Develop Reference

Matplotlib using Wedge() in polar plots

TL/DR: How to use Wedge() in polar coordinates?
I'm generating a 2D histogram plot in polar coordinates (r, theta). At various values of r there can be different numbers of theta values (to preserve equal area sized bins). To draw the color coded bins I'm currently using pcolormesh() calls for each radial ring. This works ok, but near the center of the plot where there may be only 3 bins (each 120 degrees "wide" in theta space), pcolormesh() draws triangles that don't "sweep" out full arc (just connecting the two outer arc points with a straight line).
I've found a workaround using ax.bar() call, one for each radial ring and passing in arrays of theta values (each bin rendering as an individual bar). But when doing 90 rings with 3 to 360 theta bins in each, it's incredibly slow (minutes).
I tried using Wedge() patches, but can't get them to render correctly in the polar projection. Here is sample code showing both approaches:
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.patches import Wedge
from matplotlib.collections import PatchCollection
# Theta coordinates in degrees
theta1=45
theta2=80
# Radius coordinates
r1 = 0.4
r2 = 0.5
# Plot using bar()
fig, ax = plt.subplots(figsize=[6,6], subplot_kw={'projection': 'polar'})
theta_mid = np.deg2rad((theta1 + theta2)/2)
theta_width = np.deg2rad(theta2 - theta1)
height = r2 - r1
ax.bar(x=theta_mid, height = height, width=theta_width, bottom=r1)
ax.set_rlim(0, 1)
plt.savefig('bar.png')
# Plot using Wedge()
fig, ax = plt.subplots(figsize=[6,6], subplot_kw={'projection': 'polar'})
patches = []
patches.append( Wedge(center=(0, 0), r = r1, theta1=theta1, theta2=theta2, width = r2-r1, color='blue'))
p = PatchCollection(patches)
ax.add_collection(p)
ax.set_rlim(0, 1)
plt.savefig('wedge.png')
The outputs of each are:
Bar
Wedge
I've tried using radians for the wedge (because polar plots usually want their angle values in radians). That didn't help.
Am I missing something in how I'm using the Wedge? If I add thousands of Wedges to my Patch collection should I have any expectation it will be faster than bar()?

Thinking this was an actual bug, I opened this issue https://github.com/matplotlib/matplotlib/issues/22717 on matplotlib where one of the maintainers nicely pointed out that I should be using Rectangle() instead of Wedge().
The solution they provided is
from matplotlib.patches import Rectangle
fig, ax = plt.subplots(figsize=[6,6], subplot_kw={'projection': 'polar'})
p = PatchCollection([Rectangle((np.deg2rad(theta1), r1), theta_width, height, color='blue')])
ax.add_collection(p)
ax.set_rlim(0, 1)
plt.savefig('wedge.png')

Matplotlib line width based on axis, not on points

If you set a line width in Matplotlib, you have to give the line width in points. In my case, I have two circles, both with radius R and I want to connect them with a line. I want this line to be 2*R wide in order to get a rod-shape. But when I say myLines[i].set_linewidth(2*R) this makes the lines always a specific thickness, regardless of how much I have zoomed in.
Is there a way to make lines a specific thickness not based on the number of pixels or points, but scaling with the axis? How can I make my line have the same width as the diameter of my circles?
I hope I explained myself well enough and I am looking forward to an answer.

Line in Data units
In order to draw a line with the linewidth in data units, you may want to have a look at this answer.
It uses a class data_linewidth_plot which closely resembles the plt.plot() command's signature.
l = data_linewidth_plot( x, y, ax=ax, label='some line', linewidth = 1, alpha = 0.4)
The linewidth argument is interpreted in (y-)data units.
Using this solution there is not even any need for drawing circles, since one may simply use the solid_capstyle="round" argument.
R=0.5
l = data_linewidth_plot( [0,3], [0.7,1.4], ax=ax, solid_capstyle="round",
linewidth = 2*R, alpha = 0.4)
Rod shape
A rod is much more easily produced using a rectange and two circles.

As you already figured out, linewidths are specified in axis space, not data space. To draw a line in data space, draw a rectangle instead:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Rectangle, Circle
r = 5 # rod radius
x1, y1 = (0,0) # left end of rod
x2, y2 = (10,0) # right end of rod
# create 2 circles and a joining rectangle
c1 = Circle((x1, y1), r, color='r')
c2 = Circle((x2, y2), r)
rect = Rectangle((x1, y1-r), width=x2-x1, height=2*r)
# plot artists
fig, ax = plt.subplots(1,1)
for artist in [c2, rect, c1]:
ax.add_artist(artist)
# need to set axis limits manually
ax.set_xlim(x1-r-1, x2+r+1)
ax.set_ylim(y1-r-1, y2+r+1)
# set aspect so circle don't become oval
ax.set_aspect('equal')
plt.show()

How to draw "two directions widths line" in matplotlib

How to use matplotlib or pyqtgraph draw plot like this:
Line AB is a two-directions street, green part represents the direction from point A to point B, red part represents B to A, width of each part represents the traffic volume. Widths are measured in point, will not changed at different zoom levels or dpi settings.
This is only an example, in fact I have hunderds of streets. This kind of plot is very common in many traffic softwares. I tried to use matplotlib's patheffect but result is frustrated:
from matplotlib import pyplot as plt
import matplotlib.patheffects as path_effects
x=[0,1,2,3]
y=[1,0,0,-1]
ab_width=20
ba_width=30
fig, axes= plt.subplots(1,1)
center_line, = axes.plot(x,y,color='k',linewidth=2)
center_line.set_path_effects(
[path_effects.SimpleLineShadow(offset=(0, -ab_width/2),shadow_color='g', alpha=1, linewidth=ab_width),
path_effects.SimpleLineShadow(offset=(0, ba_width/2), shadow_color='r', alpha=1, linewidth=ba_width),
path_effects.SimpleLineShadow(offset=(0, -ab_width), shadow_color='k', alpha=1, linewidth=2),
path_effects.SimpleLineShadow(offset=(0, ba_width), shadow_color='k', alpha=1, linewidth=2),
path_effects.Normal()])
axes.set_xlim(-1,4)
axes.set_ylim(-1.5,1.5)
One idea came to me is to take each part of the line as a standalone line, and recalculate it's position when changing zoom level, but it's too complicated and slow.
If there any easy way to use matplotlib or pyqtgraph draw what I want? Any suggestion will be appreciated!

If you can have each independent line, this can be done easily with the fill_between function.
from matplotlib import pyplot as plt
import numpy as np
x=np.array([0,1,2,3])
y=np.array([1,0,0,-1])
y1width=-1
y2width=3
y1 = y + y1width
y2 = y + y2width
fig = plt.figure()
ax = fig.add_subplot(111)
plt.plot(x,y, 'k', x,y1, 'k',x,y2, 'k',linewidth=2)
ax.fill_between(x, y1, y, color='g')
ax.fill_between(x, y2, y, color='r')
plt.xlim(-1,4)
plt.ylim(-3,6)
plt.show()
Here I considered the center line as the reference (thus the negative y1width), but could be done differently. The result is then:
If the lines are 'complicated', eventually intersecting at some point, then the keyword argument interpolate=True must be used to fill the crossover regions properly. Another interesting argument probably useful for your use case is where, to condition the region, for instance, where=y1 < 0. For more information you can check out the documentation.

One way of solving your issue is using filled polygons, some linear algebra and some calculus. The main idea is to draw a polygon along your x and y coordinates and along shifted coordinates to close and fill the polygon.
These are my results:
And here is the code:
from __future__ import division
import numpy
from matplotlib import pyplot, patches
def road(x, y, w, scale=0.005, **kwargs):
# Makes sure input coordinates are arrays.
x, y = numpy.asarray(x, dtype=float), numpy.asarray(y, dtype=float)
# Calculate derivative.
dx = x[2:] - x[:-2]
dy = y[2:] - y[:-2]
dy_dx = numpy.concatenate([
[(y[1] - y[0]) / (x[1] - x[0])],
dy / dx,
[(y[-1] - y[-2]) / (x[-1] - x[-2])]
])
# Offsets the input coordinates according to the local derivative.
offset = -dy_dx + 1j
offset = w * scale * offset / abs(offset)
y_offset = y + w * scale
#
AB = zip(
numpy.concatenate([x + offset.real, x[::-1]]),
numpy.concatenate([y + offset.imag, y[::-1]]),
)
p = patches.Polygon(AB, **kwargs)
# Returns polygon.
return p
if __name__ == '__main__':
# Some plot initializations
pyplot.close('all')
pyplot.ion()
# This is the list of coordinates of each point
x = [0, 1, 2, 3, 4]
y = [1, 0, 0, -1, 0]
# Creates figure and axes.
fig, ax = pyplot.subplots(1,1)
ax.axis('equal')
center_line, = ax.plot(x, y, color='k', linewidth=2)
AB = road(x, y, 20, color='g')
BA = road(x, y, -30, color='r')
ax.add_patch(AB)
ax.add_patch(BA)
The first step in calculating how to offset each data point is by calculating the discrete derivative dy / dx. I like to use complex notation to handle vectors in Python, i.e. A = 1 - 1j. This makes life easier for some mathematical operations.
The next step is to remember that the derivative gives the tangent to the curve and from linear algebra that the normal to the tangent is n=-dy_dx + 1j, using complex notation.
The final step in determining the offset coordinates is to ensure that the normal vector has unity size n_norm = n / abs(n) and multiply by the desired width of the polygon.
Now that we have all the coordinates for the points in the polygon, the rest is quite straightforward. Use patches.Polygon and add them to the plot.
This code allows you also to define if you want the patch on top of your route or below it. Just give a positive or negative value for the width. If you want to change the width of the polygon depending on your zoom level and/or resolution, you adjust the scale parameter. It also gives you freedom to add additional parameters to the patches such as fill patterns, transparency, etc.

Interpolation differences on polar contour plots in Matplotlib

I am trying to generate contour plots on a polar plot and did some quick scripting in matlab to get some results. Out of curiosity I also wanted to try out the same thing in python using the matplotlib but somehow I am seeing different sets of contour plots for the same input data. I am trying to figure out whats going on and if there is anything I could tweak in my python code to get similar results in both cases.
A screenshot of the matlab results is here:
In the matlab code I used the scatteredinterpolant function to get the interpolated data, I am assuming the differences are occurring due to the interpolation function used?
The input data is -
Angles = [-180, -90, 0 , 90, 180, -135, -45,45, 135, 180,-90, 0, 90, 180 ]
Radii = [0,0.33,0.33,0.33,0.33,0.5,0.5,0.5,0.5,0.5,0.6,0.6,0.6,0.6]
Values = [30.42,24.75, 32.23, 34.26, 26.31, 20.58, 23.38, 34.15,27.21, 22.609, 16.013, 22.75, 27.062, 18.27]
This was done using python 2.7, on spyder. I have tried both scipy.interpolate.griddata as well as matplotlib.mlab.griddata and the results are similar. I was unable to get the nn method working in mlab.griddata because it kept giving me masked data.
Apologies if I am missing anything relevant - please let me know if anyother info is required I will update my post.
Edit:
The linear scipt griddata image looks like:
And the cubic scipy image looks like
As for the code, here is the code - I pass the interpolation type string into the function where this code is present. So 'linear' and 'cubic' are the 2 inputs.
val = np.array(list(values[i]))
radius = np.array(list(gamma[i]))
ang = [math.radians(np.array(list(theta[i]))[x]) for x in xrange(0,len(theta[i]))]
radiiGrid = np.linspace(min(radius),max(radius),100)
anglesGrid = np.linspace(min(ang),max(ang),100)
radiiGrid, anglesGrid = np.meshgrid(radiiGrid, anglesGrid)
zgrid = griddata((ang,radius),val,(anglesGrid,radiiGrid), method=interpType)
The angle input is what comes out of np.array(list(theta[i]))[x] - this is because the angle information is stored in a list of tuples (this is because I am reading in and sorting data). I took a look at the code to make sure the data is correct and it seems to line up. gamma corresponds to radii and values are the values in the sample data I provided.
Hope this helps!

