The below code block (A) solves my overall problem of being able to re-use plots, but i'm wondering if there is a better way (one that doesn't involve creating a function for each plt.plot, as is shown below.
Code block A:
import maptplotlib.pyplot as plt
#create a function just to call the plt.plot
def first_plot(): plt.plot(x,y)
first_plot()
# Now i can just show the first plot
plt.show()
def second_plot(): plt.plot(x,z)
first_plot() # instead of plt.plot(x,y)
second_plot() # instead of plt.plot(x,z)
# now i can show both plots
plt.show()
if the plots are complicated:
plot.plot(lots of details)
and their are many:
plots = [first,second,third,fourth,...]
I would think this would be advantageous because it avoids code re-use.
However, creating a function just to call plt.plot() indicates to me their might be a better way.what i would like to be doing is something like
first_plot = plt.plot(x,y)
first_plot()
#now i want to show just the first plot
plt.show() # first call
second_plot = plt.plot(x,z)
# now i want to show them together combined
first_plot()
second_plot()
plt.show() # second call
But this doesn't seem to work/e.g the second call to plt.show() wouldn't produce the first plot. (even if you unpack first_plot (which in reality, from code block B, is actual a list):
In [17]: first_plot
Out[17]: [<matplotlib.lines.Line2D at 0x7fd33ac8ff50>]
This cannot be used to produce the plot again in anyway I can see. This might be because of the relationship between plt.show and plt.plot, which i don't fully understand. Plt.plot seems to put the plot in a que, that then plt.show pops out. However,
plt.shows description talks about a blocking and unblocking mode and interactive and non-interactive mode:
show(*args, **kw)
When running in ipython with its pylab mode, display all
figures and return to the ipython prompt.
In non-interactive mode, display all figures and block until
the figures have been closed; in interactive mode it has no
effect unless figures were created prior to a change from
non-interactive to interactive mode (not recommended). In
that case it displays the figures but does not block.
A single experimental keyword argument, *block*, may be
set to True or False to override the blocking behavior
described above.
Which i don't understand. But regardless of how i call plt.show() vs plt.show(False) (blocking?) it doesn't seem to have an impact e.g the output is the same in the context of code block A and B.
So put another way, is there a way to select which plots created to show/overlay at different points in the code?
However awkward, it seems the best way to "re-use" plots as described in my question is to do as I original suggested and put it inside a function.
def some_plot(): return plot(x,y)
and then if you want to re-use the plot simply call the function again:
some_plot()
Related
Even though I set the plotting style in every plotting script manually right at the beginning of the code, directly after importing the modules, the figure-plotting-style to seaborn-whitegrid, the resulting figures comprise the plain matplotlib.pyplot standard white background without grid display, e.g. like this graph:
Therefore, I assume that setting the style has no effect on my scripts, but I can't tell where it goes wrong, or where the style changes back to default. It'd be practical to print out the currently active plotting style at any given point, e.g. during a debugging session.
This is how I set my plotting style:
import .....
import matplotlib.pyplot as plt
import .....
# * Set the matplotlib.pyplot default template (plotting style and background) *
plt.style.use('seaborn-whitegrid')
# Start of script #
...
I even inserted the line plt.style.use('seaborn-whitegrid') in my plotting-submodules, which the main script calls, but it doesn't seem to have an effect on the output.
EDIT on additional trials:
According to what was suggested below by #Bernardo Trindade, I replaced plt.style.use with mpl.style.use:
# * Set the matplotlib.pyplot default template (plotting style and background) *
mpl_plt_default_template = 'seaborn-whitegrid'
import matplotlib as mpl
mpl.style.use(mpl_plt_default_template)
What I also tried is temporary styling, but with no avail either:
# Set plotting style
with plt.style.context('seaborn-whitegrid'):
# * Instantiate figure
fig, axs = plt.subplots(rows, cols, figsize=(width, height))
....
All of the above has not lead to any difference in the output graphs; still showing the plain standard matplotlib background without grid etc.
System specifics:
Lubuntu 20.04 LTS,
python 3.9x,
VS Code
I don't think there's a variable that stores the name of the stylesheet currently used, as all the style information is kept in a large dictionary whose values are simply updated when calling matplotlib.style.use().
Have you tried:
import matplotlib as mpl
mpl.style.use('seaborn-whitegrid')
Finally I found the culprit:
ax.grid(which='both', alpha=1)
Previously, this line never caused an issue, the grids were displayed just fine.
By contrast, now this setting the grid like this renders the grid invisible.
1) CONCLUSION (short answer)
The easiest way to make it work, i.e. safeguard that the grid appears, is setting either the parameter b to True like so:
ax.grid(b=True, which='both', alpha=1)
Alternatively, the kwarg visible can be set to True like so:
ax.grid(which='both', alpha=1, visible=True)
I recommend using visible=True since the naming of the kwarg is more intuitive as to what it actually does, compared to the optional parameter merely called b.
