ellipse not showing up - python matplotib - python

I am trying to plot an ellipse.
ax = plt.subplot(111)
ellipse = Ellipse(mean1L, ellipse_x, ellipse_y, angle=theta)
ax.add_artist(ellipse)
plt.show()
Every argument seems fine, but it isn't showing up.
What am I doing wrong?

The ellipse is outside of the axis limits.
Instead of ax.add_artist(ellipse) you would rather use
ax.add_patch(ellipse)
to be able to adjust the axis limits easily to the added patch.
This will allow to later call ax.autoscale_view() to automatically adjust the axis limits.
import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse
ax = plt.subplot(111)
ellipse = Ellipse((2,2), 1,1.5 , angle=60)
ax.add_patch(ellipse)
ax.autoscale_view()
plt.show()

Related

Scatter Plot Points overlapping axis

For some reason when I use a zorder with my scatter plot the edges of the points overlap the axis. I tried some of the solutions from [here] (matplotlib axis tick labels covered by scatterplot (using spines)) but they didn't work for me. Is there a way from preventing this from happening?
I understand I could also add an ax.axvline() at my boundaries but that would be an annoying workaround for lots of plots.
xval = np.array([0,0,0,3,3,3,0,2,3,0])
yval = np.array([0,2,3,5,1,0,1,0,4,5])
zval = yval**2-4
fig = plt.figure(figsize=(6,6))
ax = plt.subplot(111)
ax.scatter(xval,yval,cmap=plt.cm.rainbow,c=zval,s=550,zorder=20)
ax.set_ylim(0,5)
ax.set_xlim(0,3)
#These don't work
ax.tick_params(labelcolor='k', zorder=100)
ax.tick_params(direction='out', length=4, color='k', zorder=100)
#This will work but I don't want to have to do this for the plot edges every time
ax.axvline(0,c='k',zorder=100)
plt.show()
For me the solution you linked to works; that is, setting the z-order of the scatter plot to a negative number. E.g.
xval = np.array([0,0,0,3,3,3,0,2,3,0])
yval = np.array([0,2,3,5,1,0,1,0,4,5])
zval = yval**2-4
fig = plt.figure(figsize=(6,6))
ax = plt.subplot(111)
ax.scatter(xval,yval,cmap=plt.cm.rainbow,c=zval,s=550,zorder=-1)
ax.set_ylim(0,5)
ax.set_xlim(0,3)
plt.show()
]1
You can fix the overlap using the following code with a large number for the zorder. This will work on both the x- and y-axis.
for k,spine in ax.spines.items():
spine.set_zorder(1000)
This works for me
import numpy as np
import matplotlib.pyplot as plt
xval = np.array([0,0,0,3,3,3,0,2,3,0])
yval = np.array([0,2,3,5,1,0,1,0,4,5])
zval = yval**2-4
fig = plt.figure(figsize=(6,6))
ax = plt.subplot(111)
ax.scatter(xval,yval,cmap=plt.cm.rainbow,c=zval,s=550,zorder=20)
ax.set_ylim(-1,6)
ax.set_xlim(-1,4)
#These don't work
ax.tick_params(labelcolor='k', zorder=100)
ax.tick_params(direction='out', length=4, color='k', zorder=100)
#This will work but I don't want to have to do this for the plot edges every time
ax.axvline(0,c='k',zorder=100)
plt.show()
Your circle sizes are big enough that they go beyond the axis scope. So we simply change the ylim and xlim
Changed
ax.set_ylim(0,5)
ax.set_xlim(0,3)
to
ax.set_ylim(-1,6)
ax.set_xlim(-1,4)
Also, zorder doesn't play a role in pushing the points to edges.