Polar plots in matplotlib can get tricky. When that happens, a quick solution is to convert radii and angle to x,y, plot in a normal projection. Then make a empty polar axis to superimpose on it:
from scipy.interpolate import griddata
Angles = [-180, -90, 0 , 90, 180, -135,
-45,45, 135, 180,-90, 0, 90, 180 ]
Radii = [0,0.33,0.33,0.33,0.33,0.5,0.5,
0.5,0.5,0.5,0.6,0.6,0.6,0.6]
Angles = np.array(Angles)/180.*np.pi
x = np.array(Radii)*np.sin(Angles)
y = np.array(Radii)*np.cos(Angles)
Values = [30.42,24.75, 32.23, 34.26, 26.31, 20.58,
23.38, 34.15,27.21, 22.609, 16.013, 22.75, 27.062, 18.27]
Xi = np.linspace(-1,1,100)
Yi = np.linspace(-1,1,100)
#make the axes
f = plt.figure()
left, bottom, width, height= [0,0, 1, 0.7]
ax = plt.axes([left, bottom, width, height])
pax = plt.axes([left, bottom, width, height],
projection='polar',
axisbg='none')
cax = plt.axes([0.8, 0, 0.05, 1])
ax.set_aspect(1)
ax.axis('Off')
# grid the data.
Vi = griddata((x, y), Values, (Xi[None,:], Yi[:,None]), method='cubic')
cf = ax.contour(Xi,Yi,Vi, 15, cmap=plt.cm.jet)
#make a custom colorbar, because the default is ugly
gradient = np.linspace(1, 0, 256)
gradient = np.vstack((gradient, gradient))
cax.xaxis.set_major_locator(plt.NullLocator())
cax.yaxis.tick_right()
cax.imshow(gradient.T, aspect='auto', cmap=plt.cm.jet)
cax.set_yticks(np.linspace(0,256,len(cf1.get_array())))
cax.set_yticklabels(map(str, cf.get_array())[::-1])

color matplotlib map using bicubic interpolation

I know that matplotlib and scipy can do bicubic interpolation:
http://matplotlib.org/examples/pylab_examples/image_interp.html
http://docs.scipy.org/doc/scipy/reference/tutorial/interpolate.html
http://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.interp2d.html
I also know that it is possible to draw a map of the world with matplotlib:
http://matplotlib.org/basemap/users/geography.html
http://matplotlib.org/basemap/users/examples.html
http://matplotlib.org/basemap/api/basemap_api.html
But can I do a bicubic interpolation based on 4 data points and only color the land mass?
For example using these for 4 data points (longitude and latitude) and colors:
Lagos: 6.453056, 3.395833; red HSV 0 100 100 (or z = 0)
Cairo: 30.05, 31.233333; green HSV 90 100 100 (or z = 90)
Johannesburg: -26.204444, 28.045556; cyan HSV 180 100 100 (or z = 180)
Mogadishu: 2.033333, 45.35; purple HSV 270 100 100 (or z = 270)
I am thinking that it must be possible to do the bicubic interpolation across the range of latitudes and longitudes and then add oceans, lakes and rivers on top of that layer? I can do this with drawmapboundary. Actually there is an option maskoceans for this:
http://matplotlib.org/basemap/api/basemap_api.html#mpl_toolkits.basemap.maskoceans
I can interpolate the data like this:
xnew, ynew = np.mgrid[-1:1:70j, -1:1:70j]
tck = interpolate.bisplrep(x, y, z, s=0)
znew = interpolate.bisplev(xnew[:,0], ynew[0,:], tck)
Or with scipy.interpolate.interp2d:
http://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.interp2d.html
Here it is explained how to convert to map projection coordinates:
http://matplotlib.org/basemap/users/mapcoords.html
But I need to figure out how to do this for a calculated surface instead of individual points. Actually there is an example of such a topographic map using external data, which I should be able to replicate:
http://matplotlib.org/basemap/users/examples.html
P.S. I am not looking for a complete solution. I would much prefer to solve this myself. Rather I am looking for suggestions and hints. I have been using gnuplot for more than 10 years and only switched to matplotlib within the past few weeks, so please don't assume I know even the simplest things about matplotlib.

I think this is what you are looking for (roughly). Note the crucial things are masking the data array before you plot the pcolor and passing in the hsv colormap (Docs: cmap parameter for pcolormesh and available colormaps).
I've kept the code for plotting the maps quite close to the examples so it should be easy to follow. I've kept your interpolation code for the same reason. Note that the interpolation is linear rather than cubic - kx=ky=1 - because you don't give enough points to do cubic interpolation (you'd need at least 16 - scipy will complain with less saying that "m must be >= (kx+1)(ky+1)", although the constraint is not mentioned in the documentation).
I've also extended the range of your meshgrid and kept in lat / lon for x and y throughout.
Code
from mpl_toolkits.basemap import Basemap,maskoceans
import matplotlib.pyplot as plt
import numpy as np
from scipy import interpolate
# set up orthographic map projection with
# perspective of satellite looking down at 0N, 20W (Africa in main focus)
# use low resolution coastlines.
map = Basemap(projection='ortho',lat_0=0,lon_0=20,resolution='l')
# draw coastlines, country boundaries
map.drawcoastlines(linewidth=0.25)
map.drawcountries(linewidth=0.25)
# Optionally (commented line below) give the map a fill colour - e.g. a blue sea
#map.drawmapboundary(fill_color='aqua')
# draw lat/lon grid lines every 30 degrees.
map.drawmeridians(np.arange(0,360,30))
map.drawparallels(np.arange(-90,90,30))
data = {'Lagos': (6.453056, 3.395833,0),
'Cairo': (30.05, 31.233333,90),
'Johannesburg': (-26.204444, 28.045556,180),
'Mogadishu': (2.033333, 45.35, 270)}
x,y,z = zip(*data.values())
xnew, ynew = np.mgrid[-30:60:0.1, -50:50:0.1]
tck = interpolate.bisplrep(x, y, z, s=0,kx=1,ky=1)
znew = interpolate.bisplev(xnew[:,0], ynew[0,:], tck)
znew = maskoceans(xnew, ynew, znew)
col_plot = map.pcolormesh(xnew, ynew, znew, latlon=True, cmap='hsv')
plt.show()
Output

Observe that doing the opposite, that is putting a raster on the sea and lay a mask over the continents, is easy as pie. Simply use map.fillcontinents(). So the basic idea of this solution is to modify the fillcontinents function so that it lays polygons over the oceans.
The steps are:
Create a large circle-like polygon that covers the entire globe.
Create a polygon for each shape in the map.coastpolygons array.
Cut the shape of the landmass polygon away from the circle using shapely and its difference method.
Add the remaining polygons, which have the shape of the oceans, on the top, with a high zorder.
The code:
from mpl_toolkits.basemap import Basemap
import numpy as np
from scipy import interpolate
from shapely.geometry import Polygon
from descartes.patch import PolygonPatch
def my_circle_polygon( (x0, y0), r, resolution = 50 ):
circle = []
for theta in np.linspace(0,2*np.pi, resolution):
x = r * np.cos(theta) + x0
y = r * np.sin(theta) + y0
circle.append( (x,y) )
return Polygon( circle[:-1] )
def filloceans(the_map, color='0.8', ax=None):
# get current axes instance (if none specified).
if not ax:
ax = the_map._check_ax()
# creates a circle that covers the world
r = 0.5*(map.xmax - map.xmin) # + 50000 # adds a little bit of margin
x0 = 0.5*(map.xmax + map.xmin)
y0 = 0.5*(map.ymax + map.ymin)
oceans = my_circle_polygon( (x0, y0) , r, resolution = 100 )
# for each coastline polygon, gouge it out of the circle
for x,y in the_map.coastpolygons:
xa = np.array(x,np.float32)
ya = np.array(y,np.float32)
xy = np.array(zip(xa.tolist(),ya.tolist()))
continent = Polygon(xy)
## catches error when difference with lakes
try:
oceans = oceans.difference(continent)
except:
patch = PolygonPatch(continent, color="white", zorder =150)
ax.add_patch( patch )
for ocean in oceans:
sea_patch = PolygonPatch(ocean, color="blue", zorder =100)
ax.add_patch( sea_patch )
########### DATA
x = [3.395833, 31.233333, 28.045556, 45.35 ]
y = [6.453056, 30.05, -26.204444, 2.033333]
z = [0, 90, 180, 270]
# set up orthographic map projection
map = Basemap(projection='ortho', lat_0=0, lon_0=20, resolution='l')
## Plot the cities on the map
map.plot(x,y,".", latlon=1)
# create a interpolated mesh and set it on the map
interpol_func = interpolate.interp2d(x, y, z, kind='linear')
newx = np.linspace( min(x), max(x) )
newy = np.linspace( min(y), max(y) )
X,Y = np.meshgrid(newx, newy)
Z = interpol_func(newx, newy)
map.pcolormesh( X, Y, Z, latlon=1, zorder=3)
filloceans(map, color="blue")
Voilà: