set xticks by calculation but also force 0 to be displayed

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I am setting xticks manually based on user inputs for distance. The plot will span the distance range; for example -1.3E-06 to 1.3E-06. The calculation works well to calculate the distance between tick marks, but depending on how it's done it may or may not automatically included 0 on the x axis. Is there a way I can force 0 to be displayed on the axis without altering my method for calculating the distance between tick marks.

Accepted Answer

Walter Roberson
Walter Roberson on 10 Jun 2021
format long g
%demonstration values
x = linspace(-1.4e-6,1.32e-6,29)
x = 1×29
-1.4e-06 -1.30285714285714e-06 -1.20571428571429e-06 -1.10857142857143e-06 -1.01142857142857e-06 -9.14285714285714e-07 -8.17142857142857e-07 -7.2e-07 -6.22857142857143e-07 -5.25714285714286e-07 -4.28571428571428e-07 -3.31428571428571e-07 -2.34285714285714e-07 -1.37142857142857e-07 -3.99999999999998e-08 5.71428571428574e-08 1.54285714285714e-07 2.51428571428572e-07 3.48571428571429e-07 4.45714285714286e-07 5.42857142857143e-07 6.4e-07 7.37142857142857e-07 8.34285714285715e-07 9.31428571428571e-07 1.02857142857143e-06 1.12571428571429e-06 1.22285714285714e-06 1.32e-06
y = sinpi(2e6*x)
y = 1×29
-0.587785252292474 -0.945356108439569 -0.961536119504149 -0.630482294044714 -0.0717461367613788 0.512899277405907 0.912324417231455 0.982287250728688 0.697522514745702 0.160865609613698 -0.433883739117559 -0.871947226253863 -0.995129582585911 -0.758946694294555 -0.248689887164854 0.351374824081344 0.824549628907141 0.999959716170871 0.814260281314378 0.334511859971844 -0.266036845566676 -0.770513242775789 -0.996738762088015 -0.863017923326349 -0.417640539972131 0.178556894798636 0.710273137068058 0.985492653565727 0.904827052466019
array_of_tick_values = -1.3e-6:.24e-6:1.3e-6
array_of_tick_values = 1×11
-1.3e-06 -1.06e-06 -8.2e-07 -5.8e-07 -3.4e-07 -1e-07 1.4e-07 3.8e-07 6.2e-07 8.6e-07 1.1e-06
%force a 0
xticks(union(array_of_tick_values, 0))
However, this could result in two ticks being close together. For example if your calculated tick was at -1e-10 then you would get a tick at -1e-10 and at 0. In the above example plot you can see that the -0.1 tick is close to the 0 tick
xtrange = max(abs(array_of_tick_values));
array_of_tick_values = linspace(-xtrange,xtrange,length(array_of_tick_values));
Robert Demyanovich
Robert Demyanovich on 10 Jun 2021
That is true. I tried your linspace solution and it didn't work. But the union solution did and that is good enough for me.

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More Answers (1)

Joel Lynch
Joel Lynch on 10 Jun 2021
Edited: Joel Lynch on 10 Jun 2021
Walter Roberson
Walter Roberson on 10 Jun 2021
ismember(0, vector_of_values)
but that leaves the searching open in that form. But if the vector is sorted you could
idx = find(vector_of_values <= 0, 1, 'last')
if vector_of_values(idx) == 0
exact match for 0
vector_of_values = [vector_of_values(1:idx), 0, vector_of_values(idx+1:end)]
assuming row vector
... But the union() approach I posted is much more compact and has the same effect.

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