Develop Reference

Change x- and y-numbering in imshow

I would like to plot a function of two variables in python. Similar to this article, we can obtain an output like
using this code:
from numpy import exp,arange
from pylab import meshgrid,cm,imshow,contour,clabel,colorbar,axis,title,show
from matplotlib import pyplot
# the function that I'm going to plot
def z_func(x,y):
return (1-(x**2+y**3))*exp(-(x**2+y**2)/2)
x = arange(-3.0,3.0,0.1)
y = arange(-3.0,3.0,0.1)
z = [[0] * y.__len__() for i in range(x.__len__())]
for i in range(0, x.__len__()):
for j in range(0, y.__len__()):
z[j][i] = z_func(x[i], y[j])
im = imshow(z,cmap=cm.RdBu, extent = [-3, 3, -3, 3], interpolation = "none", origin='lower') # drawing the function
# adding the Contour lines with labels
cset = contour(z,arange(-1,1.5,0.2),linewidths=2,cmap=cm.Set2)
clabel(cset,inline=True,fmt='%1.1f',fontsize=10)
colorbar(im) # adding the colobar on the right
# latex fashion title
title('$z=(1-x^2+y^3) e^{-(x^2+y^2)/2}$')
show()
As you can see, the x- and y-labels go from 0 to 59 (which is the count of elements in x and y). How can I correct these values such that they range from -3 to 3?
A minor sub-question: Why do I need to "transpose" in z[j][i] = z_func(x[i], y[j])? Does Python treat the first dimension as "column" and the second as "row"?

You're trying to plot both the z-function and the countour plots. You need to add the "extent" parameter to matplotlib.pyplot.countour plot too.
cset = contour(z, arange(-1,1.5,0.2),
extent = [-3, 3, -3, 3],
linewidths = 2,
cmap = cm.Set2)

How do I specify the number of axis points in matplotlib and how do I extract theese points?

I have a small script that creates a matplotlib graph with 2000 random points following a random walk.
I'm wondering if there is a simple way to change the number of points on the y-axis as well as how I can extract these values?
When I run the code below, I get 5 points on the Y-axis but I'm looking for a way to expand this to 20 points as well as creating an array or series with these values. Many thanks in advance.
import matplotlib.pyplot as plt
dims = 1
step_n = 2000
step_set = [-1, 0, 1]
origin = np.zeros((1,dims))
random.seed(30)
step_shape = (step_n,dims)
steps = np.random.choice(a=step_set, size=step_shape)
path = np.concatenate([origin, steps]).cumsum(0)
plt.plot(path)

import matplotlib.pyplot as plt
import numpy as np
import random
dims = 1
step_n = 2000
step_set = [-1, 0, 1]
origin = np.zeros((1,dims))
random.seed(30)
step_shape = (step_n,dims)
steps = np.random.choice(a=step_set, size=step_shape)
path = np.concatenate([origin, steps]).cumsum(0)
#first variant
plt.plot(path)
plt.locator_params(axis='x', nbins=20)
plt.locator_params(axis='y', nbins=20)
You can use locator_params in order to specify the number of ticks. Of course you can retrieve these points. For this you must create a subplot with ax, and then you can get the y_ticks with get_yticks.
#second variant
# create subplot
fig, ax = plt.subplots(1,1, figsize=(20, 11))
img = ax.plot(path)
plt.locator_params(axis='y', nbins=20)
y_values = ax.get_yticks() # y_values is a numpy array with your y values

