I'm trying to adapt the following resources to this question:
Python conversion between coordinates
https://matplotlib.org/gallery/pie_and_polar_charts/polar_scatter.html
I can't seem to get the coordinates to transfer the dendrogram shape over to polar coordinates.
Does anyone know how to do this? I know there is an implementation in networkx but that requires building a graph and then using pygraphviz backend to get the positions.
Is there a way to convert dendrogram cartesian coordinates to polar coordinates with matplotlib and numpy?
import requests
from ast import literal_eval
import matplotlib.pyplot as plt
import numpy as np
def read_url(url):
r = requests.get(url)
return r.text
def cartesian_to_polar(x, y):
rho = np.sqrt(x**2 + y**2)
phi = np.arctan2(y, x)
return(rho, phi)
def plot_dendrogram(icoord,dcoord,figsize, polar=False):
if polar:
icoord, dcoord = cartesian_to_polar(icoord, dcoord)
with plt.style.context("seaborn-white"):
fig = plt.figure(figsize=figsize)
ax = fig.add_subplot(111, polar=polar)
for xs, ys in zip(icoord, dcoord):
ax.plot(xs,ys, color="black")
ax.set_title(f"Polar= {polar}", fontsize=15)
# Load the dendrogram data
string_data = read_url("https://pastebin.com/raw/f953qgdr").replace("\r","").replace("\n","").replace("\u200b\u200b","")
# Convert it to a dictionary (a subset of the output from scipy.hierarchy.dendrogram)
dendrogram_data = literal_eval(string_data)
icoord = np.asarray(dendrogram_data["icoord"], dtype=float)
dcoord = np.asarray(dendrogram_data["dcoord"], dtype=float)
# Plot the cartesian version
plot_dendrogram(icoord,dcoord, figsize=(8,3), polar=False)
# Plot the polar version
plot_dendrogram(icoord,dcoord, figsize=(5,5), polar=True)
I just tried this and it's closer but still not correct:
import matplotlib.transforms as mtransforms
with plt.style.context("seaborn-white"):
fig, ax = plt.subplots(figsize=(5,5))
for xs, ys in zip(icoord, dcoord):
ax.plot(xs,ys, color="black",transform=trans_offset)
ax_polar = plt.subplot(111, projection='polar')
trans_offset = mtransforms.offset_copy(ax_polar.transData, fig=fig)
for xs, ys in zip(icoord, dcoord):
ax_polar.plot(xs,ys, color="black",transform=trans_offset)
You can make the "root" of the tree start in the middle and have the leaves outside. You also have to add more points to the "bar" part for it to look nice and round.
We note that each element of icoord and dcoord (I will call this seg) has four points:
seg[1] seg[2]
+-------------+
| |
+ seg[0] + seg[3]
The vertical bars are fine as straight lines between the two points, but we need more points between seg[1] and seg[2] (the horizontal bar, which will need to become an arc).
This function will add more points in those positions and can be called on both xs and ys in the plotting function:
def smoothsegment(seg, Nsmooth=100):
return np.concatenate([[seg[0]], np.linspace(seg[1], seg[2], Nsmooth), [seg[3]]])
Now we must modify the plotting function to calculate the radial coordinates. Some experimentation has led to the log formula I am using, based on the other answer which also uses log scale. I've left a gap open on the right for the radial labels and done a very rudimentary mapping of the "icoord" coordinates to the radial ones so that the labels correspond to the ones in the rectangular plot. I don't know exactly how to handle the radial dimension. The numbers are correct for the log, but we probably want to map them as well.