2) ELABORATE ANSWER
The following information is based on the official matplotlib docs:
The allegedly optional parameter b must be set manually to True or False, because otherwise, it'll be set to None in this context,
even though the current docs state:
b : bool or None, optional
Whether to show the grid lines. If any kwargs are supplied, it is assumed you want the grid on and b will be set to True.
If b is None and there are no kwargs, this toggles the visibility of the lines.
NOTE on boolean parameter "b":
I've tried both True and False to find out that even False delivers a proper grid, which is counterintuitive.
On the contrary, when put explicitely to None, which is equivalent to passing nothing, the grid won't appear:
ax.grid(b=None, which='both', alpha=1)
As mentioned before, the grid is invisible in this context also with the following, which was my original code line:
ax.grid(which='both', alpha=1)
Skimming through the **kwargs revealed an extra visibility kwarg called visible:
NOTE on naming of the visibility kwarg in its docs:
Artist.set_visible(self, b)[source]
It is called b, just like the
optional parameter
of ax.grid().
I confirmed that they actually do the same thing.
The grid appears, when putting the visible-kwarg to True:
ax.grid(b=None, which='both', alpha=1, visible=True)
When put to contradictory values, the following error is thrown:
ValueError: 'b' and 'visible' specify inconsistent grid visibilities
This error is caused when employing one of the following two possibilities:
ax.grid(b=False, which='both', alpha=1, visible=True)
ax.grid(b=True, which='both', alpha=1, visible=False)
Eventually, to make the grid appear successfully, set either the parameter b to True like so:
ax.grid(b=True, which='both', alpha=1)
Alternatively, the kwarg visible can be set to True like so:
ax.grid(which='both', alpha=1, visible=True)
Often I need to optimize a complex plot in many iterations, an example would be finding a good position for the legend. Let's say my code looks something like this:
def main()
...
plot(x, y)
def plot(data, legends)
fig, ax = subplots(1,2)
...
ax[0].legend(frameon=False,loc="lower right")
Now I need to run the last line multiple times with different locations and look at the resulting figure. Currently, how I do this would be to set a breakpoint after that last line, start debugging until it pauses at the breakpoint, and then I will iteratively run ax[0].legend() with different arguments followed by fig.savefig(path) in the debug console. The result can be seen in an image viewer. Is there a quicker, "real time" way to do this task? Is there a VS Code shortcut for "Run selected code in Debug Console" maybe? Edit: it is possible to assign this shortcut, its name in the list of key bindings is Evaluate in Debug Console.
I am using matplotlib 1.4.3 with python 3.3. I would like to draw multiple figures with multiples sub-plot in it. I am facing some kind of bug that is really boring me:
When I use fill() and boxplot() methods whithin a figure, results of those functions are hidden if the figure is not the first one created.
This bug seems to be related somehow to polygon display and matplotlib environment state.
When a parse only one single figure, everything is working fine. When I parse multiple figures, the first one is ok. But, in every other subsequent figures, everything is all-right except wiskerbox and polygons that are hidden.
Each plot code is wrapped into a function, which accepts positional arguments, *args and **kwargs. Lets say signature are:
def myplot(t, x, *args, *kwargs):
# [...]
hFig = plt.figure()
# [...]
return hFig
As far as I understand python mechanisms, after the function call is resolved, there must be nothing alive (I do not use global variables) except what matplotlib environment has stored into its global namespace variables.
In every call, I close() my figure, I also have tried hFig.clf() in addition before leaving function, but it does not solve the problem.
Each plot is wrapped into printer (decorator) to add generic functionalities:
def myprint(func):
def inner(*args, **kwargs)
# [...]
hFig = func(*args, **kwargs)
# [...]
return inner
What I have tried so far:
Increased zscore of wiskerbox and polygons, not working;
Execute plot generation in different threads, not working;
Execute plot generation in different processes, working but I have to change my function signature because it can be pickled.
I do not want use dill and pathos, even if I would I cannot.
It looks like it is a matplotlib environment bug, because when I run different processes, this environment is recreated from scratch and it works the way it should. I would like to know if there is a way to reset matplotlib environment state within a python script. If not, what can I do for solving this issue.
Obs.: I am using GridSpecs object and subplot() method to create my figures. The problem was not present when I computed boxes myself and used add_axes() method.
