Why doesn't matplotlib operate on objects directly? - matplotlib

Why does matplotlib work the way it does? For example, why do you do this:
x = plt.bar([1,2,3,4],[1,2,2,1])
plt.title('Title')
Instead of:
x = plt.bar([1,2,3,4],[1,2,2,1])
x.title('Title')
It seems like plots should be objects that have methods like "title", "xlabel", etc.
It feels unpythonic to me to have functions which don't even take in the object that they operate on as a parameter, but I'm assuming there's a good reason that it works this way?

Matplotlib actually has an object oriented API. So the example would be
fig, ax = plt.subplots()
ax.bar([1,2,3,4],[1,2,2,1])
ax.set_title('Title')
ax.set_xlabel("X-label")
fig.set_facecolor("pink")
The object to work on is mostly the matplotlib.Axes.axes, called ax here. It has the plotting methods, as well as all kinds of other methods to tweak the graph.
The matplotlib API is very well documented. You will find the respecitive methods and return types that you need for OO-programming in the documentation.
You can then e.g. look at the return type of ax.bar, which is a matplotlib.container.BarContainer and colorize the second bar like
bars = ax.bar([1,2,3,4],[1,2,2,1])
bars[1].set_color("crimson")
Pyplot itself is not very pythonic, because it has been designed to resemble the MATLAB language. But under the hood, it normally just calls the respective class methods from the API.

Related

Pass existing matplotlib axis object to wxmplot / wxpython

Lets assume i have a class which handels my data processing and this class contains a function that delivers my a standardize plot with a function that retunrs an matplotlib axis object.... now someone wants a GUI for this...
def get_plot(self, ax=None,.... *lkwargs)
# This function does some stuff and returns "ax": a matplotlib axis object
return ax
now want i want to do is to use the ax object with wxmplot. Ideally i want to pass the existing ax object with all its handels / content to wxmplot. Is this possible?
Thanks in advance
The simplest answer is "No". Wxmplot creates matplotlib figure and axes objects, but none of the main interface routines accept matplotlib objects.
OTOH, one can access the matplotlib axis after the wxmplot plotting objects are created. That is, it is possible to "do some stuff" (depending on the details) with the underlying axes.

save pyplot figure "as figure" (not as image)

How can I save a figure using PyPlot in Julia, so that the figure can be reloaded as a figure later in Julia? (not as an image)
You can use serialize to store any Julia object. This beautifully works for plots as well.
Let us start by generating a plot:
using Plots
pyplot()
p = plot(rand(10));
using Serialization
Serialization.serialize("myfile.jld", p);
Note that you need a semicolon after plot command so it does not appear on the screen.
Let us now read the plot (to have a full test I ended the previous Julia session and started a new one):
using Plots
pyplot();
using Serialization
p2 = Serialization.deserialize("myfile.jld");
In order to display it now it is enough to type in REPL:
julia> p2
You might want also want to use plain PyPlot (I strongly recommend Plots for flexibility). In that case your best bet is to follow rules described in object-oriented API of Matplotlib:
using PyPlot
ioff()
fig = subplot()
fig.plot(rand(10))
fig.set_title("Hello world")
using Serialization
serialize("pp.jld", fig)
In order to plot de-serialize back the object:
using PyPlot
ioff()
using Serialization
fig = deserialize("pp.jld")
show()
Finally, note that the serialization is good only for short term storage. If anything changes (e.g. you update Julia packages) you might not be able to de-serialize the plot.
Hence another good alternative for processable plots are saving them to LaTeX or SVG format - both is possible in Julia.

