Related
I have programmed plt.quiver(x,y,u,v,color), where there are arrows that start at x,y and the direction is determined by u,v. My question is how can I know exactly where the arrow ends?
In general, the arrows are of length length as described in the Quiver documentation and are auto-calculated by the matplotlib. I don't know which kwarg may help to return the length.
Another approach could be to define the exact position by scaling the plot with the help of scale=1, units='xy'.
import numpy as np
import matplotlib.pyplot as plt
# define arrow
x = np.linspace(0,1,11)
y = np.linspace(1,0,11)
u = v = np.zeros((11,11))
u[5,5] = 0.3
v[5,5] = 0.3
plt.quiver(x, y, u, v, scale=1, units='xy')
plt.axis('equal')
plt.xlim(0,1)
plt.ylim(0,1)
plt.show()
Color arrows that end at a specific point
Applying the principles above could result in:
import numpy as np
import matplotlib.pyplot as plt
import random
n = 11
cx = 0.7 #x-position of specific end point
cy = 0.5 #y-position of specific end point
# define random arrows
x = np.linspace(0,1,n)
y = np.linspace(0,1,n)
u = np.zeros((n,n))
v = np.zeros((n,n))
# color everything black
colors = [(0, 0, 0)]*n*n
# make sure at least some points end at the same point
u[5][5] = 0.2
u[5][8] = -0.1
v[2][7] = 0.3
# search for specific point
for i in range(len(x)):
for j in range(len(y)):
endPosX = x[i] + u[j][i]
endPosY = y[j] + v[j][i]
if np.isclose(endPosX, cx) and np.isclose(endPosY, cy):
#found specific point -> color it red
colors[j*n+i] = (1,0,0)
# plot data
plt.quiver(x, y, u, v, color=colors, scale=1, units='xy')
plt.axis('equal')
plt.show()
I want to plot colored pie charts at specific positions without distorting their circular aspect ratio. I'm using Wedge patches because I could not find a better solution. Here is the code
import matplotlib.pyplot as plt
import numpy as np
from matplotlib import patches, collections
fig, axes = plt.subplots()
for i in range(20):
x = np.random.uniform(low=0, high=1, size=10).cumsum()
axes.scatter(x=x, y=np.repeat(i, x.shape[0]), c='gray', s=1)
pies = []
N = 4
cmap = plt.cm.get_cmap("hsv", N + 1)
colors = list(map(cmap, range(N)))
print(colors)
for i in range(2, 2 + N):
thetas = np.linspace(0, 360, num=i)
assert len(thetas) - 1 <= len(colors)
for theta1, theta2, c in zip(thetas[:-1], thetas[1:], colors):
wedge = patches.Wedge((i, i), r=i / 10, theta1=theta1, theta2=theta2,
color=c)
pies.append(wedge)
axes.add_collection(collections.PatchCollection(pies,
match_original=True))
plt.show()
How to preserve the aspect ratio of pie charts? Setting axes.set_aspect("equal") is NOT an option because it squeezes the plot completely when I have more data points.
I've been looking at how to draw circles and preserve the aspect ratio but the solution cannot be adopted here - I'm plotting Wedges/pie charts, not Circles.
I also looked at matplotlib transforms but couldn't find the answer there either.
I tried the same thing, and matplotlib really doesn't try to make this easy for you, but I found a solution that you should be able to use.
You need to separate the centers from the wedges and add them to the PatchCollection as offsets. Then you can apply different transforms to the offsets (transOffset) and shape (transform).
Notice that I have changed the r-value (radius). This value is no longer in data coordinates, so it should always be the same size, regardless of how much you zoom, but it is too small to be visible at i/10.
from matplotlib import patches, collections, transforms
offsets = []
for i in range(2, 2 + N):
thetas = np.linspace(0, 360, num=i)
assert len(thetas) - 1 <= len(colors)
for theta1, theta2, c in zip(thetas[:-1], thetas[1:], colors):
wedge = patches.Wedge((0, 0), r=10, theta1=theta1, theta2=theta2,
color=c)
offsets.append((i, i))
pies.append(wedge)
coll = collections.PatchCollection(
pies, match_original=True, offsets=offsets,
transform=transforms.IdentityTransform(),
transOffset=axes.transData
)
It works fine for me when I set set_aspect('equal'):
Image is narrowed because y-range is longer than x-range I think.
If you'd set y_lim between 0 and a number lower than y_max, you'd see it better:
import matplotlib.pyplot as plt
import numpy as np
from matplotlib import patches, collections
fig, axes = plt.subplots()
for i in range(20):
x = np.random.uniform(low=0, high=1, size=10).cumsum()
axes.scatter(x=x, y=np.repeat(i, x.shape[0]), c='gray', s=1)
pies = []
N = 4
cmap = plt.cm.get_cmap("hsv", N + 1)
colors = list(map(cmap, range(N)))
print(colors)
for i in range(2, 2 + N):
thetas = np.linspace(0, 360, num=i)
assert len(thetas) - 1 <= len(colors)
for theta1, theta2, c in zip(thetas[:-1], thetas[1:], colors):
wedge = patches.Wedge((i, i), r=i / 10, theta1=theta1, theta2=theta2,
color=c)
pies.append(wedge)
axes.add_collection(collections.PatchCollection(pies,
match_original=True))
axes.set_aspect('equal')
axes.set_ylim(0,7.5)
plt.show()
In Matplotlib, it's not too tough to make a legend (example_legend(), below), but I think it's better style to put labels right on the curves being plotted (as in example_inline(), below). This can be very fiddly, because I have to specify coordinates by hand, and, if I re-format the plot, I probably have to reposition the labels. Is there a way to automatically generate labels on curves in Matplotlib? Bonus points for being able to orient the text at an angle corresponding to the angle of the curve.
