Develop Reference

A smart way of plotting a density map in galactic coordinates with astropy with z axis showing density of sources?

I am using matplotlib.pyplot and astropy to build a plot in galactic coordinates and my goal is to show the density of stars in the sky.
For that, the only data I have is a two-column table with the coordinates of the stars in Right Ascension (RA) and Declination (Dec).
Right now my code is doing the following:
import astropy.coordinates as coord
import matplotlib.pyplot as plt
import astropy.units as u
coordinates = coord.SkyCoord(ra=RA*u.deg, dec=DEC*u.deg)
fig = plt.figure(figsize=(8, 6))
ax = fig.add_subplot(111, projection="aitoff")
ax.plot(coordinates.galactic.l.wrap_at('180d').radian,
coordinates.galactic.b.radian, 'k.', alpha=0.01, ms=1)
ax.grid(True)
So for now I am basically using plt.plot to plot all datapoints (which in the case is half-million datapoints) using a very low alpha and symbol size and the plot looks like this:
However, this isn't the plot I want, as the colour scale quickly saturates.
My question is: Is there a way of making a similar plot but properly reflecting the density of datapoint in the z-axis (color)? For example, I want to be able of controling the color table for a given number-density of sources.
I've seen some answers to similar questions are available.
For example, this question (Plotting a heatmap in galactic coordinates) does a similar thing, but for a specific z-axis described by some data.
I am also aware of this question (How can I make a scatter plot colored by density in matplotlib?) and I tried each solution in this post, but they all failed since I am using a subplot which already has a projection.
Any ideas?

Matplotlib: plot arbitrary vectors on polar axes

There are previous questions about quiver plots on polar axes in matplotlib, however they concern vector fields. I'm interested in drawing arbitrary vectors on polar axes. If there is a genuine duplicate, please link it.
I'm writing some software which concerns a circular world. I'm plotting an agent's trajectory from the centre of a circular arena to the edge. This is visualised by drawing a vector from the centre of the circle to the edge. I'm trying to use matplotlib's quiver plot to plot vectors on a set of polar axes. Here's a minimum working example:
import matplotlib.pyplot as plt
import numpy as np
if __name__ == '__main__':
fig = plt.figure()
ax = fig.add_subplot(111, projection='polar')
# Plot origin (agent's start point)
ax.plot(0, 0, color='black', marker='o', markersize=5)
# Plot agent's path
ax.quiver((0, 0), (0, 1), color='black')
# Example of where (0, 1) should be
ax.plot(0, 1, color='black', marker='o', markersize=5)
# Plot configuration
ax.set_rticks([])
ax.set_rmin(0)
ax.set_rmax(1)
ax.set_thetalim(-np.pi, np.pi)
ax.set_xticks(np.linspace(np.pi, -np.pi, 4, endpoint=False))
ax.grid(False)
ax.set_theta_direction(-1)
ax.set_theta_zero_location("N")
plt.show()
If you run the code, you get this plot
The plot shows the origin plotted correctly, an example point at (0, 1) to show where the vector should end, then the vector itself which appears far too short (though the direction is correct). From the docs, I understand that quiver takes cartesian coordinates (x,y) denoting the start point of the vector and (u,v) denoting the vector's direction. In my previous experience with quiver (u,v) essentially denotes where the vector's tip will be, so in this case we'd expect the vector to be drawn from (0,0) to (0,1) which isn't the case and I don't know why.
In short, I want to be able to draw arbitrary vectors on a set of polar axes and quiver isn't working as I expected. Three questions:
Is my code actually sensible given my goal? I want to draw a unit vector from the origin to the edge of the polar plot.
Am I completely misunderstanding how to use quiver?
How can I draw arbitrary vectors on polar axes in matplotlib? I know about arrow and I'm going to give that a try though initial attempts were unsuccessful.
Short of using a standard plot and just defining my own polar system within it I'm completely stumped.

You did not specify u and v in ax.quiver(x,y,u,v). To make sure the arrow is 1 unit long you will need to set the scale und units as well.
ax.quiver(0,0,0,1, color='black', angles="xy", scale_units='xy', scale=1.)

Python: Matplotlib Surface_plot

I'm trying to Plot a high resolution surface_plot, but I would also really like some nice grid lines on top of it. If i use the gridlines in the same argument
ax.plot_surface(x_itp, y_itp, z_itp, rstride=1, cstride=1, facecolors=facecolors, linewidth=0.1)
I get a LOT of grid lines. If I, on the other hand, set "rstride" and "cstride" to higher values, my sphere will become ugly.
I then tried to smash a
ax.plot_wireframe(x_itp, y_itp, z_itp, rstride=3, cstride=3)
in afterwards, but it just lies on top of the colored sphere.. meaning that I can see the backside of the wireframe and then the surface_plot behind it all.
Have anyone tried this?
Another option was to use "Basemap" which can create a nice grid, but then I will have to adapt my colored surface to that.?!
My plot looks like this:
If I add edges to the map with a higher "rstride" and "cstride" then it looks like this:
code :
norm = plt.Normalize()
facecolors = plt.cm.jet(norm(d_itp))
# surface plot
fig, ax = plt.subplots(1, 1, subplot_kw={'projection':'3d', 'aspect':'equal'})
ax.hold(True)
surf = ax.plot_surface(x_itp, y_itp, z_itp, rstride=4, cstride=4, facecolors=facecolors)
surf.set_edgecolors("black")
I want to show the \theta and \phi angles around the sphere.. maybe with 30 degrees apart.
Cheers!
Morten

It looks like you may need to use basemap. With plot_surface() you can either have high resolution plot or low resolution with good grid on top. But not both. I just made a simple basemap with contour plot. I think you can do easily apply pcolor on it. Just do not draw continent and country boundary. Then, you have a nice sphere which gives more control. After making your plot, you can easily add grid on it.
from mpl_toolkits.basemap import Basemap
import matplotlib.pyplot as plt
import numpy as np
map = Basemap(projection='ortho',lat_0=45,lon_0=-150)
map.drawmapboundary(fill_color='aquamarine')
map.drawmeridians(np.arange(0,360,30)) # grid every 30 deg
map.drawparallels(np.arange(-90,90,30))
nlats = 73; nlons = 145; delta = 2.*np.pi/(nlons-1)
lats = (0.5*np.pi-delta*np.indices((nlats,nlons))[0,:,:])
lons = (delta*np.indices((nlats,nlons))[1,:,:])
wave = 0.6*(np.sin(2.*lats)**6*np.cos(4.*lons))
mean = 0.5*np.cos(2.*lats)*((np.sin(2.*lats))**2 + 2.)
x, y = map(lons*180./np.pi, lats*180./np.pi) # projection from lat, lon to sphere
cs = map.contour(x,y,wave+mean,15,linewidths=1.5) # contour data. You can use pcolor() for your project
plt.title('test1')
plt.show()

Fine-tuning of pcolor() polar plot

I have a 64x360 Matrix of values belonging to radial and azimuthal coordinates. I want to visualize them in two plots: a cartesian and a polar plot.
I visualized the heatmap in cartesian coordinates using imshow():
import numpy as np
import matplotlib.pyplot as plt
P=np.loadtxt('Pdata.csv')
print np.shape(P)
plt.imshow(P)
plt.xlabel('radius')
plt.ylabel('theta')
plt.show()
This gives me the desired plot:
The same plot in polar coordinates was also pretty straigh forward using pcolor():
r=np.arange(0,np.shape(P)[1],1)
t=np.arange(0,np.shape(P)[0],1)
R,T = np.meshgrid(r,t)
fig = plt.figure()
ax = fig.add_subplot(111, polar = True)
ax.pcolor(T,R,P)
plt.show()
However, I am not really satisfied with the result:
The resolution of the plot seems to be pretty limited so that it's not possible to distinguish between angles with higher intensity and lower intensity, as it is in the cartesian plot. The whole solid angle seems to be divided into six or seven "cake wedges" only. Is there an easy and pythonic way to enhance the angular resolution?

Ok, I found out something. It works with:
t = np.radians(np.linspace(0, np.shape(P)[0],np.shape(P)[0]))
r = np.arange(0, np.shape(P)[1], 1)
Just as seen here: Polar contour plot in matplotlib - best (modern) way to do it?

plot circle on unequal axes with pyplot

I would like to plot a circle on an auto-scaled pyplot-generated graphic. When I run
ax.get_aspect()
hoping for a value with which I could manipulate the axes of a ellipse, pyplot returns:
auto
which is less than useful. What methods would you suggest for plotting a circle on a pyplot plot with unequal axes?

This question is more than one year old, but I too just had this question. I needed to add circles to a matplotlib plot and I wanted to be able to specify the circle's location in the plot using data coordinates, and I didn't want the circle radius to change with panning/zooming (or worse the circle turning into an ellipse).
The best and most simple solution that I've found is simply plot a curve with a single point and include a circle marker:
ax.plot(center_x,center_y,'bo',fillstyle='none',markersize=5)
which gives a nice, fixed-size blue circle with no fill!

It really does depend what you want it for.
The problem with defining a circle in data coordinates when aspect ratio is auto, is that you will be able to resize the figure (or its window), and the data scales will stretch nicely. Unfortunately, this would also mean that your circle is no longer a circle, but an ellipse.
There are several ways of addressing this. Firstly, and most simply, you could fix your aspect ratio and then put a circle on the plot in data coordinates:
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = plt.axes()
ax.set_aspect(1)
theta = np.linspace(-np.pi, np.pi, 200)
plt.plot(np.sin(theta), np.cos(theta))
plt.show()
With this, you will be able to zoom and pan around as per usual, but the shape will always be a circle.
If you just want to put a circle on a figure, independent of the data coordinates, such that panning and zooming of an axes did not effect the position and zoom on the circle, then you could do something like:
import matplotlib.patches as mpatches
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = plt.axes()
patch = mpatches.Circle((325, 245), 180, alpha=0.5, transform=None)
fig.artists.append(patch)
plt.show()
This is fairly advanced mpl, but even so, I think it is fairly readable.
HTH,

Building on #user3208430, if you want the circle to always appear at the same place in the axes (regardless of data ranges), you can position it using axes coordinates via transform:
ax.plot(.94, .94, 'ro', fillstyle='full', markersize=5, transform=ax.transAxes)
Where x and y are between [0 and 1]. This example places the marker in the upper right-hand corner of the axes.