Update: Here you can find a MCVE of my problem. By doing this simple example, I found the line which makes my bug happens (looks like I have old bad Matlab behaviours). It seems that plt.hold(False) alters the way of polygons and boxplot are displayed. And, as I pointed out, it was related to matplotlib global namespace variable. I just misunderstood the way this method works, and in each sub-process, it was reset.
import datetime
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.gridspec as gspec
def bloodyplotit(t_, x_):
hFig = plt.figure()
gs = gspec.GridSpec(1, 4, height_ratios=[1], width_ratios=[15, 2, 3, 1])
gs.update(left=0.10, right=0.90, top=0.90, bottom=0.25, hspace=0.05, wspace=0.05)
plt.hold(True)
hAxe = plt.subplot(gs[0,0])
hAxe.plot(t_, x_)
#plt.hold(False) # <------------------------ This line make the scirpt bug
hAxe = plt.subplot(gs[0,1])
hAxe.hist(x_, orientation='horizontal')
hAxe = plt.subplot(gs[0,3])
hAxe.boxplot(x_)
plt.show()
n = 1000
t = datetime.datetime.utcnow() + np.arange(n)*datetime.timedelta(minutes=1)
x = np.random.randn(1000,1)
for i in range(10):
bloodyplotit(t, x)
Here's an even more minimal script that produces the error:
x = np.random.randn(1000)
fig, ax = plt.subplots(1, 2)
ax[0].hold(True)
ax[0].boxplot(x);
ax[1].hold(False)
ax[1].boxplot(x);
As far as I can tell, this is expected behavior. According to the documentation of plt.hold,
When hold is True, subsequent plot commands will be added to the current axes. When hold is False, the current axes and figure will be cleared on the next plot command.
Boxplot is a compound object: it is created by calling multiple plotting commands. If hold is False, the axes are cleared between each of those commands, so parts of the boxplot don't show up.
Personally, I've never found a reason to toggle the hold state in 10+ years of using matplotlib. My experience is that doing it (especially globally) just causes confusion, and I'd recommend avoiding it.
When plotting with Python ggplot, every single plot command causes a GUI pane to be displayed and suspend execution ("interactive plotting"). But I want to:
avoid/ turn off this GUI and save the plot object some where in runtime (I will be displaying it in some other C# forms control).
find a Python equivalent to dev.off() command in R language which turns off the GUI for plotting.
Example:
print ggplot(data, aes('Age', 'Weight')) + geom_point(colour='steelblue')
When I execute this, it opens up a new GUI (like below) displaying the plot.
You can do the following, which returns a matplotlib figure:
g = ggplot(...) + geom_xxx(...)
fig = g.draw()
ggplots __repr__() method (what is called by print(g) is basically self.draw() then use matplotlibs plt.show() to show the plot...
You can also use ggsave(g) to save the plot somewhere.
Since plotting is triggered by __repr__ method the obvious approach is to avoid situations when it is called. Since you want to use this plot in some other place there is no reason to call print or even executing statements which will be discarded like this:
ggplot(data, aes('Age', 'Weight')) + geom_point(colour='steelblue')
Instead you can simply assign it to the variable
p = ggplot(data, aes('Age', 'Weight')) + geom_point(colour='steelblue')
what is exactly the same thing one would do in R. Using graphic device to redirect output and discarding it doesn't really make sense.
If for some reason that's not enough you switch to non-interactive matplotlib backend:
import matplotlib
matplotlib.use('Agg')
from ggplot import *
ggplot(aes(x='date', y='beef'), data=meat)
<ggplot: (...)>
I've kept a set of references to figures in a dictionary so that I could save them later if desired. I am troubled that the saved figures are blank if invoke a show() command and look at them first. Since the show() command blocks and I am not using a spyder-like interpreter, I have to close the figures before I get to savefig()
figures['myfig_1'] = figure()
...
figures['myfig_n'] = figure()
...
#show() #disabling this makes the problem go away
print "Saving:"
for fig in figures:
figure(figures[fig].number)
savefig(fig)
print "Figure " + str(figures[fig].number) + ": " + fig
The print statement here has given me the indication that the dictionary is still intact, which I think means that I have not lost the figure references (they are still returning meaningful numbers in their .number attribute.)
Another twist I have noticed is that when I have done a similar thing in a class, storing the dictionary as a member and dividing the store and save functions into their own methods, this does not happen. Is there something about the way I am closing the figures or storing the data which is making the figures loose their data?
Generally speaking, in cases like this don't use the interactive matlab-ish state machine interface to matplotlib. It's meant for interactive use.
You're trying to make a figure "active", and creating a new figure instead. It doesn't matter which figure is active, if you just retain the returned figure and/or axis objects and use them directly. (Also, don't use wildcard imports! You will regret it at some later point when you're maintaining your code!)
Just do something like this:
import matplotlib.pyplot as plt
figures = {}
figures['a'] = plt.figure()
ax = figures['a'].add_subplot(111)
ax.plot(range(10), 'ro-')
figures['b'] = plt.figure()
ax = figures['b'].add_subplot(111)
ax.plot(range(10), 'bo-')
plt.show()
for name, fig in figures.iteritems():
fig.savefig('figure-%s.png' % name)
From the documentation, whether or not the drawing elements are destroyed from show() depends on the backend, and the version of matplotlib. Not having the figures destroyed seems to be available with version 1.1.0. To figure out which backend is in use, use the get_backend() function. In my case, I was using the Qt4Agg backend. By invoking the TkAgg backend, with the call matplotlib.use('TkAgg') the figures were not destroyed before the save. Now to find out how to change the behavior of the Qt4Agg...