import numpy as np
import matplotlib.pyplot as plt
def example_legend():
plt.clf()
x = np.linspace(0, 1, 101)
y1 = np.sin(x * np.pi / 2)
y2 = np.cos(x * np.pi / 2)
plt.plot(x, y1, label='sin')
plt.plot(x, y2, label='cos')
plt.legend()
def example_inline():
plt.clf()
x = np.linspace(0, 1, 101)
y1 = np.sin(x * np.pi / 2)
y2 = np.cos(x * np.pi / 2)
plt.plot(x, y1, label='sin')
plt.plot(x, y2, label='cos')
plt.text(0.08, 0.2, 'sin')
plt.text(0.9, 0.2, 'cos')
Update: User cphyc has kindly created a Github repository for the code in this answer (see here), and bundled the code into a package which may be installed using pip install matplotlib-label-lines.
Pretty Picture:
In matplotlib it's pretty easy to label contour plots (either automatically or by manually placing labels with mouse clicks). There does not (yet) appear to be any equivalent capability to label data series in this fashion! There may be some semantic reason for not including this feature which I am missing.
Regardless, I have written the following module which takes any allows for semi-automatic plot labelling. It requires only numpy and a couple of functions from the standard math library.
Description
The default behaviour of the labelLines function is to space the labels evenly along the x axis (automatically placing at the correct y-value of course). If you want you can just pass an array of the x co-ordinates of each of the labels. You can even tweak the location of one label (as shown in the bottom right plot) and space the rest evenly if you like.
In addition, the label_lines function does not account for the lines which have not had a label assigned in the plot command (or more accurately if the label contains '_line').
Keyword arguments passed to labelLines or labelLine are passed on to the text function call (some keyword arguments are set if the calling code chooses not to specify).
Issues
Annotation bounding boxes sometimes interfere undesirably with other curves. As shown by the 1 and 10 annotations in the top left plot. I'm not even sure this can be avoided.
It would be nice to specify a y position instead sometimes.
It's still an iterative process to get annotations in the right location
It only works when the x-axis values are floats
Gotchas
By default, the labelLines function assumes that all data series span the range specified by the axis limits. Take a look at the blue curve in the top left plot of the pretty picture. If there were only data available for the x range 0.5-1 then then we couldn't possibly place a label at the desired location (which is a little less than 0.2). See this question for a particularly nasty example. Right now, the code does not intelligently identify this scenario and re-arrange the labels, however there is a reasonable workaround. The labelLines function takes the xvals argument; a list of x-values specified by the user instead of the default linear distribution across the width. So the user can decide which x-values to use for the label placement of each data series.
Also, I believe this is the first answer to complete the bonus objective of aligning the labels with the curve they're on. :)
label_lines.py:
from math import atan2,degrees
import numpy as np
#Label line with line2D label data
def labelLine(line,x,label=None,align=True,**kwargs):
ax = line.axes
xdata = line.get_xdata()
ydata = line.get_ydata()
if (x < xdata[0]) or (x > xdata[-1]):
print('x label location is outside data range!')
return
#Find corresponding y co-ordinate and angle of the line
ip = 1
for i in range(len(xdata)):
if x < xdata[i]:
ip = i
break
y = ydata[ip-1] + (ydata[ip]-ydata[ip-1])*(x-xdata[ip-1])/(xdata[ip]-xdata[ip-1])
if not label:
label = line.get_label()
if align:
#Compute the slope
dx = xdata[ip] - xdata[ip-1]
dy = ydata[ip] - ydata[ip-1]
ang = degrees(atan2(dy,dx))
#Transform to screen co-ordinates
pt = np.array([x,y]).reshape((1,2))
trans_angle = ax.transData.transform_angles(np.array((ang,)),pt)[0]
else:
trans_angle = 0
#Set a bunch of keyword arguments
if 'color' not in kwargs:
kwargs['color'] = line.get_color()
if ('horizontalalignment' not in kwargs) and ('ha' not in kwargs):
kwargs['ha'] = 'center'
if ('verticalalignment' not in kwargs) and ('va' not in kwargs):
kwargs['va'] = 'center'
if 'backgroundcolor' not in kwargs:
kwargs['backgroundcolor'] = ax.get_facecolor()
if 'clip_on' not in kwargs:
kwargs['clip_on'] = True
if 'zorder' not in kwargs:
kwargs['zorder'] = 2.5
ax.text(x,y,label,rotation=trans_angle,**kwargs)
def labelLines(lines,align=True,xvals=None,**kwargs):
