Develop Reference

How to evenly spread annotation imageboxes around a scatterplot?

I would like to annotate a scatterplot with images corresponding to each datapoint. With standard parameters the images end up clashing with each other and other important features such as legend axis, etc. Thus, I would like the images to form a circle or a rectangle around the main scatter plot.
My code looks like this for now and I am struggling to modify it to organise the images around the center point of the plot.
import matplotlib.cbook as cbook
import pandas as pd
import numpy as np
from matplotlib import pyplot as plt
from matplotlib.offsetbox import OffsetImage, AnnotationBbox
import seaborn as sns
#Generate n points around a 2d circle
def generate_circle_points(n, centre_x, center_y, radius=1):
"""Generate n points around a circle.
Args:
n (int): Number of points to generate.
centre_x (float): x-coordinate of circle centre.
center_y (float): y-coordinate of circle centre.
radius (float): Radius of circle.
Returns:
list: List of points.
"""
points = []
for i in range(n):
angle = 2 * np.pi * i / n
x = centre_x + radius * np.cos(angle)
y = center_y + radius * np.sin(angle)
points.append([x, y])
return points
fig, ax = plt.subplots(1, 1, figsize=(7.5, 7.5))
data = pd.DataFrame(data={'x': np.random.uniform(0.5, 2.5, 20),
'y': np.random.uniform(10000, 50000, 20)})
with cbook.get_sample_data('grace_hopper.jpg') as image_file:
image = plt.imread(image_file)
# Set logarithmic scale for x and y axis
ax.set(xscale="log", yscale='log')
# Add grid
ax.grid(True, which='major', ls="--", c='gray')
coordianates = generate_circle_points(n=len(data),
centre_x=0, center_y=0, radius=10)
# Plot the scatter plot
scatter = sns.scatterplot(data=data, x='x', y='y', ax=ax)
for index, row in data.iterrows():
imagebox = OffsetImage(image, zoom=0.05)
imagebox.image.axes = ax
xy = np.array([row['x'], row['y']])
xybox = np.array(coordianates[index])
ab = AnnotationBbox(imagebox, xy,
xycoords='data',
boxcoords="offset points",
xybox=xybox,
pad=0)
ax.add_artist(ab)
for the moment the output looks like this:enter image description here
Ideally I would like the output to look to something like this:
enter image description here
Many thanks in advance for your help

Not an answer but a long comment:
You can control the location of the arrows, but sometimes it is easier to export figures as SVGs and edit them in Adobe Illustrator or Inkscape.
R has a dodge argument which is really nice, but even then is not always perfect. Solutions in Python exist but are laborious.
The major issue is that this needs to be done last as alternations to the plot would make it problematic. A few points need mentioning.
Your figures will have to have a fixed size (57mm / 121mm / 184mm for Science, 83mm / 171mm for RSC, 83mm / 178mm for ACS etc.), if you need to scale the figure in Illustrator keep note of the scaling factor, adding it as a textbox outside of the canvas —as the underlying plot will need to be replaced at least once due to Murphy's law. Exporting at the right size the SVG is ideal. Sounds silly, but it helps. Likewise, make sure the font size does not go under the minimum spec (7-9 points).

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')

Modify matplotlib colormap

I'm trying to produce a similar version of this image using Python:
I'm close but can't quite figure out how to modify a matplotlib colormap to make values <0.4 go to white. I tried masking those values and using set_bad but I ended up with a real blocky appearance, losing the nice smooth contours seen in the original image.
Result with continuous colormap (problem: no white):
Result with set_bad (problem: no smooth transition to white):
Code so far:
from netCDF4 import Dataset as NetCDFFile
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.basemap import Basemap
nc = NetCDFFile('C:/myfile1.nc')
nc1 = NetCDFFile('C:/myfile2.nc')
lat = nc.variables['lat'][:]
lon = nc.variables['lon'][:]
time = nc.variables['time'][:]
uwnd = nc.variables['uwnd'][:]
vwnd = nc1.variables['vwnd'][:]
map = Basemap(llcrnrlon=180.,llcrnrlat=0.,urcrnrlon=340.,urcrnrlat=80.)
lons,lats = np.meshgrid(lon,lat)
x,y = map(lons,lats)
speed = np.sqrt(uwnd*uwnd+vwnd*vwnd)
#speed = np.ma.masked_where(speed < 0.4, speed)
#cmap = plt.cm.jet
#cmap.set_bad(color='white')
levels = np.arange(0.0,3.0,0.1)
ticks = np.arange(0.0,3.0,0.2)
cs = map.contourf(x,y,speed[0],levels, cmap='jet')
vw = plt.quiver(x,y,speed)
cbar = plt.colorbar(cs, orientation='horizontal', cmap='jet', spacing='proportional',ticks=ticks)
cbar.set_label('850 mb Vector Wind Anomalies (m/s)')
map.drawcoastlines()
map.drawparallels(np.arange(20,80,20),labels=[1,1,0,0], linewidth=0.5)
map.drawmeridians(np.arange(200,340,20),labels=[0,0,0,1], linewidth=0.5)
#plt.show()
plt.savefig('phase8_850wind_anom.png',dpi=600)