Fitting a line through 3D x,y,z scatter plot data

I have a handful of data points that cluster along a line in 3d space. I have the x,y,z data in a csv file that I want to import. I would like to find an equation that represents that line, or the plane perpendicular to that line, or whatever is mathematically correct. These data are independent of each other. Maybe there are better ways to do this than what I tried to do but...
I attempted to replicate an old post here that seemed to be doing exactly what I'm trying to do
Fitting a line in 3D
but it seems that maybe updates over the past decade have left the second part of the code not working? Or maybe I'm just doing something wrong. I've included the entire thing that I frankensteined together from this at the bottom. There are two lines that seem to be giving me a problem.
I've snippeted them out here...
import numpy as np
pts = np.add.accumulate(np.random.random((10,3)))
x,y,z = pts.T
# this will find the slope and x-intercept of a plane
# parallel to the y-axis that best fits the data
A_xz = np.vstack((x, np.ones(len(x)))).T
m_xz, c_xz = np.linalg.lstsq(A_xz, z)[0]
# again for a plane parallel to the x-axis
A_yz = np.vstack((y, np.ones(len(y)))).T
m_yz, c_yz = np.linalg.lstsq(A_yz, z)[0]
# the intersection of those two planes and
# the function for the line would be:
# z = m_yz * y + c_yz
# z = m_xz * x + c_xz
# or:
def lin(z):
x = (z - c_xz)/m_xz
y = (z - c_yz)/m_yz
return x,y
#verifying:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
ax = Axes3D(fig)
zz = np.linspace(0,5)
xx,yy = lin(zz)
ax.scatter(x, y, z)
ax.plot(xx,yy,zz)
plt.savefig('test.png')
plt.show()
They return this, but no values...
FutureWarning: rcond parameter will change to the default of machine precision times max(M, N) where M and N are the input matrix dimensions.
To use the future default and silence this warning we advise to pass rcond=None, to keep using the old, explicitly pass rcond=-1.
m_xz, c_xz = np.linalg.lstsq(A_xz, z)[0]
FutureWarning: rcond parameter will change to the default of machine precision times max(M, N) where M and N are the input matrix dimensions.
To use the future default and silence this warning we advise to pass rcond=None, to keep using the old, explicitly pass rcond=-1.
m_yz, c_yz = np.linalg.lstsq(A_yz, z)[0]
I don't know where to go from here. I don't even actually need the plot, I just needed an equation and am ill-equipped to move forward. If anyone knows an easier way to do this, or can point me in the right direction, I'm willing to learn, but I'm very, very lost. Thank you in advance!!
Here is my entire frankensteined code in case that is what is causing the issue.
import pandas as pd
import numpy as np
mydataset = pd.read_csv('line1.csv')
x = mydataset.iloc[:,0]
y = mydataset.iloc[:,1]
z = mydataset.iloc[:,2]
data = np.concatenate((x[:, np.newaxis],
y[:, np.newaxis],
z[:, np.newaxis]),
axis=1)
# Calculate the mean of the points, i.e. the 'center' of the cloud
datamean = data.mean(axis=0)
# Do an SVD on the mean-centered data.
uu, dd, vv = np.linalg.svd(data - datamean)
# Now vv[0] contains the first principal component, i.e. the direction
# vector of the 'best fit' line in the least squares sense.
# Now generate some points along this best fit line, for plotting.
# we want it to have mean 0 (like the points we did
# the svd on). Also, it's a straight line, so we only need 2 points.
linepts = vv[0] * np.mgrid[-100:100:2j][:, np.newaxis]
# shift by the mean to get the line in the right place
linepts += datamean
# Verify that everything looks right.
import matplotlib.pyplot as plt
import mpl_toolkits.mplot3d as m3d
ax = m3d.Axes3D(plt.figure())
ax.scatter3D(*data.T)
ax.plot3D(*linepts.T)
plt.show()
# this will find the slope and x-intercept of a plane
# parallel to the y-axis that best fits the data
A_xz = np.vstack((x, np.ones(len(x)))).T
m_xz, c_xz = np.linalg.lstsq(A_xz, z)[0]
# again for a plane parallel to the x-axis
A_yz = np.vstack((y, np.ones(len(y)))).T
m_yz, c_yz = np.linalg.lstsq(A_yz, z)[0]
# the intersection of those two planes and
# the function for the line would be:
# z = m_yz * y + c_yz
# z = m_xz * x + c_xz
# or:
def lin(z):
x = (z - c_xz)/m_xz
y = (z - c_yz)/m_yz
return x,y
print(x,y)
#verifying:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
fig = plt.figure()
ax = Axes3D(fig)
zz = np.linspace(0,5)
xx,yy = lin(zz)
ax.scatter(x, y, z)
ax.plot(xx,yy,zz)
plt.savefig('test.png')
plt.show()

As was proposed in the old post you refer to, you could also make use of principal component analysis instead of a least squares approach. For that I suggest sklearn.decomposition.PCA from the sklearn package.
An example can be found below using the csv-file you provided.
import pandas as pd
import numpy as np
from sklearn.decomposition import PCA
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
mydataset = pd.read_csv('line1.csv')
x = mydataset.iloc[:,0]
y = mydataset.iloc[:,1]
z = mydataset.iloc[:,2]
coords = np.array((x, y, z)).T
pca = PCA(n_components=1)
pca.fit(coords)
direction_vector = pca.components_
print(direction_vector)
# Create plot
origin = np.mean(coords, axis=0)
euclidian_distance = np.linalg.norm(coords - origin, axis=1)
extent = np.max(euclidian_distance)
line = np.vstack((origin - direction_vector * extent,
origin + direction_vector * extent))
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(coords[:, 0], coords[:, 1], coords[:,2])
ax.plot(line[:, 0], line[:, 1], line[:, 2], 'r')