ax = lines[0].axes
labLines = []
labels = []
#Take only the lines which have labels other than the default ones
for line in lines:
label = line.get_label()
if "_line" not in label:
labLines.append(line)
labels.append(label)
if xvals is None:
xmin,xmax = ax.get_xlim()
xvals = np.linspace(xmin,xmax,len(labLines)+2)[1:-1]
for line,x,label in zip(labLines,xvals,labels):
labelLine(line,x,label,align,**kwargs)
Test code to generate the pretty picture above:
from matplotlib import pyplot as plt
from scipy.stats import loglaplace,chi2
from labellines import *
X = np.linspace(0,1,500)
A = [1,2,5,10,20]
funcs = [np.arctan,np.sin,loglaplace(4).pdf,chi2(5).pdf]
plt.subplot(221)
for a in A:
plt.plot(X,np.arctan(a*X),label=str(a))
labelLines(plt.gca().get_lines(),zorder=2.5)
plt.subplot(222)
for a in A:
plt.plot(X,np.sin(a*X),label=str(a))
labelLines(plt.gca().get_lines(),align=False,fontsize=14)
plt.subplot(223)
for a in A:
plt.plot(X,loglaplace(4).pdf(a*X),label=str(a))
xvals = [0.8,0.55,0.22,0.104,0.045]
labelLines(plt.gca().get_lines(),align=False,xvals=xvals,color='k')
plt.subplot(224)
for a in A:
plt.plot(X,chi2(5).pdf(a*X),label=str(a))
lines = plt.gca().get_lines()
l1=lines[-1]
labelLine(l1,0.6,label=r'$Re=${}'.format(l1.get_label()),ha='left',va='bottom',align = False)
labelLines(lines[:-1],align=False)
plt.show()
#Jan Kuiken's answer is certainly well-thought and thorough, but there are some caveats:
it does not work in all cases
it requires a fair amount of extra code
it may vary considerably from one plot to the next
A much simpler approach is to annotate the last point of each plot. The point can also be circled, for emphasis. This can be accomplished with one extra line:
import matplotlib.pyplot as plt
for i, (x, y) in enumerate(samples):
plt.plot(x, y)
plt.text(x[-1], y[-1], f'sample {i}')
A variant would be to use the method matplotlib.axes.Axes.annotate.
Nice question, a while ago I've experimented a bit with this, but haven't used it a lot because it's still not bulletproof. I divided the plot area into a 32x32 grid and calculated a 'potential field' for the best position of a label for each line according the following rules:
white space is a good place for a label
Label should be near corresponding line
Label should be away from the other lines
The code was something like this:
import matplotlib.pyplot as plt
import numpy as np
from scipy import ndimage
def my_legend(axis = None):
if axis == None:
axis = plt.gca()
N = 32
Nlines = len(axis.lines)
print Nlines
xmin, xmax = axis.get_xlim()
ymin, ymax = axis.get_ylim()
# the 'point of presence' matrix
pop = np.zeros((Nlines, N, N), dtype=np.float)
for l in range(Nlines):
# get xy data and scale it to the NxN squares
xy = axis.lines[l].get_xydata()
xy = (xy - [xmin,ymin]) / ([xmax-xmin, ymax-ymin]) * N
xy = xy.astype(np.int32)
# mask stuff outside plot
mask = (xy[:,0] >= 0) & (xy[:,0] < N) & (xy[:,1] >= 0) & (xy[:,1] < N)
xy = xy[mask]
# add to pop
for p in xy:
pop[l][tuple(p)] = 1.0
# find whitespace, nice place for labels
ws = 1.0 - (np.sum(pop, axis=0) > 0) * 1.0
# don't use the borders
ws[:,0] = 0
ws[:,N-1] = 0
ws[0,:] = 0
ws[N-1,:] = 0
# blur the pop's
for l in range(Nlines):
pop[l] = ndimage.gaussian_filter(pop[l], sigma=N/5)
for l in range(Nlines):
# positive weights for current line, negative weight for others....
w = -0.3 * np.ones(Nlines, dtype=np.float)
w[l] = 0.5
# calculate a field
p = ws + np.sum(w[:, np.newaxis, np.newaxis] * pop, axis=0)
plt.figure()
plt.imshow(p, interpolation='nearest')
plt.title(axis.lines[l].get_label())
pos = np.argmax(p) # note, argmax flattens the array first
best_x, best_y = (pos / N, pos % N)
x = xmin + (xmax-xmin) * best_x / N
y = ymin + (ymax-ymin) * best_y / N
axis.text(x, y, axis.lines[l].get_label(),
horizontalalignment='center',
verticalalignment='center')
plt.close('all')
x = np.linspace(0, 1, 101)
y1 = np.sin(x * np.pi / 2)
y2 = np.cos(x * np.pi / 2)
y3 = x * x
plt.plot(x, y1, 'b', label='blue')
plt.plot(x, y2, 'r', label='red')
plt.plot(x, y3, 'g', label='green')
my_legend()
plt.show()
And the resulting plot:
matplotx (which I wrote) has line_labels() which plots the labels to the right of the lines. It's also smart enough to avoid overlaps when too many lines are concentrated in one spot. (See stargraph for examples.) It does that by solving a particular non-negative-least-squares problem on the target positions of the labels. Anyway, in many cases where there's no overlap to begin with, such as the example below, that's not even necessary.