Matplotlib scatter plot - Remove white padding

I'm working with matplotlib to plot a variable in latitude longitude coordinates. The problem is that this image cannot include axes or borders. I have been able to remove axis, but the white padding around my image has to be completely removed (see example images from code below here: http://imgur.com/a/W0vy9) .
I have tried several methods from Google searches, including these StackOverflow methodologies:
Remove padding from matplotlib plotting
How to remove padding/border in a matplotlib subplot (SOLVED)
Matplotlib plots: removing axis, legends and white spaces
but nothing has worked in removing the white space. If you have any advice (even if it is to ditch matplotlib and to try another plotting library instead) I would appreciate it!
Here is a basic form of the code I'm using that shows this behavior:
import numpy as np
import matplotlib
from mpl_toolkits.basemap import Basemap
from scipy import stats
lat = np.random.randint(-60.5, high=60.5, size=257087)
lon = np.random.randint(-179.95, high=180, size=257087)
maxnsz = np.random.randint(12, 60, size=257087)
percRange = np.arange(100,40,-1)
percStr=percRange.astype(str)
val_percentile=np.percentile(maxnsz, percRange, interpolation='nearest')
#Rank all values
all_percentiles=stats.rankdata(maxnsz)/len(maxnsz)
#Figure setup
fig = matplotlib.pyplot.figure(frameon=False, dpi=600)
#Basemap code can go here
x=lon
y=lat
cmap = matplotlib.cm.get_cmap('cool')
h=np.where(all_percentiles >= 0.999)
hl=np.where((all_percentiles < 0.999) & (all_percentiles > 0.90))
mh=np.where((all_percentiles > 0.75) & (all_percentiles < 0.90))
ml=np.where((all_percentiles >= 0.4) & (all_percentiles < 0.75))
l=np.where(all_percentiles < 0.4)
all_percentiles[h]=0
all_percentiles[hl]=0.25
all_percentiles[mh]=0.5
all_percentiles[ml]=0.75
all_percentiles[l]=1
rgba_low=cmap(1)
rgba_ml=cmap(0.75)
rgba_mh=cmap(0.51)
rgba_hl=cmap(0.25)
rgba_high=cmap(0)
matplotlib.pyplot.axis('off')
matplotlib.pyplot.scatter(x[ml],y[ml], c=rgba_ml, s=3, marker=',',edgecolor='none', alpha=0.4)
matplotlib.pyplot.scatter(x[mh],y[mh], c=rgba_mh, s=3, marker='o', edgecolor='none', alpha=0.5)
matplotlib.pyplot.scatter(x[hl],y[hl], c=rgba_hl, s=4, marker='*',edgecolor='none', alpha=0.6)
matplotlib.pyplot.scatter(x[h],y[h], c=rgba_high, s=5, marker='^', edgecolor='none',alpha=0.75)
fig.savefig('/home/usr/code/python/testfig.jpg', bbox_inches=0, nbins=0, transparent="True", pad_inches=0.0)
fig.canvas.draw()
The problem is that all the solutions given at Matplotlib plots: removing axis, legends and white spaces are actually meant to work with imshow.
So, the following clearly works
import matplotlib.pyplot as plt
fig = plt.figure()
ax=fig.add_axes([0,0,1,1])
ax.set_axis_off()
im = ax.imshow([[2,3,4,1], [2,4,4,2]], origin="lower", extent=[1,4,2,8])
ax.plot([1,2,3,4], [2,3,4,8], lw=5)
ax.set_aspect('auto')
plt.show()
and produces
But here, you are using scatter. Adding a scatter plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax=fig.add_axes([0,0,1,1])
ax.set_axis_off()
im = ax.imshow([[2,3,4,1], [2,4,4,2]], origin="lower", extent=[1,4,2,8])
ax.plot([1,2,3,4], [2,3,4,8], lw=5)
ax.scatter([2,3,4,1], [2,3,4,8], c="r", s=2500)
ax.set_aspect('auto')
plt.show()
produces
Scatter has the particularity that matplotlib tries to make all points visible by default, which means that the axes limits are set such that all scatter points are visible as a whole.
To overcome this, we need to specifically set the axes limits:
import matplotlib.pyplot as plt
fig = plt.figure()
ax=fig.add_axes([0,0,1,1])
ax.set_axis_off()
im = ax.imshow([[2,3,4,1], [2,4,4,2]], origin="lower", extent=[1,4,2,8])
ax.plot([1,2,3,4], [2,3,4,8], lw=5)
ax.scatter([2,3,4,1], [2,3,4,8], c="r", s=2500)
ax.set_xlim([1,4])
ax.set_ylim([2,8])
ax.set_aspect('auto')
plt.show()
such that we will get the desired behaviour.