The answer to get the result smooth lies in constructing your own colormap. To do this one has to create an RGBA-matrix: a matrix with on each row the amount (between 0 and 1) of Red, Green, Blue, and Alpha (transparency; 0 means that the pixel does not have any coverage information and is transparent).
As an example the distance to some point is plotted in two dimensions. Then:
For any distance higher than some critical value, the colors will be taken from a standard colormap.
For any distance lower than some critical value, the colors will linearly go from white to the first color of the previously mentioned map.
The choices depend fully on what you want to show. The colormaps and their sizes depend on your problem. For example, you can choose different types of interpolation: linear, exponential, ...; single- or multi-color colormaps; etc..
The code:
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable
# create colormap
# ---------------
# create a colormap that consists of
# - 1/5 : custom colormap, ranging from white to the first color of the colormap
# - 4/5 : existing colormap
# set upper part: 4 * 256/4 entries
upper = mpl.cm.jet(np.arange(256))
# set lower part: 1 * 256/4 entries
# - initialize all entries to 1 to make sure that the alpha channel (4th column) is 1
lower = np.ones((int(256/4),4))
# - modify the first three columns (RGB):
# range linearly between white (1,1,1) and the first color of the upper colormap
for i in range(3):
lower[:,i] = np.linspace(1, upper[0,i], lower.shape[0])
# combine parts of colormap
cmap = np.vstack(( lower, upper ))
# convert to matplotlib colormap
cmap = mpl.colors.ListedColormap(cmap, name='myColorMap', N=cmap.shape[0])
# show some example
# -----------------
# open a new figure
fig, ax = plt.subplots()
# some data to plot: distance to point at (50,50)
x,y = np.meshgrid(np.linspace(0,99,100),np.linspace(0,99,100))
z = (x-50)**2. + (y-50)**2.
# plot data, apply colormap, set limit such that our interpretation is correct
im = ax.imshow(z, interpolation='nearest', cmap=cmap, clim=(0,5000))
# add a colorbar to the bottom of the image
div = make_axes_locatable(ax)
cax = div.append_axes('bottom', size='5%', pad=0.4)
cbar = plt.colorbar(im, cax=cax, orientation='horizontal')
# save/show the image
plt.savefig('so.png')
plt.show()
The result:

Overlap area of 2 ellipses using matplotlib

Does anyone know if it is possible to calculate the overlapping area of two ellipses using matplotlib.patches.Ellipse.
I have to ellipses like this:
And i would like to calculate the ratio between the overlap area and the are of the individual ellipses.
Is this possible using only the Ellipse from matplotlib.patches

You cannot compute the area of the intersect with matplotlib (at least not to my knowledge), but you can use shapely to do so and then use matplotlib to visualise the result. Here a quick demo:
from matplotlib import pyplot as plt
from shapely.geometry.point import Point
from shapely import affinity
from matplotlib.patches import Polygon
import numpy as np
def create_ellipse(center, lengths, angle=0):
"""
create a shapely ellipse. adapted from
https://gis.stackexchange.com/a/243462
"""
circ = Point(center).buffer(1)
ell = affinity.scale(circ, int(lengths[0]), int(lengths[1]))
ellr = affinity.rotate(ell, angle)
return ellr
fig,ax = plt.subplots()
##these next few lines are pretty important because
##otherwise your ellipses might only be displayed partly
##or may be distorted
ax.set_xlim([-5,5])
ax.set_ylim([-5,5])
ax.set_aspect('equal')
##first ellipse in blue
ellipse1 = create_ellipse((0,0),(2,4),10)
verts1 = np.array(ellipse1.exterior.coords.xy)
patch1 = Polygon(verts1.T, color = 'blue', alpha = 0.5)
ax.add_patch(patch1)
##second ellipse in red
ellipse2 = create_ellipse((1,-1),(3,2),50)
verts2 = np.array(ellipse2.exterior.coords.xy)
patch2 = Polygon(verts2.T,color = 'red', alpha = 0.5)
ax.add_patch(patch2)
##the intersect will be outlined in black
intersect = ellipse1.intersection(ellipse2)
verts3 = np.array(intersect.exterior.coords.xy)
patch3 = Polygon(verts3.T, facecolor = 'none', edgecolor = 'black')
ax.add_patch(patch3)
##compute areas and ratios
print('area of ellipse 1:',ellipse1.area)
print('area of ellipse 2:',ellipse2.area)
print('area of intersect:',intersect.area)
print('intersect/ellipse1:', intersect.area/ellipse1.area)
print('intersect/ellipse2:', intersect.area/ellipse2.area)
plt.show()
The resulting plot looks like this:
And the computed areas (printed out to the terminal) are:
area of ellipse 1: 25.09238792436751
area of ellipse 2: 18.81929094327563
area of intersect: 13.656608779925698
intersect/ellipse1: 0.5442530547945023
intersect/ellipse2: 0.7256707397260032
Note that I adapted the code to generate the ellipse-shaped polygon from this post. Hope this helps.