import matplotlib.pyplot as plt
import matplotx
import numpy as np
# create data
rng = np.random.default_rng(0)
offsets = [1.0, 1.50, 1.60]
labels = ["no balancing", "CRV-27", "CRV-27*"]
x0 = np.linspace(0.0, 3.0, 100)
y = [offset * x0 / (x0 + 1) + 0.1 * rng.random(len(x0)) for offset in offsets]
# plot
with plt.style.context(matplotx.styles.dufte):
for yy, label in zip(y, labels):
plt.plot(x0, yy, label=label)
plt.xlabel("distance [m]")
matplotx.ylabel_top("voltage [V]") # move ylabel to the top, rotate
matplotx.line_labels() # line labels to the right
plt.show()
# plt.savefig("out.png", bbox_inches="tight")
A simpler approach like the one Ioannis Filippidis do :
import matplotlib.pyplot as plt
import numpy as np
# evenly sampled time at 200ms intervals
tMin=-1 ;tMax=10
t = np.arange(tMin, tMax, 0.1)
# red dashes, blue points default
plt.plot(t, 22*t, 'r--', t, t**2, 'b')
factor=3/4 ;offset=20 # text position in view
textPosition=[(tMax+tMin)*factor,22*(tMax+tMin)*factor]
plt.text(textPosition[0],textPosition[1]+offset,'22 t',color='red',fontsize=20)
textPosition=[(tMax+tMin)*factor,((tMax+tMin)*factor)**2+20]
plt.text(textPosition[0],textPosition[1]+offset, 't^2', bbox=dict(facecolor='blue', alpha=0.5),fontsize=20)
plt.show()
code python 3 on sageCell
I'm new to Python (was an IDL user before hand) so I hope that I'm asking this in an understandable way. I've been trying to create a polar plot with x number of bins where the data in the bin is averaged and given a colour associated with that value. This seems to work fine while using the plt.fill command where I can define the bin and then the fill colour. The problem comes when I then try to make a colour bar to go with it. I keep getting errors that state AttributeError: 'Figure' object has no attribute 'autoscale_None'
Any advice would be helpful thanks.
import matplotlib.pyplot as plt
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from matplotlib.pyplot import figure, show, rc, grid
import pylab
r = np.arange(50)/5.
rstep = r[1] - r[0]
theta = np.arange(50)/50.*2.*np.pi
tstep = theta[1] - theta[0]
colorv = np.arange(50)/50.
# force square figure and square axes looks better for polar, IMO
width, height = mpl.rcParams['figure.figsize']
size = min(width, height)
# make a square figure
fig = figure(figsize=(size, size))
ax = fig.add_axes([0.1, 0.1, .8, .8])#, polar=True)
my_cmap = cm.jet
for j in range(len(r)):
rbox = np.array([r[j], r[j], r[j]+ rstep, r[j] + rstep])
for i in range(len(theta)):
thetabox = np.array([theta[i], theta[i] + tstep, theta[i] + tstep, theta[i]])
x = rbox*np.cos(thetabox)
y = rbox*np.sin(thetabox)
plt.fill(x,y, facecolor = my_cmap(colorv[j]))
# Add colorbar, make sure to specify tick locations to match desired ticklabels
cbar = fig.colorbar(fig, ticks=[np.min(colorv), np.max(colorv)])
cb = plt.colorbar()
plt.show()
* here is a slightly better example of my real data, there are holes missing everywhere, so in this example I've just made a big one in a quarter of the circle. When I've tried meshing, the code seems to try to interpolate over these regions.
r = np.arange(50)/50.*7. + 3.
rstep = r[1] - r[0]
theta = np.arange(50)/50.*1.5*np.pi - np.pi
tstep = theta[1] - theta[0]
colorv = np.sin(r/10.*np.pi)
# force square figure and square axes looks better for polar, IMO
width, height = mpl.rcParams['figure.figsize']
size = min(width, height)
# make a square figure
fig = figure(figsize=(size, size))
ax = fig.add_axes([0.1, 0.1, .8, .8])#, polar=True)
my_cmap = cm.jet
for j in range(len(r)):
rbox = np.array([r[j], r[j], r[j]+ rstep, r[j] + rstep])
for i in range(len(theta)):
thetabox = np.array([theta[i], theta[i] + tstep, theta[i] + tstep, theta[i]])
x = rbox*np.cos(thetabox)
y = rbox*np.sin(thetabox)
plt.fill(x,y, facecolor = my_cmap(colorv[j]))
# Add colorbar, make sure to specify tick locations to match desired ticklabels
#cbar = fig.colorbar(fig, ticks=[np.min(colorv), np.max(colorv)])
#cb = plt.colorbar()
plt.show()
And then with a meshing involved...
from matplotlib.mlab import griddata
r = np.arange(50)/5.
rstep = r[1] - r[0]
theta = np.arange(50)/50.*1.5*np.pi - np.pi
tstep = theta[1] - theta[0]
colorv = np.sin(r/10.*np.pi)
# force square figure and square axes looks better for polar, IMO
width, height = mpl.rcParams['figure.figsize']
size = min(width, height)
# make a square figure
fig = figure(figsize=(size, size))
ax = fig.add_axes([0.1, 0.1, .8, .8])#, polar=True)
my_cmap = cm.jet
x = r*np.cos(theta)
y = r*np.sin(theta)
X,Y = np.meshgrid(x,y)
data = griddata(x,y,colorv,X,Y)
cax = plt.contourf(X,Y, data)
plt.colorbar()
# Add colorbar, make sure to specify tick locations to match desired ticklabels
#cbar = fig.colorbar(fig, ticks=[np.min(colorv), np.max(colorv)])
#cb = plt.colorbar()
plt.show()
colorbar needs things to be an instance of ScalarMappable in order to make a colorbar from them.
Because you're manually setting each tile, there's nothing that essentially has a colorbar.