Accessing backend specific functionality with Julia Plots

Plots is simple and powerful but sometimes I would like to have a little bit more control over individual elements of the plot to fine-tune its appearance.
Is it possible to update the plot object of the backend directly?
E.g., for the default pyplot backend, I tried
using Plots
p = plot(sin)
p.o[:axes][1][:xaxis][:set_ticks_position]("top")
but the plot does not change. Calling p.o[:show]() afterwards does not help, either.
In other words: Is there a way to use the PyPlot interface for a plot that was initially created with Plots?
Edit:
The changes to the PyPlot object become visible (also in the gui) when saving the figure:
using Plots
using PyPlot
p = Plots.plot(sin, top_margin=1cm)
gui() # not needed when using the REPL
gca()[:xaxis][:set_ticks_position]("top")
PyPlot.savefig("test.png")
Here, I used p.o[:axes][1] == gca(). One has to set top_margin=1cm because the plot area is not adjusted automatically (for my actual fine-tuning, this doesn't matter).
This also works for subsequent updates as long as only the PyPlot interface is used. E.g., after the following commands, the plot will have a red right border in addition to labels at the top:
gca()[:spines]["right"][:set_color]("red")
PyPlot.savefig("test.png")
However, when a Plots command like plot!(xlabel="foo") is used, all previous changes made with PyPlot are overwritten (which is not suprising).
The remaining question is how to update the gui interactively without having to call PyPlot.savefig explicitly.
No - the plot is a Plots object, not a PyPlot object. In your specific example you can do plot(sin, xmirror = true).
I'm trying to do the same but didn't find a solution to update an existing plot. But here is a partial answer: you can query information from the PyPlot axes object
julia> Plots.plot(sin, 1:4)
julia> Plots.PyPlot.plt[:xlim]()
(1.0,4.0)
julia> Plots.plot(sin, 20:24)
julia> ax = Plots.PyPlot.plt[:xlim]()
(20.0,24.0)
and it gets updated.

Logarithmic scaling / colorbar in Julia using PyPlot (matplotlib)

I am using Julia 0.5 and the latest version of PyPlot.
I am printing an 2D-Array using plot.pcolorand it works pretty good. But now I have data that needs a logarithmic scaling. I searched on the web and what I found was an example using
plt.pcolor(X, Y, Z1, norm=LogNorm(vmin=Z1.min(), vmax=Z1.max()), cmap='PuBu_r')
But since LogNorm seems to be a python function ist doesn't work in Julia. Does anyone have an idea what I can hand over to norm=to get a logarithmic scaling?
An example would be:
using PyPlot
A = rand(20,20)
figure()
PyPlot.pcolor(A, cmap="PuBu_r")
colorbar()
Matplotlib fields and methods can be accessed using the
matplotlib[:colors][:LogNorm]
syntax (i.e. for the corresponding matplotlib.colors.LogNorm object).
UPDATE: Thank you for your mwe. Based on that example, I managed to make it work like this:
PyPlot.pcolor(A, norm=matplotlib[:colors][:LogNorm](vmin=minimum(A), vmax=maximum(A)), cmap="PuBu_r")

Multiplot with matplotlib without knowing the number of plots before running

I have a problem with Matplotlib's subplots. I do not know the number of subplots I want to plot beforehand, but I know that I want them in two rows. so I cannot use
plt.subplot(212)
because I don't know the number that I should provide.
It should look like this:
Right now, I plot all the plots into a folder and put them together with illustrator, but there has to be a better way with Matplotlib. I can provide my code if I was unclear somewhere.
My understanding is that you only know the number of plots at runtime and hence are struggling with the shorthand syntax, e.g.:
plt.subplot(121)
Thankfully, to save you having to do some awkward math to figure out this number programatically, there is another interface which allows you to use the form:
plt.subplot(n_cols, n_rows, plot_num)
So in your case, given you want n plots, you can do:
n_plots = 5 # (or however many you programatically figure out you need)
n_cols = 2
n_rows = (n_plots + 1) // n_cols
for plot_num in range(n_plots):
ax = plt.subplot(n_cols, n_rows, plot_num)
# ... do some plotting
Alternatively, there is also a slightly more pythonic interface which you may wish to be aware of:
fig, subplots = plt.subplots(n_cols, n_rows)
for ax in subplots:
# ... do some plotting
(Notice that this was subplots() not the plain subplot()). Although I must admit, I have never used this latter interface.
HTH