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How to convert a matrix to heatmap image in torch [duplicate]

Using Matplotlib, I want to plot a 2D heat map. My data is an n-by-n Numpy array, each with a value between 0 and 1. So for the (i, j) element of this array, I want to plot a square at the (i, j) coordinate in my heat map, whose color is proportional to the element's value in the array.
How can I do this?

The imshow() function with parameters interpolation='nearest' and cmap='hot' should do what you want.
Please review the interpolation parameter details, and see Interpolations for imshow and Image antialiasing.
import matplotlib.pyplot as plt
import numpy as np
a = np.random.random((16, 16))
plt.imshow(a, cmap='hot', interpolation='nearest')
plt.show()

Seaborn is a high-level API for matplotlib, which takes care of a lot of the manual work.
seaborn.heatmap automatically plots a gradient at the side of the chart etc.
import numpy as np
import seaborn as sns
import matplotlib.pylab as plt
uniform_data = np.random.rand(10, 12)
ax = sns.heatmap(uniform_data, linewidth=0.5)
plt.show()
You can even plot upper / lower left / right triangles of square matrices. For example, a correlation matrix, which is square and is symmetric, so plotting all values would be redundant.
corr = np.corrcoef(np.random.randn(10, 200))
mask = np.zeros_like(corr)
mask[np.triu_indices_from(mask)] = True
with sns.axes_style("white"):
ax = sns.heatmap(corr, mask=mask, vmax=.3, square=True, cmap="YlGnBu")
plt.show()

I would use matplotlib's pcolor/pcolormesh function since it allows nonuniform spacing of the data.
Example taken from matplotlib:
import matplotlib.pyplot as plt
import numpy as np
# generate 2 2d grids for the x & y bounds
y, x = np.meshgrid(np.linspace(-3, 3, 100), np.linspace(-3, 3, 100))
z = (1 - x / 2. + x ** 5 + y ** 3) * np.exp(-x ** 2 - y ** 2)
# x and y are bounds, so z should be the value *inside* those bounds.
# Therefore, remove the last value from the z array.
z = z[:-1, :-1]
z_min, z_max = -np.abs(z).max(), np.abs(z).max()
fig, ax = plt.subplots()
c = ax.pcolormesh(x, y, z, cmap='RdBu', vmin=z_min, vmax=z_max)
ax.set_title('pcolormesh')
# set the limits of the plot to the limits of the data
ax.axis([x.min(), x.max(), y.min(), y.max()])
fig.colorbar(c, ax=ax)
plt.show()

For a 2d numpy array, simply use imshow() may help you:
import matplotlib.pyplot as plt
import numpy as np
def heatmap2d(arr: np.ndarray):
plt.imshow(arr, cmap='viridis')
plt.colorbar()
plt.show()
test_array = np.arange(100 * 100).reshape(100, 100)
heatmap2d(test_array)
This code produces a continuous heatmap.
You can choose another built-in colormap from here.

Here's how to do it from a csv:
import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import griddata
# Load data from CSV
dat = np.genfromtxt('dat.xyz', delimiter=' ',skip_header=0)
X_dat = dat[:,0]
Y_dat = dat[:,1]
Z_dat = dat[:,2]
# Convert from pandas dataframes to numpy arrays
X, Y, Z, = np.array([]), np.array([]), np.array([])
for i in range(len(X_dat)):
X = np.append(X, X_dat[i])
Y = np.append(Y, Y_dat[i])
Z = np.append(Z, Z_dat[i])
# create x-y points to be used in heatmap
xi = np.linspace(X.min(), X.max(), 1000)
yi = np.linspace(Y.min(), Y.max(), 1000)
# Interpolate for plotting
zi = griddata((X, Y), Z, (xi[None,:], yi[:,None]), method='cubic')
# I control the range of my colorbar by removing data
# outside of my range of interest
zmin = 3
zmax = 12
zi[(zi<zmin) | (zi>zmax)] = None
# Create the contour plot
CS = plt.contourf(xi, yi, zi, 15, cmap=plt.cm.rainbow,
vmax=zmax, vmin=zmin)
plt.colorbar()
plt.show()
where dat.xyz is in the form
x1 y1 z1
x2 y2 z2
...

Use matshow() which is a wrapper around imshow to set useful defaults for displaying a matrix.
a = np.diag(range(15))
plt.matshow(a)
https://matplotlib.org/stable/api/_as_gen/matplotlib.axes.Axes.matshow.html
This is just a convenience function wrapping imshow to set useful defaults for displaying a matrix. In particular:
Set origin='upper'.
Set interpolation='nearest'.
Set aspect='equal'.
Ticks are placed to the left and above.
Ticks are formatted to show integer indices.

Here is a new python package to plot complex heatmaps with different kinds of row/columns annotations in Python: https://github.com/DingWB/PyComplexHeatmap

Matplotlib contour hatching not working if only two levels was used

I am trying to plot hatches over contours lines that
statisfy certian criteria folliwng the example found here. Yet, I got regular contours (the yellow lines) instead of the hatches. Any ideas how to resolve that. Thanks
import matplotlib.pyplot as plt
import numpy as np
# invent some numbers, turning the x and y arrays into simple
# 2d arrays, which make combining them together easier.
x = np.linspace(-3, 5, 150).reshape(1, -1)
y = np.linspace(-3, 5, 120).reshape(-1, 1)
z = np.cos(x) + np.sin(y)
# we no longer need x and y to be 2 dimensional, so flatten them.
x, y = x.flatten(), y.flatten()
fig2, ax2 = plt.subplots()
n_levels = 6
a=ax2.contourf(x, y, z, n_levels)
fig2.colorbar(a)
[m,n]=np.where(z > 0.5)
z1=np.zeros(z.shape)
z1[m,n]=99
cs = ax2.contour(x, y, z1,2,hatches=['','.'])
plt.show()enter code here

Use contourf() with proper parameters to get useful plot with hatching. See important comment within the working code below:
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(-3, 5, 150).reshape(1, -1)
y = np.linspace(-3, 5, 120).reshape(-1, 1)
z = np.cos(x) + np.sin(y)
x, y = x.flatten(), y.flatten()
fig2, ax2 = plt.subplots()
n_levels = 6
a = ax2.contourf(x, y, z, n_levels)
fig2.colorbar(a)
[m,n] = np.where(z > 0.5)
z1=np.zeros(z.shape)
z1[m, n] = 99
# use contourf() with proper hatch pattern and alpha value
cs = ax2.contourf(x, y, z1 ,3 , hatches=['', '..'], alpha=0.25)
plt.show()
The output plot:

How to change the length of axes for 3D plots in matplotlib [duplicate]

I have this so far:
x,y,z = data.nonzero()
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")
Which creates:
What I'd like to do is stretch this out to make the Z axis 9 times taller and keep X and Y the same. I'd like to keep the same coordinates though.
So far I tried this guy:
fig = plt.figure(figsize=(4.,35.))
But that just stretches out the plot.png image.

The code example below provides a way to scale each axis relative to the others. However, to do so you need to modify the Axes3D.get_proj function. Below is an example based on the example provided by matplot lib: http://matplotlib.org/1.4.0/mpl_toolkits/mplot3d/tutorial.html#line-plots
(There is a shorter version at the end of this answer)
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
#Make sure these are floating point values:
scale_x = 1.0
scale_y = 2.0
scale_z = 3.0
#Axes are scaled down to fit in scene
max_scale=max(scale_x, scale_y, scale_z)
scale_x=scale_x/max_scale
scale_y=scale_y/max_scale
scale_z=scale_z/max_scale
#Create scaling matrix
scale = np.array([[scale_x,0,0,0],
[0,scale_y,0,0],
[0,0,scale_z,0],
[0,0,0,1]])
print scale
def get_proj_scale(self):
"""
Create the projection matrix from the current viewing position.
elev stores the elevation angle in the z plane
azim stores the azimuth angle in the x,y plane
dist is the distance of the eye viewing point from the object
point.
"""
relev, razim = np.pi * self.elev/180, np.pi * self.azim/180
xmin, xmax = self.get_xlim3d()
ymin, ymax = self.get_ylim3d()
zmin, zmax = self.get_zlim3d()
# transform to uniform world coordinates 0-1.0,0-1.0,0-1.0
worldM = proj3d.world_transformation(
xmin, xmax,
ymin, ymax,
zmin, zmax)
# look into the middle of the new coordinates
R = np.array([0.5, 0.5, 0.5])
xp = R[0] + np.cos(razim) * np.cos(relev) * self.dist
yp = R[1] + np.sin(razim) * np.cos(relev) * self.dist
zp = R[2] + np.sin(relev) * self.dist
E = np.array((xp, yp, zp))
self.eye = E
self.vvec = R - E
self.vvec = self.vvec / proj3d.mod(self.vvec)
if abs(relev) > np.pi/2:
# upside down
V = np.array((0, 0, -1))
else:
V = np.array((0, 0, 1))
zfront, zback = -self.dist, self.dist
viewM = proj3d.view_transformation(E, R, V)
perspM = proj3d.persp_transformation(zfront, zback)
M0 = np.dot(viewM, worldM)
M = np.dot(perspM, M0)
return np.dot(M, scale);
Axes3D.get_proj=get_proj_scale
"""
You need to include all the code above.
From here on you should be able to plot as usual.
"""
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
ax.plot(x, y, z, label='parametric curve')
ax.legend()
plt.show()
Standard output:
Scaled by (1, 2, 3):
Scaled by (1, 1, 3):
The reason I particularly like this method,
Swap z and x, scale by (3, 1, 1):
Below is a shorter version of the code.
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
"""
Scaling is done from here...
"""
x_scale=1
y_scale=1
z_scale=2
scale=np.diag([x_scale, y_scale, z_scale, 1.0])
scale=scale*(1.0/scale.max())
scale[3,3]=1.0
def short_proj():
return np.dot(Axes3D.get_proj(ax), scale)
ax.get_proj=short_proj
"""
to here
"""
ax.plot(z, y, x, label='parametric curve')
ax.legend()
plt.show()