There are a number of ways to fake it from your colormap, but in this case there's a much simpler solution.
pcolormesh does exactly what you want, and will be much faster.
As an example:
import numpy as np
import matplotlib.pyplot as plt
# Linspace makes what you're doing _much_ easier (and includes endpoints)
r = np.linspace(0, 10, 50)
theta = np.linspace(0, 2*np.pi, 50)
fig = plt.figure()
ax = fig.add_subplot(111, projection='polar')
# "Grid" r and theta into 2D arrays (see the docs for meshgrid)
r, theta = np.meshgrid(r, theta)
cax = ax.pcolormesh(theta, r, r, edgecolors='black', antialiased=True)
# We could just call `plt.colorbar`, but I prefer to be more explicit
# and pass in the artist that I want it to extract colors from.
fig.colorbar(cax)
plt.show()
Or, if you'd prefer non-polar axes, as in your example code:
import numpy as np
import matplotlib.pyplot as plt
r = np.linspace(0, 10, 50)
theta = np.linspace(0, 2*np.pi, 50)
# "Grid" r and theta and convert them to cartesian coords...
r, theta = np.meshgrid(r, theta)
x, y = r * np.cos(theta), r * np.sin(theta)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.axis('equal')
cax = ax.pcolormesh(x, y, r, edgecolors='black', antialiased=True)
fig.colorbar(cax)
plt.show()
Note: If you'd prefer the boundary lines a bit less dark, just specify linewidth=0.5 or something similar to pcolormesh.
Finally, if you did want to directly make the colorbar from the colormap in your original code, you'd create an instance of ScalarMappable from it and pass this to colorbar. It's easier than it sounds, but it's a bit verbose.
As an example, in your original code, if you do something like the following:
cax = cm.ScalarMappable(cmap=my_cmap)
cax.set_array(colorv)
fig.colorbar(cax)
It should do what you want.
So I've found a workaround. Since I know of a region where I definitely won't have data, I've plotted some there. I've made sure that the data covers the entire range of what I'm potting. I then cover it up (this region was going to be covered anyway, it shows where the "earth" is located). Now I can go ahead and use plt.fill as I had originally and use the colour bar from the randomly potted data. I know this isn't probably the correct way, but it works and doesn't try to interpolate my data.
Thanks so much for helping get this sorted. and if you know of a better way, I'd be happy to hear it!
hid = plt.pcolormesh(X,Y, data, antialiased=True)
#here we cover up the region that we just plotted in
r3 = [1 for i in range(360)]
theta3 = np.arange(360)*np.pi/180.
plt.fill(theta3, r3, 'w')
#now we can go through and fill in all the regions
for j in range(len(r)):
rbox = np.array([r[j], r[j], r[j]+ rstep, r[j] + rstep])
for i in range(len(theta)):
thetabox = np.array([theta[i], theta[i] + tstep, theta[i] + tstep, theta[i]])
x = rbox*np.cos(thetabox)
y = rbox*np.sin(thetabox)
colorv = np.sin(r[j]/10.*np.pi)
plt.fill(thetabox,rbox, facecolor = my_cmap(colorv))
#And now we can plot the color bar that fits the data Tada :)
plt.colorbar()
plt.show()
I have a set of X,Y data points (about 10k) that are easy to plot as a scatter plot but that I would like to represent as a heatmap.
I looked through the examples in Matplotlib and they all seem to already start with heatmap cell values to generate the image.
Is there a method that converts a bunch of x, y, all different, to a heatmap (where zones with higher frequency of x, y would be "warmer")?
If you don't want hexagons, you can use numpy's histogram2d function:
import numpy as np
import numpy.random
import matplotlib.pyplot as plt
# Generate some test data
x = np.random.randn(8873)
y = np.random.randn(8873)
heatmap, xedges, yedges = np.histogram2d(x, y, bins=50)
extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]
plt.clf()
plt.imshow(heatmap.T, extent=extent, origin='lower')
plt.show()
This makes a 50x50 heatmap. If you want, say, 512x384, you can put bins=(512, 384) in the call to histogram2d.
Example:
In Matplotlib lexicon, i think you want a hexbin plot.
If you're not familiar with this type of plot, it's just a bivariate histogram in which the xy-plane is tessellated by a regular grid of hexagons.
So from a histogram, you can just count the number of points falling in each hexagon, discretiize the plotting region as a set of windows, assign each point to one of these windows; finally, map the windows onto a color array, and you've got a hexbin diagram.
Though less commonly used than e.g., circles, or squares, that hexagons are a better choice for the geometry of the binning container is intuitive:
hexagons have nearest-neighbor symmetry (e.g., square bins don't,
e.g., the distance from a point on a square's border to a point
inside that square is not everywhere equal) and
hexagon is the highest n-polygon that gives regular plane
tessellation (i.e., you can safely re-model your kitchen floor with hexagonal-shaped tiles because you won't have any void space between the tiles when you are finished--not true for all other higher-n, n >= 7, polygons).
