I have a 3 1D arrays: radius, angle and temperature. Together they form a 2D temperature map of a ring.
The arrays take the form:
r = [0,0,0,1,1,1,2,2,2]
th = [0.,0.78539816,1.57079633,2.35619449,3.14159265,3.92699082,4.71238898,5.49778714,6.28318531]
z = [-1,2,5,2,4,-1,3,2,3]
I don't understand how I can make those z data fall on the right coordinates.
I can make it work, with random numbers, using the following simple code:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
fig = plt.figure()
ax = Axes3D(fig)
rad = np.linspace(.2, 1, 4)
azm = np.linspace(0, 2 * np.pi, 9)
r, th = np.meshgrid(rad, azm)
z = np.random.rand(9,4) ** th * r
ax0 = plt.subplot(projection="polar")
im = plt.pcolormesh(th, r, z, cmap='bwr')
plt.plot(azm, r, color='k', ls='none')
plt.axis('off')
cbar = fig.colorbar(im)
ax0.set_title('3 radii polar heat map')
This is how my example code comes out
I ended splitting the list z into 3 lists and made a matrix out of them using z_mat = np.array([z1,z2,z3]). I took care that the lists for rad and azm contained one item more than z, which is a requirement for pcolormesh. After that I transposed the matrix to have the same dimensions as r and th using z_mat_trans = z_mat.transpose()
Related
I have a 3d plot made using matplotlib. I now want to fill the vertical space between the drawn line and the x,y axis to highlight the height of the line on the z axis. On a 2d plot this would be done with fill_between but there does not seem to be anything similar for a 3d plot. Can anyone help?
here is my current code
from stravalib import Client
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
... code to get the data ....
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure()
ax = fig.gca(projection='3d')
zi = alt
x = df['x'].tolist()
y = df['y'].tolist()
ax.plot(x, y, zi, label='line')
ax.legend()
plt.show()
and the current plot
just to be clear I want a vertical fill to the x,y axis intersection NOT this...
You're right. It seems that there is no equivalent in 3D plot for the 2D plot function fill_between. The solution I propose is to convert your data in 3D polygons. Here is the corresponding code:
import math as mt
import matplotlib.pyplot as pl
import numpy as np
import random as rd
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
# Parameter (reference height)
h = 0.0
# Code to generate the data
n = 200
alpha = 0.75 * mt.pi
theta = [alpha + 2.0 * mt.pi * (float(k) / float(n)) for k in range(0, n + 1)]
xs = [1.0 * mt.cos(k) for k in theta]
ys = [1.0 * mt.sin(k) for k in theta]
zs = [abs(k - alpha - mt.pi) * rd.random() for k in theta]
# Code to convert data in 3D polygons
v = []
for k in range(0, len(xs) - 1):
x = [xs[k], xs[k+1], xs[k+1], xs[k]]
y = [ys[k], ys[k+1], ys[k+1], ys[k]]
z = [zs[k], zs[k+1], h, h]
#list is necessary in python 3/remove for python 2
v.append(list(zip(x, y, z)))
poly3dCollection = Poly3DCollection(v)
# Code to plot the 3D polygons
fig = pl.figure()
ax = Axes3D(fig)
ax.add_collection3d(poly3dCollection)
ax.set_xlim([min(xs), max(xs)])
ax.set_ylim([min(ys), max(ys)])
ax.set_zlim([min(zs), max(zs)])
ax.set_xlabel("x")
ax.set_ylabel("y")
ax.set_zlabel("z")
pl.show()
It produces the following figure:
I hope this will help you.
I have a spreadsheet file that I would like to input to create a 3D surface graph using Matplotlib in Python.
I used plot_trisurf and it worked, but I need the projections of the contour profiles onto the graph that I can get with the surface function, like this example.
I'm struggling to arrange my Z data in a 2D array that I can use to input in the plot_surface method. I tried a lot of things, but none seems to work.
Here it is what I have working, using plot_trisurf
import matplotlib
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import pandas as pd
df=pd.read_excel ("/Users/carolethais/Desktop/Dissertação Carol/Códigos/Resultados/res_02_0.5.xlsx")
fig = plt.figure()
ax = fig.gca(projection='3d')
# I got the graph using trisurf
graf=ax.plot_trisurf(df["Diametro"],df["Comprimento"], df["temp_out"], cmap=matplotlib.cm.coolwarm)
ax.set_xlim(0, 0.5)
ax.set_ylim(0, 100)
ax.set_zlim(25,40)
fig.colorbar(graf, shrink=0.5, aspect=15)
ax.set_xlabel('Diâmetro (m)')
ax.set_ylabel('Comprimento (m)')
ax.set_zlabel('Temperatura de Saída (ºC)')
plt.show()
This is a part of my df, dataframe:
Diametro Comprimento temp_out
0 0.334294 0.787092 34.801994
1 0.334294 8.187065 32.465551
2 0.334294 26.155976 29.206090
3 0.334294 43.648591 27.792126
4 0.334294 60.768219 27.163233
... ... ... ...