Please note that the answer below simplifies the patch, but uses the same underlying principle as the answer by #ChristianSarofeen.
Solution
As already indicated in other answers, it is not a feature that is currently implemented in matplotlib. However, since what you are requesting is simply a 3D transformation that can be applied to the existing projection matrix used by matplotlib, and thanks to the wonderful features of Python, this problem can be solved with a simple oneliner:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([scale_x, scale_y, scale_z, 1]))
where scale_x, scale_y and scale_z are values from 0 to 1 that will re-scale your plot along each of the axes accordingly. ax is simply the 3D axes which can be obtained with ax = fig.gca(projection='3d')
Explanation
To explain, the function get_proj of Axes3D generates the projection matrix from the current viewing position. Multiplying it by a scaling matrix:
scale_x, 0, 0
0, scale_y, 0
0, 0, scale_z
0, 0, 1
includes the scaling into the projection used by the renderer. So, what we are doing here is substituting the original get_proj function with an expression taking the result of the original get_proj and multiplying it by the scaling matrix.
Example
To illustrate the result with the standard parametric function example:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z ** 2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
# OUR ONE LINER ADDED HERE:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([0.5, 0.5, 1, 1]))
ax.plot(x, y, z)
plt.show()
for values 0.5, 0.5, 1, we get:
while for values 0.2, 1.0, 0.2, we get:

In my case I wanted to stretch z-axis 2 times for better point visibility
from mpl_toolkits import mplot3d
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
# plt.rcParams["figure.figsize"] = (10,200)
# plt.rcParams["figure.autolayout"] = True
ax = plt.axes(projection='3d')
ax.set_box_aspect(aspect = (1,1,2))
ax.plot(dataX,dataY,dataZ)

I looks like by default, mplot3d will leave quite a bit of room at the top and bottom of a very tall plot. But, you can trick it into filling that space using fig.subplots_adjust, and extending the top and bottom out of the normal plotting area (i.e. top > 1 and bottom < 0). Some trial and error here is probably needed for your particular plot.
I've created some random arrays for x, y, and z with limits similar to your plot, and have found the parameters below (bottom=-0.15, top = 1.2) seem to work ok.
You might also want to change ax.view_init to set a nice viewing angle.
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
from numpy import random
# Make some random data with similar limits to the OP's example
x,y,z=random.rand(3,100)
z*=250
y*=800
y+=900
x*=350
x+=1200
fig=plt.figure(figsize=(4,35))
# Set the bottom and top outside the actual figure limits,
# to stretch the 3D axis
fig.subplots_adjust(bottom=-0.15,top=1.2)
ax = fig.add_subplot(111, projection='3d')
# Change the viewing angle to an agreeable one
ax.view_init(2,None)
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")

Sounds like you're trying to adjust the scale of the plot. I don't think there's a way to stretch a linear scale to user specifications, but you can use set_yscale(), set_xscale(), set_zscale() to alter the scales with respect to each other.
Intuitively, set_yscale(log), set_xscale(log), set_zscale(linear) might solve your problems.
A likely better option: specify a stretch, set them all to symlog with the same log base and then specify the Z-axis's symlog scale with the linscalex/linscaley kwargs to your specifications.
More here:
http://matplotlib.org/mpl_toolkits/mplot3d/api.html

I found this while searching on a similar problem. After experimenting a bit, perhaps I can share some of my prelim findings here..matplotlib library is VAST!! (am a newcomer). Note that quite akin to this question, all i wanted was to 'visually' stretch the chart without distorting it.
Background story (only key code snippets are shown to avoid unnecessary clutter for those who know the library, and if you want a run-able code please drop a comment):
I have three 1-d ndarrays representing the X,Y and Z data points respectively. Clearly I can't use plot_surface (as it requires 2d ndarrays for each dim) so I went for the extremely useful plot_trisurf:
fig = plt.figure()
ax = Axes3D(fig)
3d_surf_obj = ax.plot_trisurf(X, Y, Z_defl, cmap=cm.jet,linewidth=0,antialiased=True)
You can think of the plot like a floating barge deforming in waves...As you can see, the axes stretch make it pretty deceiving visually (note that x is supposed to be at x6 times longer than y and >>>>> z). While the plot points are correct, I wanted something more visually 'stretched' at the very least. Was looking for A QUICK FIX, if I may. Long story cut short, I found a bit of success with...'figure.figsize' general setting (see snippet below).
matplotlib.rcParams.update({'font.serif': 'Times New Roman',
'font.size': 10.0,
'axes.labelsize': 'Medium',
'axes.labelweight': 'normal',
'axes.linewidth': 0.8,
###########################################
# THIS IS THE IMPORTANT ONE FOR STRETCHING
# default is [6,4] but...i changed it to
'figure.figsize':[15,5] # THIS ONE #
})
For [15,5] I got something like...
Pretty neat!!
So I started to push it.... and got up to [20,6] before deciding to settle there..
If you want to try for visually stretching the vertical axis, try with ratios like... [7,10], which in this case gives me ...
Not too shabby !
Should do it for visual prowess.

Multiply all your z values by 9,
ax.scatter(x, y, 9*z, zdir='z', c= 'red')
And then give the z-axis custom plot labels and spacing.
ax.ZTick = [0,-9*50, -9*100, -9*150, -9*200];
ax.ZTickLabel = {'0','-50','-100','-150','-200'};

How to draw circle by data with matplotlib + python?

I can draw a circle by scatter, which has been shown in the image. But I want to draw them buy a line, because there are many circles in total, I need to link nodes together for a certain circle. Thanks.

I the order of the points is random, you can change X-Y to polar, and sort the data by angle:
create some random order points first:
import pylab as pl
import numpy as np
angle = np.arange(0, np.pi*2, 0.05)
r = 50 + np.random.normal(0, 2, angle.shape)
x = r * np.cos(angle)
y = r * np.sin(angle)
idx = np.random.permutation(angle.shape[0])
x = x[idx]
y = y[idx]
Then use arctan2() to calculate the angle, and sort the data by it:
angle = np.arctan2(x, y)
order = np.argsort(angle)
x = x[order]
y = y[order]
fig, ax = pl.subplots()
ax.set_aspect(1.0)
x2 = np.r_[x, x[0]]
y2 = np.r_[y, y[0]]
ax.plot(x, y, "o")
ax.plot(x2, y2, "r", lw=2)
here is the output:

Here is one way to do it. This answer uses different methods than the linked possible duplicate, so may be worth keeping.
import matplotlib.pyplot as plt
from matplotlib import patches
fig = plt.figure(figsize=plt.figaspect(1.0))
ax = fig.add_subplot(111)
cen = (2.0,1.0); r = 3.0
circle = patches.Circle(cen, r, facecolor='none')
ax.add_patch(circle)
ax.set_xlim(-6.0, 6.0)
ax.set_ylim(-6.0, 6.0)
If all you have are the x and y points, you can use PathPatch. Here's a tutorial

If your data points are already in order, the plot command should work fine. If you're looking to generate a circle from scratch, you can use a parametric equation.
>>> import numpy as np
>>> import matplotlib.pyplot as plt
>>> t = np.linspace(0,2*np.pi, 100)
>>> x = np.cos(t)
>>> y = np.sin(t)
>>> plt.plot(x,y)

How to visualize scalar 2D data with Matplotlib?

So i have a meshgrid (matrices X and Y) together with scalar data (matrix Z), and i need to visualize this. Preferably some 2D image with colors at the points showing the value of Z there.
I've done some research but haven't found anything which does exactly what i want.
pyplot.imshow(Z) has a good look, but it doesn't take my X and Y matrices, so the axes are wrong and it is unable to handle non-linearly spaced points given by X and Y.
pyplot.pcolor(X,Y,Z) makes colored squares with colors corresponding to the data at one of its corners, so it kind of misrepresents the data (it should show the data in its center or something). In addition it ignores two of the edges from the data matrix.
I pretty sure there must exist some better way somewhere in Matplotlib, but the documentation makes it hard to get an overview. So i'm asking if someone else knows of a better way. Bonus if it allows me to refresh the matrix Z to make an animation.

This looks nice, but it's inefficient:
from pylab import *
origin = 'lower'
delta = 0.025
x = y = arange(-3.0, 3.01, delta)
X, Y = meshgrid(x, y)
Z1 = bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
Z = 10 * (Z1 - Z2)
nr, nc = Z.shape
CS = contourf(
X, Y, Z,
levels = linspace(Z.min(), Z.max(), len(x)),
ls = '-',
cmap=cm.bone,
origin=origin)
CS1 = contour(
CS,
levels = linspace(Z.min(), Z.max(), len(x)),
ls = '-',
cmap=cm.bone,
origin=origin)
show()
It it were me, I'd re-interpolate (using scipy.interpolate) the data to a regular grid and use imshow(), setting the extents to fix the axes.
Edit (per comment):
Animating a contour plot can be accomplished like this, but, like I said, the above is inefficient just plain abuse of the contour plot function. The most efficient way to do what you want is to employ SciPy. Do you have that installed?
import matplotlib
matplotlib.use('TkAgg') # do this before importing pylab
import time
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)
def animate():
origin = 'lower'
delta = 0.025
x = y = arange(-3.0, 3.01, delta)
X, Y = meshgrid(x, y)
Z1 = bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
Z = 10 * (Z1 - Z2)
CS1 = ax.contourf(
X, Y, Z,
levels = linspace(Z.min(), Z.max(), 10),
cmap=cm.bone,
origin=origin)
for i in range(10):
tempCS1 = contourf(
X, Y, Z,
levels = linspace(Z.min(), Z.max(), 10),
cmap=cm.bone,
origin=origin)
del tempCS1
fig.canvas.draw()
time.sleep(0.1)
Z += x/10
win = fig.canvas.manager.window
fig.canvas.manager.window.after(100, animate)
plt.show()

If your meshgrid has uniform spacing, you could continue to use pcolor, but just shift X and Y for the purposes of centering the data at the particular values rather than at the corners.
You could also use a scatter plot to explicitly place points of some size at the exact X and Y points and then set the color to Z:
x = numpy.arange(10)
y = numpy.arange(10)
X,Y = numpy.meshgrid(x,y)
Z = numpy.arange(100).reshape((10,10))
scatter(X,Y,c=Z,marker='s',s=1500)
#I picked a marker size that basically overlapped the symbols at the edges
axis('equal')
or:
pcolor(X+0.5,Y+0.5,Z)
axis('equal')
or as Paul suggested, using one of the contour functions

In case anyone comes across this article looking for what I was looking for, I took the above example and modified it to use imshow with an input stack of frames, instead of generating and using contours on the fly. Starting with a 3D array of images of shape (nBins, nBins, nBins), called frames.
def animate_frames(frames):
nBins = frames.shape[0]
frame = frames[0]
tempCS1 = plt.imshow(frame, cmap=plt.cm.gray)
for k in range(nBins):
frame = frames[k]
tempCS1 = plt.imshow(frame, cmap=plt.cm.gray)
del tempCS1
fig.canvas.draw()
#time.sleep(1e-2) #unnecessary, but useful
fig.clf()
fig = plt.figure()
ax = fig.add_subplot(111)
win = fig.canvas.manager.window
fig.canvas.manager.window.after(100, animate_frames, frames)
I also found a much simpler way to go about this whole process, albeit less robust:
fig = plt.figure()
for k in range(nBins):
plt.clf()
plt.imshow(frames[k],cmap=plt.cm.gray)
fig.canvas.draw()
time.sleep(1e-6) #unnecessary, but useful
Note that both of these only seem to work with ipython --pylab=tk, a.k.a.backend = TkAgg
Thank you for the help with everything.

The following function creates boxes of half the size at the boundary (as shown in the attached picture).
import matplotlib.pyplot as plt
import numpy as np
from scipy.ndimage.filters import convolve
def pcolor_all(X, Y, C, **kwargs):
X = np.concatenate([X[0:1,:], X], axis=0)
X = np.concatenate([X[:,0:1], X], axis=1)
Y = np.concatenate([Y[0:1,:], Y], axis=0)
Y = np.concatenate([Y[:,0:1], Y], axis=1)
X = convolve(X, [[1,1],[1,1]])/4
Y = convolve(Y, [[1,1],[1,1]])/4
plt.pcolor(X, Y, C, **kwargs)
X, Y = np.meshgrid(
[-1,-0.5,0,0.5,1],
[-2,-1,0,1,2])
C = X**2-Y**2
plt.figure(figsize=(4,4))
pcolor_all(X, Y, C, cmap='gray')
plt.savefig('plot.png')