(Matplotlib uses the term hexbin plot; so do (AFAIK) all of the plotting libraries for R; still i don't know if this is the generally accepted term for plots of this type, though i suspect it's likely given that hexbin is short for hexagonal binning, which is describes the essential step in preparing the data for display.)
from matplotlib import pyplot as PLT
from matplotlib import cm as CM
from matplotlib import mlab as ML
import numpy as NP
n = 1e5
x = y = NP.linspace(-5, 5, 100)
X, Y = NP.meshgrid(x, y)
Z1 = ML.bivariate_normal(X, Y, 2, 2, 0, 0)
Z2 = ML.bivariate_normal(X, Y, 4, 1, 1, 1)
ZD = Z2 - Z1
x = X.ravel()
y = Y.ravel()
z = ZD.ravel()
gridsize=30
PLT.subplot(111)
# if 'bins=None', then color of each hexagon corresponds directly to its count
# 'C' is optional--it maps values to x-y coordinates; if 'C' is None (default) then
# the result is a pure 2D histogram
PLT.hexbin(x, y, C=z, gridsize=gridsize, cmap=CM.jet, bins=None)
PLT.axis([x.min(), x.max(), y.min(), y.max()])
cb = PLT.colorbar()
cb.set_label('mean value')
PLT.show()
Edit: For a better approximation of Alejandro's answer, see below.
I know this is an old question, but wanted to add something to Alejandro's anwser: If you want a nice smoothed image without using py-sphviewer you can instead use np.histogram2d and apply a gaussian filter (from scipy.ndimage.filters) to the heatmap:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy.ndimage.filters import gaussian_filter
def myplot(x, y, s, bins=1000):
heatmap, xedges, yedges = np.histogram2d(x, y, bins=bins)
heatmap = gaussian_filter(heatmap, sigma=s)
extent = [xedges[0], xedges[-1], yedges[0], yedges[-1]]
return heatmap.T, extent
fig, axs = plt.subplots(2, 2)
# Generate some test data
x = np.random.randn(1000)
y = np.random.randn(1000)
sigmas = [0, 16, 32, 64]
for ax, s in zip(axs.flatten(), sigmas):
if s == 0:
ax.plot(x, y, 'k.', markersize=5)
ax.set_title("Scatter plot")
else:
img, extent = myplot(x, y, s)
ax.imshow(img, extent=extent, origin='lower', cmap=cm.jet)
ax.set_title("Smoothing with $\sigma$ = %d" % s)
plt.show()
Produces:
The scatter plot and s=16 plotted on top of eachother for Agape Gal'lo (click for better view):
One difference I noticed with my gaussian filter approach and Alejandro's approach was that his method shows local structures much better than mine. Therefore I implemented a simple nearest neighbour method at pixel level. This method calculates for each pixel the inverse sum of the distances of the n closest points in the data. This method is at a high resolution pretty computationally expensive and I think there's a quicker way, so let me know if you have any improvements.
Update: As I suspected, there's a much faster method using Scipy's scipy.cKDTree. See Gabriel's answer for the implementation.
Anyway, here's my code:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
def data_coord2view_coord(p, vlen, pmin, pmax):
dp = pmax - pmin
dv = (p - pmin) / dp * vlen
return dv
def nearest_neighbours(xs, ys, reso, n_neighbours):
im = np.zeros([reso, reso])
extent = [np.min(xs), np.max(xs), np.min(ys), np.max(ys)]
xv = data_coord2view_coord(xs, reso, extent[0], extent[1])
yv = data_coord2view_coord(ys, reso, extent[2], extent[3])
for x in range(reso):
for y in range(reso):
xp = (xv - x)
yp = (yv - y)
d = np.sqrt(xp**2 + yp**2)
im[y][x] = 1 / np.sum(d[np.argpartition(d.ravel(), n_neighbours)[:n_neighbours]])
return im, extent
n = 1000
xs = np.random.randn(n)
ys = np.random.randn(n)
resolution = 250
fig, axes = plt.subplots(2, 2)
for ax, neighbours in zip(axes.flatten(), [0, 16, 32, 64]):
if neighbours == 0:
ax.plot(xs, ys, 'k.', markersize=2)
ax.set_aspect('equal')
ax.set_title("Scatter Plot")
else:
im, extent = nearest_neighbours(xs, ys, resolution, neighbours)
ax.imshow(im, origin='lower', extent=extent, cmap=cm.jet)
ax.set_title("Smoothing over %d neighbours" % neighbours)
ax.set_xlim(extent[0], extent[1])
ax.set_ylim(extent[2], extent[3])
plt.show()
Result:
Instead of using np.hist2d, which in general produces quite ugly histograms, I would like to recycle py-sphviewer, a python package for rendering particle simulations using an adaptive smoothing kernel and that can be easily installed from pip (see webpage documentation). Consider the following code, which is based on the example:
import numpy as np
import numpy.random
import matplotlib.pyplot as plt
import sphviewer as sph
def myplot(x, y, nb=32, xsize=500, ysize=500):
xmin = np.min(x)
xmax = np.max(x)
ymin = np.min(y)
ymax = np.max(y)
x0 = (xmin+xmax)/2.
y0 = (ymin+ymax)/2.
pos = np.zeros([len(x),3])
pos[:,0] = x
pos[:,1] = y
w = np.ones(len(x))
P = sph.Particles(pos, w, nb=nb)
S = sph.Scene(P)
S.update_camera(r='infinity', x=x0, y=y0, z=0,
xsize=xsize, ysize=ysize)
R = sph.Render(S)
R.set_logscale()
img = R.get_image()
extent = R.get_extent()
for i, j in zip(xrange(4), [x0,x0,y0,y0]):
extent[i] += j
print extent
return img, extent
fig = plt.figure(1, figsize=(10,10))
ax1 = fig.add_subplot(221)
ax2 = fig.add_subplot(222)
ax3 = fig.add_subplot(223)
ax4 = fig.add_subplot(224)
# Generate some test data
x = np.random.randn(1000)
y = np.random.randn(1000)
#Plotting a regular scatter plot
ax1.plot(x,y,'k.', markersize=5)
ax1.set_xlim(-3,3)
ax1.set_ylim(-3,3)
heatmap_16, extent_16 = myplot(x,y, nb=16)
heatmap_32, extent_32 = myplot(x,y, nb=32)
heatmap_64, extent_64 = myplot(x,y, nb=64)
ax2.imshow(heatmap_16, extent=extent_16, origin='lower', aspect='auto')
ax2.set_title("Smoothing over 16 neighbors")
ax3.imshow(heatmap_32, extent=extent_32, origin='lower', aspect='auto')
ax3.set_title("Smoothing over 32 neighbors")
#Make the heatmap using a smoothing over 64 neighbors
ax4.imshow(heatmap_64, extent=extent_64, origin='lower', aspect='auto')
ax4.set_title("Smoothing over 64 neighbors")
plt.show()
which produces the following image:
As you see, the images look pretty nice, and we are able to identify different substructures on it. These images are constructed spreading a given weight for every point within a certain domain, defined by the smoothing length, which in turns is given by the distance to the closer nb neighbor (I've chosen 16, 32 and 64 for the examples). So, higher density regions typically are spread over smaller regions compared to lower density regions.
The function myplot is just a very simple function that I've written in order to give the x,y data to py-sphviewer to do the magic.
If you are using 1.2.x
import numpy as np
import matplotlib.pyplot as plt
x = np.random.randn(100000)
y = np.random.randn(100000)
plt.hist2d(x,y,bins=100)
plt.show()
Seaborn now has the jointplot function which should work nicely here:
import numpy as np
import seaborn as sns
import matplotlib.pyplot as plt
# Generate some test data
x = np.random.randn(8873)
y = np.random.randn(8873)
sns.jointplot(x=x, y=y, kind='hex')
plt.show()
Here's Jurgy's great nearest neighbour approach but implemented using scipy.cKDTree. In my tests it's about 100x faster.
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from scipy.spatial import cKDTree
def data_coord2view_coord(p, resolution, pmin, pmax):
dp = pmax - pmin
dv = (p - pmin) / dp * resolution
return dv
n = 1000
xs = np.random.randn(n)
ys = np.random.randn(n)
resolution = 250
extent = [np.min(xs), np.max(xs), np.min(ys), np.max(ys)]
xv = data_coord2view_coord(xs, resolution, extent[0], extent[1])
yv = data_coord2view_coord(ys, resolution, extent[2], extent[3])
def kNN2DDens(xv, yv, resolution, neighbours, dim=2):
"""
"""
# Create the tree
tree = cKDTree(np.array([xv, yv]).T)
# Find the closest nnmax-1 neighbors (first entry is the point itself)
grid = np.mgrid[0:resolution, 0:resolution].T.reshape(resolution**2, dim)
dists = tree.query(grid, neighbours)
# Inverse of the sum of distances to each grid point.
inv_sum_dists = 1. / dists[0].sum(1)
# Reshape
im = inv_sum_dists.reshape(resolution, resolution)
return im
fig, axes = plt.subplots(2, 2, figsize=(15, 15))
for ax, neighbours in zip(axes.flatten(), [0, 16, 32, 63]):
if neighbours == 0:
ax.plot(xs, ys, 'k.', markersize=5)
ax.set_aspect('equal')
ax.set_title("Scatter Plot")
else:
im = kNN2DDens(xv, yv, resolution, neighbours)
ax.imshow(im, origin='lower', extent=extent, cmap=cm.Blues)
ax.set_title("Smoothing over %d neighbours" % neighbours)
ax.set_xlim(extent[0], extent[1])
ax.set_ylim(extent[2], extent[3])
plt.savefig('new.png', dpi=150, bbox_inches='tight')
and the initial question was... how to convert scatter values to grid values, right?
histogram2d does count the frequency per cell, however, if you have other data per cell than just the frequency, you'd need some additional work to do.
x = data_x # between -10 and 4, log-gamma of an svc
y = data_y # between -4 and 11, log-C of an svc
z = data_z #between 0 and 0.78, f1-values from a difficult dataset
So, I have a dataset with Z-results for X and Y coordinates. However, I was calculating few points outside the area of interest (large gaps), and heaps of points in a small area of interest.
Yes here it becomes more difficult but also more fun. Some libraries (sorry):
from matplotlib import pyplot as plt
from matplotlib import cm
import numpy as np
from scipy.interpolate import griddata
pyplot is my graphic engine today,
cm is a range of color maps with some initeresting choice.
numpy for the calculations,
and griddata for attaching values to a fixed grid.
The last one is important especially because the frequency of xy points is not equally distributed in my data. First, let's start with some boundaries fitting to my data and an arbitrary grid size. The original data has datapoints also outside those x and y boundaries.
#determine grid boundaries
gridsize = 500
x_min = -8
x_max = 2.5
y_min = -2
y_max = 7
So we have defined a grid with 500 pixels between the min and max values of x and y.
In my data, there are lots more than the 500 values available in the area of high interest; whereas in the low-interest-area, there are not even 200 values in the total grid; between the graphic boundaries of x_min and x_max there are even less.
So for getting a nice picture, the task is to get an average for the high interest values and to fill the gaps elsewhere.
I define my grid now. For each xx-yy pair, i want to have a color.
xx = np.linspace(x_min, x_max, gridsize) # array of x values
yy = np.linspace(y_min, y_max, gridsize) # array of y values
grid = np.array(np.meshgrid(xx, yy.T))
grid = grid.reshape(2, grid.shape[1]*grid.shape[2]).T
Why the strange shape? scipy.griddata wants a shape of (n, D).
Griddata calculates one value per point in the grid, by a predefined method.
I choose "nearest" - empty grid points will be filled with values from the nearest neighbor. This looks as if the areas with less information have bigger cells (even if it is not the case). One could choose to interpolate "linear", then areas with less information look less sharp. Matter of taste, really.
points = np.array([x, y]).T # because griddata wants it that way
z_grid2 = griddata(points, z, grid, method='nearest')
# you get a 1D vector as result. Reshape to picture format!
z_grid2 = z_grid2.reshape(xx.shape[0], yy.shape[0])
And hop, we hand over to matplotlib to display the plot
fig = plt.figure(1, figsize=(10, 10))
ax1 = fig.add_subplot(111)
ax1.imshow(z_grid2, extent=[x_min, x_max,y_min, y_max, ],
origin='lower', cmap=cm.magma)
ax1.set_title("SVC: empty spots filled by nearest neighbours")
ax1.set_xlabel('log gamma')
ax1.set_ylabel('log C')
plt.show()
Around the pointy part of the V-Shape, you see I did a lot of calculations during my search for the sweet spot, whereas the less interesting parts almost everywhere else have a lower resolution.
Make a 2-dimensional array that corresponds to the cells in your final image, called say heatmap_cells and instantiate it as all zeroes.
Choose two scaling factors that define the difference between each array element in real units, for each dimension, say x_scale and y_scale. Choose these such that all your datapoints will fall within the bounds of the heatmap array.
For each raw datapoint with x_value and y_value:
heatmap_cells[floor(x_value/x_scale),floor(y_value/y_scale)]+=1
Very similar to #Piti's answer, but using 1 call instead of 2 to generate the points:
import numpy as np
import matplotlib.pyplot as plt
pts = 1000000
mean = [0.0, 0.0]
cov = [[1.0,0.0],[0.0,1.0]]
x,y = np.random.multivariate_normal(mean, cov, pts).T
plt.hist2d(x, y, bins=50, cmap=plt.cm.jet)
plt.show()
Output:
Here's one I made on a 1 Million point set with 3 categories (colored Red, Green, and Blue). Here's a link to the repository if you'd like to try the function. Github Repo
histplot(
X,
Y,
labels,
bins=2000,
range=((-3,3),(-3,3)),
normalize_each_label=True,
colors = [
[1,0,0],
[0,1,0],
[0,0,1]],
gain=50)
I'm afraid I'm a little late to the party but I had a similar question a while ago. The accepted answer (by #ptomato) helped me out but I'd also want to post this in case it's of use to someone.
''' I wanted to create a heatmap resembling a football pitch which would show the different actions performed '''
import numpy as np
import matplotlib.pyplot as plt
import random
#fixing random state for reproducibility
np.random.seed(1234324)
fig = plt.figure(12)
ax1 = fig.add_subplot(121)
ax2 = fig.add_subplot(122)
#Ratio of the pitch with respect to UEFA standards
hmap= np.full((6, 10), 0)
#print(hmap)
xlist = np.random.uniform(low=0.0, high=100.0, size=(20))
ylist = np.random.uniform(low=0.0, high =100.0, size =(20))
#UEFA Pitch Standards are 105m x 68m
xlist = (xlist/100)*10.5
ylist = (ylist/100)*6.5
ax1.scatter(xlist,ylist)
#int of the co-ordinates to populate the array
xlist_int = xlist.astype (int)
ylist_int = ylist.astype (int)
#print(xlist_int, ylist_int)
for i, j in zip(xlist_int, ylist_int):
#this populates the array according to the x,y co-ordinate values it encounters
hmap[j][i]= hmap[j][i] + 1
#Reversing the rows is necessary
hmap = hmap[::-1]
#print(hmap)
im = ax2.imshow(hmap)
Here's the result
None of these solutions worked for my application, so this is what I came up with. Essentially I am placing a 2D Gaussian at every single point:
import cv2
import numpy as np
import matplotlib.pyplot as plt
def getGaussian2D(ksize, sigma, norm=True):
oneD = cv2.getGaussianKernel(ksize=ksize, sigma=sigma)
twoD = np.outer(oneD.T, oneD)
return twoD / np.sum(twoD) if norm else twoD
def pt2heat(pts, shape, kernel=16, sigma=5):
heat = np.zeros(shape)
k = getGaussian2D(kernel, sigma)
for y,x in pts:
x, y = int(x), int(y)
for i in range(-kernel//2, kernel//2):
for j in range(-kernel//2, kernel//2):
if 0 <= x+i < shape[0] and 0 <= y+j < shape[1]:
heat[x+i, y+j] = heat[x+i, y+j] + k[i+kernel//2, j+kernel//2]
return heat
heat = pts2heat(pts, img.shape[:2])
plt.imshow(heat, cmap='heat')
Here are the points overlayed ontop of it's associated image, along with the resulting heat map: