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.
Related
I have a parallel coordinates plot with lots of data points so I'm trying to use a continuous colour bar to represent that, which I think I have worked out. However, I haven't been able to remove the default key that is put in when creating the plot, which is very long and hinders readability. Is there a way to remove this table to make the graph much easier to read?
This is the code I'm currently using to generate the parallel coordinates plot:
parallel_coordinates(data[[' male_le','
female_le','diet','activity','obese_perc','median_income']],'median_income',colormap = 'rainbow',
alpha = 0.5)
fig, ax = plt.subplots(figsize=(6, 1))
fig.subplots_adjust(bottom=0.5)
cmap = mpl.cm.rainbow
bounds = [0.00,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0]
norm = mpl.colors.BoundaryNorm(bounds, cmap.N,)
plt.colorbar(mpl.cm.ScalarMappable(norm = norm, cmap=cmap),cax = ax, orientation = 'horizontal',
label = 'normalised median income', alpha = 0.5)
plt.show()
Current Output:
I want my legend to be represented as a color bar, like this:
Any help would be greatly appreciated. Thanks.
You can use ax.legend_.remove() to remove the legend.
The cax parameter of plt.colorbar indicates the subplot where to put the colorbar. If you leave it out, matplotlib will create a new subplot, "stealing" space from the current subplot (subplots are often referenced to by ax in matplotlib). So, here leaving out cax (adding ax=ax isn't necessary, as here ax is the current subplot) will create the desired colorbar.
The code below uses seaborn's penguin dataset to create a standalone example.
import matplotlib.pyplot as plt
import matplotlib as mpl
import seaborn as sns
import numpy as np
from pandas.plotting import parallel_coordinates
penguins = sns.load_dataset('penguins')
fig, ax = plt.subplots(figsize=(10, 4))
cmap = plt.get_cmap('rainbow')
bounds = np.arange(penguins['body_mass_g'].min(), penguins['body_mass_g'].max() + 200, 200)
norm = mpl.colors.BoundaryNorm(bounds, 256)
penguins = penguins.dropna(subset=['body_mass_g'])
parallel_coordinates(penguins[['bill_length_mm', 'bill_depth_mm', 'flipper_length_mm', 'body_mass_g']],
'body_mass_g', colormap=cmap, alpha=0.5, ax=ax)
ax.legend_.remove()
plt.colorbar(mpl.cm.ScalarMappable(norm=norm, cmap=cmap),
ax=ax, orientation='horizontal', label='body mass', alpha=0.5)
plt.show()
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()
I am playing around with volumetric data and I am trying to project a "cosmic web" like image.
I pretty much create a file path and open the data with a module that opens hdf5 files. The x and y values are denoted by indexing from a the file gas_pos and the histogram is weighted by different properties, gas_density in this case:
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.colors import LinearSegmentedColormap
from matplotlib.ticker import LogFormatter
cmap = LinearSegmentedColormap.from_list('mycmap', ['black', 'steelblue', 'mediumturquoise', 'darkslateblue'])
fig = plt.figure()
ax = fig.add_subplot(111)
H = ax.hist2d(gas_pos[:,0]/0.7, gas_pos[:,1]/0.7, bins=500, cmap=cmap, norm=matplotlib.colors.LogNorm(), weights=gas_density);
cb = fig.colorbar(H[3], ax=ax, shrink=0.8, pad=0.01, orientation="horizontal", label=r'$ \rho\ [M_{\odot}\ \mathrm{kpc}^{-3}]$')
ax.tick_params(axis=u'both', which=u'both',length=0)
ax.get_xaxis().set_visible(False)
ax.get_yaxis().set_visible(False)
plt.show()
giving me this:
which is nice, but I want to up the quality and remove the grainyness of it. When I try imshow interpolation:
cmap = LinearSegmentedColormap.from_list('mycmap', ['black', 'steelblue', 'mediumturquoise', 'darkslateblue'])
fig = plt.figure()
ax = fig.add_subplot(111)
H = ax.hist2d(gas_pos[:,0]/0.7, gas_pos[:,1]/0.7, bins=500, cmap=cmap, norm=matplotlib.colors.LogNorm(), weights=gas_density);
ax.tick_params(axis=u'both', which=u'both',length=0)
ax.get_xaxis().set_visible(False)
ax.get_yaxis().set_visible(False)
im = ax.imshow(H[0], cmap=cmap, interpolation='sinc', norm=matplotlib.colors.LogNorm())
cb = fig.colorbar(H[3], ax=ax, shrink=0.8, pad=0.01, orientation="horizontal", label=r'$ \rho\ [M_{\odot}\ \mathrm{kpc}^{-3}]$')
plt.show()
Am I using this incorrectly? or is there something better I can use to modify the pixelation?
If anyone is wanting to play with my data, I will upload the data later on today!
Using interpolation='sinc' is indeed a good method to smoothen a plot. Others would e.g. be "gaussian", "bicubic" or "spline16".
The problem you observe is that the imshow plot is plotted on top of the hist2d plot and thus takes its axes limits. Those limits seem to be smaller than the number of points in the imshow plot and therefore you only see part of the total data.
The solution is either not to plot the hist2d plot at all or at least to plot it into another subplot or figure.
Pursuing the first idea, you would calculate your histogram without plotting it, using numpy.histogram2d
H, xedges, yedges = np.histogram2d(gas_pos[:,0]/0.7, gas_pos[:,1]/0.7,
bins=500, weights=gas_density)
im = ax.imshow(H.T, cmap=cmap, interpolation='sinc', norm=matplotlib.colors.LogNorm())
I would also recommend reading the numpy.histogram2d documentation, which includes an example of plotting the histogram output in matplotlib.
You'll probably want to set interpolation='None' in the call to imshow, instead of interpolation='sinc'
I have multiple plots that have the same x-axis. I would like to stack them in a report and have everything line up. However, matplotlib seems to resize them slightly based on the y tick label length.
Is it possible to force the plot area and location to remain the same across plots, relative to the pdf canvas to which I save it?
import numpy as np
import matplotlib.pyplot as plt
xs=np.arange(0.,2.,0.00001)
ys1=np.sin(xs*10.) #makes the long yticklabels
ys2=10.*np.sin(xs*10.)+10. #makes the short yticklabels
fig=plt.figure() #this plot ends up shifted right on the canvas
plt.plot(xs,ys1,linewidth=2.0)
plt.xlabel('x')
plt.ylabel('y')
fig=plt.figure() #this plot ends up further left on the canvas
plt.plot(xs,ys2,linewidth=2.0)
plt.xlabel('x')
plt.ylabel('y')
Your problem is a little unclear, however plotting them as subplots in the same figure should gaurantee that the axes and figure size of the two subplots will be alligned with each other
import numpy as np
import matplotlib.pyplot as plt
xs=np.arange(0.,2.,0.00001)
ys1=np.sin(xs*10.) #makes the long yticklabels
ys2=10.*np.sin(xs*10.)+10. #makes the short yticklabels
fig, (ax1, ax2) = plt.subplots(2, 1)
ax1.plot(xs,ys1,linewidth=2.0)
ax1.set_xlabel('x')
ax1.set_ylabel('y')
ax2.plot(xs,ys2,linewidth=2.0)
ax2.set_xlabel('x')
ax2.set_ylabel('y')
plt.subplots_adjust(hspace=0.3) # adjust spacing between plots
plt.show()
This produces the following figure:
I had the same problem. The following works for me.
Force the same figure width for all your plots around all your python scripts, for example:
fig1 = plt.figure(figsize=(12,6))
...
fig2 = plt.figure(figsize=(12,4))
And do not use (very important!):
fig.tight_layout()
Save the figure
plt.savefig('figure.png')
Plot areas should now be the same.
using subplots with the same x-axis should do the trick.
use sharex=True when you create the subplots. The benefit of sharex is that zooming or panning on 1 subplot will also auto-update on all subplots with shared axes.
import numpy as np
import matplotlib.pyplot as plt
xs = np.arange(0., 2., 0.00001)
ys1 = np.sin(xs * 10.) # makes the long yticklabels
ys2 = 10. * np.sin(xs * 10.) + 10. # makes the short yticklabels
fig, (ax1, ax2) = plt.subplots(2, sharex=True)
ax1.plot(xs, ys1, linewidth=2.0)
ax1.xlabel('x')
ax1.ylabel('y')
ax2.plot(xs, ys2, linewidth=2.0)
ax2.xlabel('x')
ax2.ylabel('y')
plt.show()
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