scatter plot with single pixel marker in matplotlib

I am trying to plot a large dataset with a scatter plot.
I want to use matplotlib to plot it with single pixel marker.
It seems to have been solved.
https://github.com/matplotlib/matplotlib/pull/695
But I cannot find a mention of how to get a single pixel marker.
My simplified dataset (data.csv)
Length,Time
78154393,139.324091
84016477,229.159305
84626159,219.727537
102021548,225.222662
106399706,221.022827
107945741,206.760239
109741689,200.153263
126270147,220.102802
207813132,181.67058
610704756,50.59529
623110004,50.533158
653383018,52.993885
659376270,53.536834
680682368,55.97628
717978082,59.043843
My code is below.
import pandas as pd
import os
import numpy
import matplotlib.pyplot as plt
inputfile='data.csv'
iplevel = pd.read_csv(inputfile)
base = os.path.splitext(inputfile)[0]
fig = plt.figure()
plt.yscale('log')
#plt.xscale('log')
plt.title(' My plot: '+base)
plt.xlabel('x')
plt.ylabel('y')
plt.scatter(iplevel['Time'], iplevel['Length'],color='black',marker=',',lw=0,s=1)
fig.tight_layout()
fig.savefig(base+'_plot.png', dpi=fig.dpi)
You can see below that the points are not single pixel.
Any help is appreciated
The problem
I fear that the bugfix discussed at matplotlib git repository that you're citing is only valid for plt.plot() and not for plt.scatter()
import matplotlib.pyplot as plt
fig = plt.figure(figsize=(4,2))
ax = fig.add_subplot(121)
ax2 = fig.add_subplot(122, sharex=ax, sharey=ax)
ax.plot([1, 2],[0.4,0.4],color='black',marker=',',lw=0, linestyle="")
ax.set_title("ax.plot")
ax2.scatter([1,2],[0.4,0.4],color='black',marker=',',lw=0, s=1)
ax2.set_title("ax.scatter")
ax.set_xlim(0,8)
ax.set_ylim(0,1)
fig.tight_layout()
print fig.dpi #prints 80 in my case
fig.savefig('plot.png', dpi=fig.dpi)
The solution: Setting the markersize
The solution is to use a usual "o" or "s" marker, but set the markersize to be exactly one pixel. Since the markersize is given in points, one would need to use the figure dpi to calculate the size of one pixel in points. This is 72./fig.dpi.
For aplot`, the markersize is directly
ax.plot(..., marker="o", ms=72./fig.dpi)
For a scatter the markersize is given through the s argument, which is in square points,
ax.scatter(..., marker='o', s=(72./fig.dpi)**2)
Complete example:
import matplotlib.pyplot as plt
fig = plt.figure(figsize=(4,2))
ax = fig.add_subplot(121)
ax2 = fig.add_subplot(122, sharex=ax, sharey=ax)
ax.plot([1, 2],[0.4,0.4], marker='o',ms=72./fig.dpi, mew=0,
color='black', linestyle="", lw=0)
ax.set_title("ax.plot")
ax2.scatter([1,2],[0.4,0.4],color='black', marker='o', lw=0, s=(72./fig.dpi)**2)
ax2.set_title("ax.scatter")
ax.set_xlim(0,8)
ax.set_ylim(0,1)
fig.tight_layout()
fig.savefig('plot.png', dpi=fig.dpi)
For anyone still trying to figure this out, the solution I found was to specify the s argument in plt.scatter.
The s argument refers to the area of the point you are plotting.
It doesn't seem to be quite perfect, since s=1 seems to cover about 4 pixels of my screen, but this definitely makes them smaller than anything else I've been able to find.
https://matplotlib.org/devdocs/api/_as_gen/matplotlib.pyplot.scatter.html
s : scalar or array_like, shape (n, ), optional
size in points^2. Default is rcParams['lines.markersize'] ** 2.
Set the plt.scatter() parameter to linewidths=0 and figure out the right value for the parameter s.
Source: https://stackoverflow.com/a/45803960/4063622

How to avoid dataseries to fall under legend in matplotlib plots?

Is there any automatic way of manipulating legend in matplotlib to avoid overlapping data points and legend? I have lots of data points and fixed y axis range, can I instruct matplotlib move the legend to left or down if it's on datapoints. Thanks
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(1, 100)
y = -1/x
plt.plot(x,y, label='x and y', linewidth=30)
plt.legend()
plt.show()
You can manually position the legend to the position that you want it, the documentation can be found here.. You can also move it off the plotting area of the graph to avoid any chance of an overlap as shown in the code below:
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(1, 100)
y = -1/x
plt.plot(x,y, label='x and y', linewidth=30)
#adjust the plot to allow the legend to fit nicely
plt.subplots_adjust(left=0.1,right = 0.75)
plt.legend(bbox_to_anchor=(1.01, 0.5), loc=2) # move the legend
plt.show()
The resulting image is shown below:
Note: If you plan on moving the legend outside of the plotting area then you may have to adjust your plots as I have done in the code above.

matplotlib: adding second axes() with transparent background?