59995 0.437266 14.113660 31.947302
59996 0.437266 25.208851 30.317583
59997 0.437266 33.823035 29.405461
59998 0.437266 57.724209 27.891616
59999 0.437266 62.455890 27.709298
I tried this approach to use the imported data with plot_surface, but what I got was indeed a graph but it didn't work, here it's the way the graph looked with this approach:
Thank you so much
A different approach, based on re-gridding the data, that doesn't require that the original data is specified on a regular grid [deeply inspired by this example;-].
import matplotlib.pyplot as plt
import numpy as np
import matplotlib.tri as tri
from mpl_toolkits.mplot3d import Axes3D
np.random.seed(19880808)
# compute the sombrero over a cloud of random points
npts = 10000
x, y = np.random.uniform(-5, 5, npts), np.random.uniform(-5, 5, npts)
z = np.cos(1.5*np.sqrt(x*x + y*y))/(1+0.33*(x*x+y*y))
# prepare the interpolator
triang = tri.Triangulation(x, y)
interpolator = tri.LinearTriInterpolator(triang, z)
# do the interpolation
xi = yi = np.linspace(-5, 5, 101)
Xi, Yi = np.meshgrid(xi, yi)
Zi = interpolator(Xi, Yi)
# plotting
fig = plt.figure()
ax = fig.gca(projection='3d')
norm = plt.Normalize(-1,1)
ax.plot_surface(Xi, Yi, Zi,
cmap='inferno',
norm=plt.Normalize(-1,1))
plt.show()
plot_trisurf expects x, y, z as 1D arrays while plot_surface expects X, Y, Z as 2D arrays or as x, y, Z with x, y being 1D array and Z a 2D array.
Your data consists of 3 1D arrays, so plotting them with plot_trisurf is immediate but you need to use plot_surface to be able to project the isolines on the coordinate planes... You need to reshape your data.
It seems that you have 60000 data points, in the following I assume that you have a regular grid 300 points in the x direction and 200 points in y — but what is important is the idea of regular grid.
The code below shows
the use of plot_trisurf (with a coarser mesh), similar to your code;
the correct use of reshaping and its application in plot_surface;
note that the number of rows in reshaping corresponds to the number
of points in y and the number of columns to the number of points in x;
and 4. incorrect use of reshaping, the resulting subplots are somehow
similar to the plot you showed, maybe you just need to fix the number
of row and columns.
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
x, y = np.arange(30)/3.-5, np.arange(20)/2.-5
x, y = (arr.flatten() for arr in np.meshgrid(x, y))
z = np.cos(1.5*np.sqrt(x*x + y*y))/(1+0.1*(x*x+y*y))
fig, axes = plt.subplots(2, 2, subplot_kw={"projection" : "3d"})
axes = iter(axes.flatten())
ax = next(axes)
ax.plot_trisurf(x,y,z, cmap='Reds')
ax.set_title('Trisurf')
X, Y, Z = (arr.reshape(20,30) for arr in (x,y,z))
ax = next(axes)
ax.plot_surface(X,Y,Z, cmap='Reds')
ax.set_title('Surface 20×30')
X, Y, Z = (arr.reshape(30,20) for arr in (x,y,z))
ax = next(axes)
ax.plot_surface(X,Y,Z, cmap='Reds')
ax.set_title('Surface 30×20')
X, Y, Z = (arr.reshape(40,15) for arr in (x,y,z))
ax = next(axes)
ax.plot_surface(X,Y,Z, cmap='Reds')
ax.set_title('Surface 40×15')
plt.tight_layout()
plt.show()
Following the example in scikit-image doc I generate a spherical surface mesh with marching cubes algorithm. I want to center the unit sphere shell at the origin defined by the x,y,z grid. However, I cannot do that, since I don't know how to put x,y,z info with mpl_toolkits.mplot3d.art3d.Poly3DCollection. Here is the code:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
from skimage import measure
import numpy as np
x, y, z = np.ogrid[-4:4:20j, -4:4:20j, -4:4:20j]
r = np.sqrt(x ** 2 + y ** 2 + z ** 2)
verts, faces, normals, values = measure.marching_cubes_lewiner(r,level=1)
fig = plt.figure(figsize=(10, 10))
ax = fig.add_subplot(111, projection='3d')
mesh = Poly3DCollection(verts[faces])
mesh.set_edgecolor('k')
ax.add_collection3d(mesh)
plt.show()
The problem is that the marching_cubes_lewiner function does not take x,y,z into account. How can I center the resulting sphere at 0,0,0 as implied by the grid?