You can get rid of the complaint from leastsquares by adding rcond=None like this:
m_xz, c_xz = np.linalg.lstsq(A_xz, z, rcond=None)[0]
Is this the right decision for your situation? I have no idea. But there's more about it in the docs.
When I run your code with your inputs it seems to run just fine and I get values assigned to m_xz, c_xz, etc. If you don't call them explicitly with print('m_xz') (or whatever) then you won't see them.
m_xz
Out[42]: 5.186132604596112
c_xz
Out[43]: 62.5764694106141
Also, you reference your data in kind of two different ways. You get x, y, and z from your csv, but also put it into a numpy array. You can get rid of the duplication and pandas by just using numpy:
data = np.genfromtxt('line1.csv', delimiter=',', skip_header=1)
x = data[:,0]
y = data[:,1]
z = data[:,2]

Flow visualisation in python using curved (path-following) vectors

I would like to plot a vector field with curved arrows in python, as can be done in vfplot (see below) or IDL.
You can get close in matplotlib, but using quiver() limits you to straight vectors (see below left) whereas streamplot() doesn't seem to permit meaningful control over arrow length or arrowhead position (see below right), even when changing integration_direction, density, and maxlength.
So, is there a python library that can do this? Or is there a way of getting matplotlib to do it?

If you look at the streamplot.py that is included in matplotlib, on lines 196 - 202 (ish, idk if this has changed between versions - I'm on matplotlib 2.1.2) we see the following:
... (to line 195)
# Add arrows half way along each trajectory.
s = np.cumsum(np.sqrt(np.diff(tx) ** 2 + np.diff(ty) ** 2))
n = np.searchsorted(s, s[-1] / 2.)
arrow_tail = (tx[n], ty[n])
arrow_head = (np.mean(tx[n:n + 2]), np.mean(ty[n:n + 2]))
... (after line 196)
changing that part to this will do the trick (changing assignment of n):
... (to line 195)
# Add arrows half way along each trajectory.
s = np.cumsum(np.sqrt(np.diff(tx) ** 2 + np.diff(ty) ** 2))
n = np.searchsorted(s, s[-1]) ### THIS IS THE EDITED LINE! ###
arrow_tail = (tx[n], ty[n])
arrow_head = (np.mean(tx[n:n + 2]), np.mean(ty[n:n + 2]))
... (after line 196)
If you modify this to put the arrow at the end, then you could generate the arrows more to your liking.
Additionally, from the docs at the top of the function, we see the following:
*linewidth* : numeric or 2d array
vary linewidth when given a 2d array with the same shape as velocities.
The linewidth can be a numpy.ndarray, and if you can pre-calculate the desired width of your arrows, you'll be able to modify the pencil width while drawing the arrows. It looks like this part has already been done for you.
So, in combination with shortening the arrows maxlength, increasing the density, and adding start_points, as well as tweaking the function to put the arrow at the end instead of the middle, you could get your desired graph.
With these modifications, and the following code, I was able to get a result much closer to what you wanted:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec
import matplotlib.patches as pat
w = 3
Y, X = np.mgrid[-w:w:100j, -w:w:100j]
U = -1 - X**2 + Y
V = 1 + X - Y**2
speed = np.sqrt(U*U + V*V)
fig = plt.figure(figsize=(14, 18))
gs = gridspec.GridSpec(nrows=3, ncols=2, height_ratios=[1, 1, 2])
grains = 10
tmp = tuple([x]*grains for x in np.linspace(-2, 2, grains))
xs = []
for x in tmp:
xs += x
ys = tuple(np.linspace(-2, 2, grains))*grains
seed_points = np.array([list(xs), list(ys)])
# Varying color along a streamline
ax1 = fig.add_subplot(gs[0, 1])
strm = ax1.streamplot(X, Y, U, V, color=U, linewidth=np.array(5*np.random.random_sample((100, 100))**2 + 1), cmap='winter', density=10,
minlength=0.001, maxlength = 0.07, arrowstyle='fancy',
integration_direction='forward', start_points = seed_points.T)
fig.colorbar(strm.lines)
ax1.set_title('Varying Color')
plt.tight_layout()
plt.show()
tl;dr: go copy the source code, and change it to put the arrows at the end of each path, instead of in the middle. Then use your streamplot instead of the matplotlib streamplot.
Edit: I got the linewidths to vary

Starting with David Culbreth's modification, I rewrote chunks of the streamplot function to achieve the desired behaviour. Slightly too numerous to specify them all here, but it includes a length-normalising method and disables the trajectory-overlap checking. I've appended two comparisons of the new curved quiver function with the original streamplot and quiver.

Here's a way to obtain the desired output in vanilla pyplot (i.e., without modifying the streamplot function or anything that fancy). For reminder, the goal is to visualize a vector field with curved arrows whose length is proportional to the norm of the vector.
The trick is to:
make streamplot with no arrows that is traced backward from a given point (see)
plot a quiver from that point. Make the quiver small enough so that only the arrow is visible
repeat 1. and 2. in a loop for every seed and scale the length of the streamplot to be proportional to the norm of the vector.
import matplotlib.pyplot as plt
import numpy as np
w = 3
Y, X = np.mgrid[-w:w:8j, -w:w:8j]
U = -Y
V = X
norm = np.sqrt(U**2 + V**2)
norm_flat = norm.flatten()
start_points = np.array([X.flatten(),Y.flatten()]).T
plt.clf()
scale = .2/np.max(norm)
plt.subplot(121)
plt.title('scaling only the length')
for i in range(start_points.shape[0]):
plt.streamplot(X,Y,U,V, color='k', start_points=np.array([start_points[i,:]]),minlength=.95*norm_flat[i]*scale, maxlength=1.0*norm_flat[i]*scale,
integration_direction='backward', density=10, arrowsize=0.0)
plt.quiver(X,Y,U/norm, V/norm,scale=30)
plt.axis('square')
plt.subplot(122)
plt.title('scaling length, arrowhead and linewidth')
for i in range(start_points.shape[0]):
plt.streamplot(X,Y,U,V, color='k', start_points=np.array([start_points[i,:]]),minlength=.95*norm_flat[i]*scale, maxlength=1.0*norm_flat[i]*scale,
integration_direction='backward', density=10, arrowsize=0.0, linewidth=.5*norm_flat[i])
plt.quiver(X,Y,U/np.max(norm), V/np.max(norm),scale=30)
plt.axis('square')
Here's the result:

Just looking at the documentation on streamplot(), found here -- what if you used something like streamplot( ... ,minlength = n/2, maxlength = n) where n is the desired length -- you will need to play with those numbers a bit to get your desired graph
you can control for the points using start_points, as shown in the example provided by #JohnKoch
Here's an example of how I controlled the length with streamplot() -- it's pretty much a straight copy/paste/crop from the example from above.
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec
import matplotlib.patches as pat
w = 3
Y, X = np.mgrid[-w:w:100j, -w:w:100j]
U = -1 - X**2 + Y
V = 1 + X - Y**2
speed = np.sqrt(U*U + V*V)
fig = plt.figure(figsize=(14, 18))
gs = gridspec.GridSpec(nrows=3, ncols=2, height_ratios=[1, 1, 2])
grains = 10
tmp = tuple([x]*grains for x in np.linspace(-2, 2, grains))
xs = []
for x in tmp:
xs += x
ys = tuple(np.linspace(-2, 2, grains))*grains
seed_points = np.array([list(xs), list(ys)])
arrowStyle = pat.ArrowStyle.Fancy()
# Varying color along a streamline
ax1 = fig.add_subplot(gs[0, 1])
strm = ax1.streamplot(X, Y, U, V, color=U, linewidth=1.5, cmap='winter', density=10,
minlength=0.001, maxlength = 0.1, arrowstyle='->',
integration_direction='forward', start_points = seed_points.T)
fig.colorbar(strm.lines)
ax1.set_title('Varying Color')
plt.tight_layout()
plt.show()
Edit: made it prettier, though still not quite what we were looking for.

How do I create a surface plot in python using values from a text file?

I am trying to get a 3D surface plot from a text file that has four coloumns and 700 rows. I have written the code below to generate that plot, but I constantly keep getting the following error:
invalid value encountered in true_divide
z = (x*y*mlim)/(x*y)
raise ValueError("Argument Z must be 2-dimensional.")
ValueError: Argument Z must be 2-dimensional.
Here is my code
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import numpy as np
f2 = open('openfile.txt', 'r')
i6 = list()
i9 = list()
i10 = list ()
count = 0
while True:
a = f2.readline()
if not a: break
v = a.split()
i6.append(float(v[0]))
i9.append(float(v[1]))
i10.append(float(v[3]))
count = count + 1
mlim = np.array(i6)
ira = np.array(i9)
dec = np.array(i10)
x = ira
y = dec
z = (x*y*mlim)/(x*y)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_wireframe(x,y,z, rstride=10, cstride=10)
plt.savefig('filesave.png')
plt.close()
How do I solve this?
I have also attached an image of what the actual text file looks like
click here
Thank you.

To plot a 3D surface, a wire frame in your case, you must create a meshgrid (a.k.a a matrix) before so as to map this in you function (x*y*mlim)/(x*y).
The following alteration may fix your problem:
# More of your code above this
x = ira
y = dec
# Create a meshgrid
X, Y = np.meshgrid(x,y)
# Now you can build you z array using the correct mapping
# Note that mlim variable works as an scalar
z = (X*Y*mlim)/(X*Y)
The comment of #Koustav above is pertinent too. It is important check for null/zero values in case of division.