Define data
x = np.linspace(0,2*np.pi,100)
y = 2*np.sin(x)
Plot
fig = plt.figure()
ax = plt.axes()
fig.add_subplot(ax)
ax.plot(x,y)
Add second axis
newax = plt.axes(axisbg='none')
Gives me ValueError: Unknown element o, even though it does the same thing as what I am about to describe. I can also see that this works (no error) to do the same thing:
newax = plt.axes()
fig.add_subplot(newax)
newax.set_axis_bgcolor('none')
However, it turns the background color of the original figure "gray" (or whatever the figure background is)? I don't understand, as I thought this would make newax transparent except for the axes and box around the figure. Even if I switch the order, same thing:
plt.close('all')
fig = plt.figure()
newax = plt.axes()
fig.add_subplot(newax)
newax.set_axis_bgcolor('none')
ax = plt.axes()
fig.add_subplot(ax)
ax.plot(x,y)
This is surprising because I thought the background of one would be overlaid on the other, but in either case it is the newax background that appears to be visible (or at least this is the color I see).
What is going on here?
You're not actually adding a new axes.
Matplotlib is detecting that there's already a plot in that position and returning it instead of a new axes object.
(Check it for yourself. ax and newax will be the same object.)
There's probably not a reason why you'd want to, but here's how you'd do it.
(Also, don't call newax = plt.axes() and then call fig.add_subplot(newax) You're doing the same thing twice.)
Edit: With newer (>=1.2, I think?) versions of matplotlib, you can accomplish the same thing as the example below by using the label kwarg to fig.add_subplot. E.g. newax = fig.add_subplot(111, label='some unique string')
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(1,1,1)
# If you just call `plt.axes()` or equivalently `fig.add_subplot()` matplotlib
# will just return `ax` again. It _won't_ create a new axis unless we
# call fig.add_axes() or reset fig._seen
newax = fig.add_axes(ax.get_position(), frameon=False)
ax.plot(range(10), 'r-')
newax.plot(range(50), 'g-')
newax.axis('equal')
plt.show()
Of course, this looks awful, but it's what you're asking for...
I'm guessing from your earlier questions that you just want to add a second x-axis? If so, this is a completely different thing.
If you want the y-axes linked, then do something like this (somewhat verbose...):
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
newax = ax.twiny()
# Make some room at the bottom
fig.subplots_adjust(bottom=0.20)
# I'm guessing you want them both on the bottom...
newax.set_frame_on(True)
newax.patch.set_visible(False)
newax.xaxis.set_ticks_position('bottom')
newax.xaxis.set_label_position('bottom')
newax.spines['bottom'].set_position(('outward', 40))
ax.plot(range(10), 'r-')
newax.plot(range(21), 'g-')
ax.set_xlabel('Red Thing')
newax.set_xlabel('Green Thing')
plt.show()
If you want to have a hidden, unlinked y-axis, and an entirely new x-axis, then you'd do something like this:
import matplotlib.pyplot as plt
import numpy as np
fig, ax = plt.subplots()
fig.subplots_adjust(bottom=0.2)
newax = fig.add_axes(ax.get_position())
newax.patch.set_visible(False)
newax.yaxis.set_visible(False)
for spinename, spine in newax.spines.iteritems():
if spinename != 'bottom':
spine.set_visible(False)
newax.spines['bottom'].set_position(('outward', 25))
ax.plot(range(10), 'r-')
x = np.linspace(0, 6*np.pi)
newax.plot(x, 0.001 * np.cos(x), 'g-')
plt.show()
Note that the y-axis values for anything plotted on newax are never shown.
If you wanted, you could even take this one step further, and have independent x and y axes (I'm not quite sure what the point of it would be, but it looks neat...):
import matplotlib.pyplot as plt
import numpy as np
fig, ax = plt.subplots()
fig.subplots_adjust(bottom=0.2, right=0.85)
newax = fig.add_axes(ax.get_position())
newax.patch.set_visible(False)
newax.yaxis.set_label_position('right')
newax.yaxis.set_ticks_position('right')
newax.spines['bottom'].set_position(('outward', 35))
ax.plot(range(10), 'r-')
ax.set_xlabel('Red X-axis', color='red')
ax.set_ylabel('Red Y-axis', color='red')
x = np.linspace(0, 6*np.pi)
newax.plot(x, 0.001 * np.cos(x), 'g-')
newax.set_xlabel('Green X-axis', color='green')
newax.set_ylabel('Green Y-axis', color='green')
plt.show()
You can also just add an extra spine at the bottom of the plot. Sometimes this is easier, especially if you don't want ticks or numerical things along it. Not to plug one of my own answers too much, but there's an example of that here: How do I plot multiple X or Y axes in matplotlib?
As one last thing, be sure to look at the parasite axes examples if you want to have the different x and y axes linked through a specific transformation.

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