The measure.marching_cubes_lewiner takes the indices of the points in the grid for calculating the topology. It does not seem to have a way to specify the actual grid and neither any offset to it.
Hence you may manipulate the resulting verts in the desired way. I.e. one can first multiply by the difference between the grid points, effectively scaling the output, and then add the offset of the grid. In this case the transformation would be newverts = 0.42105 * oldverts - 4.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
from skimage import measure
x, y, z = np.ogrid[-4:4:20j, -4:4:20j, -4:4:20j]
r = np.sqrt(x ** 2 + y ** 2 + z ** 2)
verts, faces, normals, values = measure.marching_cubes_lewiner(r, level=1)
verts *= np.array([np.diff(ar.flat)[0] for ar in [x,y,z]])
verts += np.array([x.min(),y.min(),z.min()])
fig = plt.figure(figsize=(10, 10))
ax = fig.add_subplot(111, projection='3d')
mesh = Poly3DCollection(verts[faces])
mesh.set_edgecolor('k')
ax.add_collection3d(mesh)
ax.set_xlim(-2, 2)
ax.set_ylim(-2, 2)
ax.set_zlim(-2, 2)
plt.show()
I would like to plot a heat map on the unit sphere using the matplotlib library of python. There are several places where this question is discussed. Just like this: Heat Map half-sphere plot
I can do this partially. I can creat the sphere and the heatplot. I have coordinate matrices X,Y and Z, which have the same size. I have another variable of the same size as X, Y and Z, which contains scalars used to creat the heat map. However in case c contains scalars differ from zero in its first and last rows, just one polar cap will be colored but not the other. The code generates the above mentioned result is the next:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
#Creating the theta and phi values.
theta = np.linspace(0,np.pi,100,endpoint=True)
phi = np.linspace(0,np.pi*2,100,endpoint=True)
#Creating the coordinate grid for the unit sphere.
X = np.outer(np.sin(theta),np.cos(phi))
Y = np.outer(np.sin(theta),np.sin(phi))
Z = np.outer(np.cos(theta),np.ones(100))
#Creating a 2D matrix contains the values used to color the unit sphere.
c = np.zeros((100,100))
for i in range(100):
c[0,i] = 100
c[99,i] = 100
#Creat the plot.
fig = plt.figure()
ax = fig.add_subplot(111,projection='3d')
ax.set_axis_off()
ax.plot_surface(X,Y,Z, rstride=1, cstride=1, facecolors=cm.plasma(c/np.amax(c)), alpha=0.22, linewidth=1)
m = cm.ScalarMappable(cmap=cm.plasma)
m.set_array(c)
plt.colorbar(m)
#Show the plot.
plt.show()
The plot which was generated:
Could somebody help me what's going on here?
Thank you for your help in advance!
There are a number of small differences with your example but an
important one, namely the shape of the values array c.
As mentioned in another
answer the grid that
defines the surface is larger (by one in both dimensions) than the
grid that defines the value in each quadrangular patch, so that by
using a smaller array for c it is possible to choose correctly the
bands to color not only with respect to the beginnings of the c
array but also with respect to its ends, as I tried to demonstrate in
the following.
import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
# Creating the theta and phi values.
intervals = 8
ntheta = intervals
nphi = 2*intervals
theta = np.linspace(0, np.pi*1, ntheta+1)
phi = np.linspace(0, np.pi*2, nphi+1)
# Creating the coordinate grid for the unit sphere.
X = np.outer(np.sin(theta), np.cos(phi))
Y = np.outer(np.sin(theta), np.sin(phi))
Z = np.outer(np.cos(theta), np.ones(nphi+1))
# Creating a 2D array to be color-mapped on the unit sphere.
# {X, Y, Z}.shape → (ntheta+1, nphi+1) but c.shape → (ntheta, nphi)
c = np.zeros((ntheta, nphi)) + 0.4
# The poles are different
c[ :1, :] = 0.8
c[-1:, :] = 0.8
# as well as the zones across Greenwich
c[:, :1] = 0.0
c[:, -1:] = 0.0
# Creating the colormap thingies.
cm = mpl.cm.inferno
sm = mpl.cm.ScalarMappable(cmap=cm)
sm.set_array([])
# Creating the plot.
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, facecolors=cm(c), alpha=0.3)
plt.colorbar(m)
# Showing the plot.
plt.show()
The values in the arrays define the edges of the grid. The color of the ith face is determined by the ith value in the color array. However, for n edges you only have n-1 faces, such that the last value is ignored.
E.g. if you have 4 grid values and 4 colors, the plot will have only the first three colors in the grid.
Thus a solution for the above would be to use a color array with one color less than gridpoints in each dimension.
c = np.zeros((99,99))
c[[0,98],:] = 100
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