matplotlib and apect ratio of geographical-data plots

I process geographical information and present the results using
matplotlib. All input is lattitude/longitude [degree]. I convert into
x/y [meter] for my calculations. And I present my results in
lattitude/longitude. The problem is to get the graphs aspect-ratio
right: All graphs are too wide. Is there a standard procedure to set the
correct aspect-ratio so I can simply draw my scatter and other diagrams
using lat/lon and the result has the correct shape? On screen and on
paper (png)?
[added this part later]
This is a bare-bone stripped version of my problem. I need actual lat/lon values
around the axes and an accurate shape (square). Right now it appears wide (2x).
import math
import matplotlib.pyplot as plt
import numpy as np
from pylab import *
w=1/math.cos(math.radians(60.0))
plt_area=[0,w,59.5,60.5] #60deg North, adjacent to the prime meridian
a=np.zeros(shape=(300,300))
matshow(a, extent=plt_area)
plt.grid(False)
plt.axis(plt_area)
fig = plt.gcf()
fig.set_size_inches(8,8)
fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)
plt.show()

It seems I found the solution.
And I found it here: How can I set the aspect ratio in matplotlib?
import math
import matplotlib.pyplot as plt
import numpy as np
w=1/math.cos(math.radians(60.0))
plt_area=[0,w,59.5,60.5] #square area
a=np.zeros(shape=(300,300))
fig = plt.figure()
ax = fig.add_subplot(111)
ax.imshow(a)
plt.grid(False)
ax.axis(plt_area)
fig = plt.gcf()
fig.set_size_inches(8,8)
ax.set_aspect(w)
fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)
plt.show()

In matplotlib I usually change the figure size like this:
import matplotlib.pyplot as plt
plt.clf()
fig = plt.figure()
fig_p = plt.gcf()
fig_p.set_size_inches(8, 8) # x, y
However this sets the dimensions for the figure outer dimensions, not the plot area. You can change the plot area relative to the figure size given in ratios of the total figure size lengths of x and y respectively:
fig.subplots_adjust(left=0.1, right=0.9, bottom=0.1, top=0.9)
As long as the the relative ratios stay symmetrically the aspect ratio should be the same for the plot are.
Example 1:
plt.clf()
fig = plt.figure()
fig_p = plt.gcf()
fig_p.set_size_inches(5, 5) # x, y for figure canvas
# Relative distance ratio between origin of the figure and max extend of canvas
fig.subplots_adjust(left=0.2, right=0.8, bottom=0.2, top=0.8)
ax1 = fig.add_subplot(111)
xdata = [rand()*10 for i in xrange(100)]
ydata = [rand()*1 for i in xrange(100)]
ax1.plot(xdata, ydata, '.b', )
ax1.set_xlabel('Very Large X-Label', size=30)
plt.savefig('squareplot.png', dpi=96)
Example 2:
fig.subplots_adjust(left=0.0, right=1.0, bottom=0.0, top=1.0)
Plot area fills the figure size completely:

Don't try to fix this by fiddling fig.set_size_inches() or fig.subplots_adjust() or by changing your data; instead use a Mercator projection.
You can get a quick and dirty Mercator projection by using an aspect ratio of the reciprocal of the cosine of the mean latitude of your data. This is "pretty good" for data contained in about 1 degree of latitude, which is about 100 km. (You have to decide if, for your application, this is "good enough". If it isn't, you really have to consider some serious geographical projection libraries...)
Example:
from math import cos, radians
import matplotlib.pyplot as plt
import numpy as np
# Helsinki 60.1708 N, 24.9375 E
# Helsinki (lng, lat)
hels = [24.9375, 60.1708]
# a point 100 km directly north of Helsinki
pt_N = [24.9375, 61.0701]
# a point 100 km east of Helsinki along its parallel
pt_E = [26.7455, 60.1708]
coords = np.array([pt_N, hels, pt_E])
plt.figure()
plt.plot(coords[:,0], coords[:,1])
# either of these will estimate the "central latitude" of your data
# 1) do the plot, then average the limits of the y-axis
central_latitude = sum(plt.axes().get_ylim())/2.
# 2) actually average the latitudes in your data
central_latitude = np.average(coords, 0)[1]
# calculate the aspect ratio that will approximate a
# Mercator projection at this central latitude
mercator_aspect_ratio = 1/cos(radians(central_latitude))
# set the aspect ratio of the axes to that
plt.axes().set_aspect(mercator_aspect_ratio)
plt.show()
I picked Helsinki for the example since at that latitude the aspect ratio is almost 2... because two degrees of longitude is the about same distance as one degree of latitude.
To really see this work: a) run the above, b) resize the window. Then comment out the call to set_aspect() and do the same. In the first case, the correct aspect ratio is maintained, in the latter you get nonsensical stretching.
The points 100km north and east of Helsinki were calculated/confirmed by the EXCELLENT page calculating distances between lat/lng points at Movable Type Scripts