB, specifically related to a "diff" operation that is causing the loss of an element. The "diff" function in MATLAB is used to calculate the differences between consecutive elements in an array or matrix. I am using this function to calculate the derivate of an array.

It is clear to me that the resulting vector must have one less element. However, I need to perform the "diff" operation twice on the same array (resulting in an array with two less elements), and the issue I'm facing is that I don't know whether the vector is losing an element at the beginning or at the end. Initially, I thought it might be at the beginning. But when I compare the result of this operation with a reference value, it seems to be shifted on the axis. How can I be sure of this?

PS: It really makes a difference in my implementation.

[Stephen23]

"... the issue I'm facing is that I don't know whether the vector is losing an element at the beginning or at the end."

Neither: differences are between the elements of your vector, not at the same locations as them.

Answer 1

[Torsten]

Use "gradient" if you don't want to loose elements:

https://uk.mathworks.com/help/matlab/ref/gradient.html

[William Rose]

I didn't know about gradient. Thanks @Torsten !

t=0:7; x=t.^3;
dxdtD=diff(x); d2xdt2D=diff(diff(x));
dxdtG=gradient(x); d2xdt2G=gradient(gradient(x));
disp(x); disp(dxdtG); disp(d2xdt2G)

     0     1     8    27    64   125   216   343

     1     4    13    28    49    76   109   127

    3.0000    6.0000   12.0000   18.0000   24.0000   30.0000   25.5000   18.0000

@Bryan Ambrósio, compare the results from gradient() to the theoretical results for this signal, and to the results using diff(). An advantage of gradient() is that it gives estimates at the same locaitons as the sample locations, including at the edges. But you may not like its estimates at the edges. Your call.

legstr={'theo','Diff','Gradient'};
subplot(211);
plot(t,3*t.^2,'-g',t(1:end-1)+.5,dxdtD,'gx',t,dxdtG,'go'); 
ylabel('dx/dt'); grid on; 
legend(legstr,'Location','best')
subplot(212); 
plot(t,6*t,'-b',t(2:end-1),d2xdt2D,'bx',t,d2xdt2G,'bo'); 
ylabel('d^2X/dt^2'); xlabel('Time'); grid on; 
legend(legstr,'Location','best')

[Walter Roberson]

And gradient allows you to specify the spacing between elements, which is important for getting the right scale on the derivatives.

Answer 2

[dpb]

diff is defined as x(i+1)-x(i) for i=2:numel(x). But, MATLAB arrays are always 1-based so the result of x(2)-x(1) is in the first location of the result vector. You don't really "lose" anything, but the location of the nth position corresponding to the initial vector does move down each successive operation...

Easy enough to illustrate numerically...

>> V=randi(20,1,10)
V =
     7     6    13     9    10    10    19     2     6     9
>> dV1=diff(V)
dV1 =
    -1     7    -4     1     0     9   -17     4     3
>> dV2=diff(dV1)
dV2 =
     8   -11     5    -1     9   -26    21    -1
>> 

Oftentimes it is convenient for coding purposes to prepend to the result to maintain the same length;

dV1=[nan diff(V)];

for example.

[Steven Lord]

And note that you can call diff with two or three inputs to avoid having to call diff with one input repeatedly.

V=randi(20,1,10)

V = 1×10

17 20 7 4 18 19 19 11 7 8

dV1=diff(V)

dV1 = 1×9

3 -13 -3 14 1 0 -8 -4 1

dV2=diff(dV1)

dV2 = 1×8

-16 10 17 -13 -1 -8 4 5

dV2_anotherWay = diff(V, 2, 2)

dV2_anotherWay = 1×8

-16 10 17 -13 -1 -8 4 5

isequal(dV2, dV2_anotherWay)

ans = logical
   1

I would advise using the three input syntax to avoid the case where you accidentally reduce the array to size 1 in the dimension over which your calls have been operating and start operating on a different dimension without intending to do so.

M = magic(4)

M = 4×4

16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1

for k = 1:4
    M = diff(M)
end

M = 3×4

-11 9 7 -5 4 -4 -4 4 -5 7 9 -11

M = 2×4

15 -13 -11 9 -9 11 13 -15

M = 1×4

-24 24 24 -24

M = 1×3

48 0 -48

Did you expect the final M to have fewer columns than the original M?

M = magic(4)

M = 4×4

16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1

M2 = diff(M, 4, 1)

M2 =

  0×4 empty double matrix

[Stephen23]

"...but the location of the nth position corresponding to the initial vector does move down each successive operation..."

This explanation, like the original question, confuses data with indices.

[dpb]

I disagree @Stephen23; it simply notes that owing to the size differential, the indices to the locations of the original data are numerically lower than those to the original data from which the difference was computed to address the specifics of the original question as to which difference is stored where.

Of course, there are no other differences available to be stored, but that's somewhat different... :)

[dpb]

Yeah, @Steven Lord, I thought about adding in the alternate syntax but opted for brevity to keep to the original Q? only...sometimes one might need both differences in which case the three argument version doesn't help so much; you've got two calls either way.

I've thought of (but have never actually submitted it as formal enhancment request) it might be a worthwhile option to have a flag to return an array of successive differences as an alternative to the one n-th difference -- the problem and argument against being it makes it hard to deal with the higher dimension cases other than vector inputs.

Answer 3

[William Rose]

@Bryan Ambrósio,

[edit: changed the code for plotting d2xdt2. The plotted result is unchanged, but the new verison of the code is more robust, because it will put d2xdt2 at the correct horizontal location, even if delta T is not unity.]

You can answer your own question by experiment.

t=0:1:7; x=t.^3;
dxdt=diff(x); d2xdt2=diff(diff(x));
disp(x); disp(dxdt); disp(d2xdt2)

     0     1     8    27    64   125   216   343

     1     7    19    37    61    91   127

     6    12    18    24    30    36

You can see that the times associated with the second derivative are offset by 1 sample at the beginning and by 1 at the end. Plot the results as shown:

subplot(311); plot(t,x,'-r.'); ylabel('X'); grid on
subplot(312); plot(t(1:end-1)+.5,dxdt,'-g.'); ylabel('dX/dt'); grid on
subplot(313); plot(t(2:end-1),d2xdt2,'-b.'); 
ylabel('d^2X/dt^2'); xlabel('Time'); xlim([0 7]); grid on

The plots agree with the analytical solution for the first and second derivatives of x=t^3.