def plot_dendrogram(icoord,dcoord,figsize, polar=False):
if polar:
dcoord = -np.log(dcoord+1)
# avoid a wedge over the radial labels
gap = 0.1
imax = icoord.max()
imin = icoord.min()
icoord = ((icoord - imin)/(imax - imin)*(1-gap) + gap/2)*2*numpy.pi
with plt.style.context("seaborn-white"):
fig = plt.figure(figsize=figsize)
ax = fig.add_subplot(111, polar=polar)
for xs, ys in zip(icoord, dcoord):
if polar:
xs = smoothsegment(xs)
ys = smoothsegment(ys)
ax.plot(xs,ys, color="black")
ax.set_title(f"Polar= {polar}", fontsize=15)
if polar:
ax.spines['polar'].set_visible(False)
ax.set_rlabel_position(0)
Nxticks = 10
xticks = np.linspace(gap/2, 1-gap/2, Nxticks)
ax.set_xticks(xticks*np.pi*2)
ax.set_xticklabels(np.round(np.linspace(imin, imax, Nxticks)).astype(int))
Which results in the following figure:
First, I think you might benefit from this question.
Then, let's break down the objective: it is not very clear to me what you want to do, but I assume you want to get something that looks like this
(source, page 14)
To render something like this, you need to be able to render horizontal lines that appear as hemi-circles in polar coordinates. Then, it's a matter of mapping your horizontal lines to polar plot.
First, note that your radius are not normalized in this line:
if polar:
icoord, dcoord = cartesian_to_polar(icoord, dcoord)
you might normalize them by simply remapping icoord to [0;2pi).
Now, let's try plotting something simpler, instead of your complex plot:
icoord, dcoord = np.meshgrid(np.r_[1:10], np.r_[1:4])
# Plot the cartesian version
plot_dendrogram(icoord, dcoord, figsize=(8, 3), polar=False)
# Plot the polar version
plot_dendrogram(icoord, dcoord, figsize=(5, 5), polar=True)
Result is the following:
as you can see, the polar code does not map horizontal lines to semi-circles, therefore that is not going to work. Let's try with plt.polar instead:
plt.polar(icoord.T, dcoord.T)
produces
which is more like what we need. We need to fix the angles first, and then we shall consider that Y coordinate goes inward (while you probably want it going from center to border). It boils down to this
nic = (icoord.T - icoord.min()) / (icoord.max() - icoord.min())
plt.polar(2 * np.pi * nic, -dcoord.T)
which produces the following
Which is similar to what you need. Note that straight lines remain straight, and are not replaced with arcs, so you might want to resample them in your for loop.
Also, you might benefit from single color and log-scale to make reading easier
plt.subplots(figsize=(10, 10))
ico = (icoord.T - icoord.min()) / (icoord.max() - icoord.min())
plt.polar(2 * np.pi * ico, -np.log(dcoord.T), 'b')
Related
Consider the following data, which is defined in polar space in theta, r, and is plotted twice; once in the the orthogonal theta-r phase space, and one in cartesian space after an inverse transformation from polar coordinates to x-y (i.e. what matplotlib's projection='polar' does):
import matplotlib.pyplot as plt
import numpy as np
theta = np.linspace(0, 2*np.pi, 50)
r = np.linspace(0, 1, 50)
THETA, R = np.meshgrid(theta, r)
Z = np.sin(R*np.pi) * np.sin(THETA+np.pi/2)
fig = plt.figure()
axpol = fig.add_subplot(121)
axcart = fig.add_subplot(122, projection='polar')
axcart.contourf(THETA, R, Z, levels=10)
axpol.contourf(THETA, R, Z, levels=10)
axcart.set_title('cartesian space')
axpol.set_title('polar space')
axpol.set_xlim([0, 2*np.pi])
axpol.set_xlabel('r')
axpol.set_ylabel('theta')
plt.show()
This produces:
(NOTE: Oops, the axis labels in the polar plots (left side) should be swapped in each of the images below)
Now, if we shift the theta array by pi:
theta = np.linspace(np.pi, 3*np.pi, 50)
and rerun the above, we see
Notice that the data plotted in the projected polar space successfully wraps the data at theta > 2*np.pi back to the beginning of the angular domain (since this is defined in the projections inverse transformation), such that it appears unchanged. In polar space, this does not happen.
Of course, this is expected; this axis has no associated transformation, and thus does know know how to wrap the data, or that it even should.
My question is, how can I enable this behavior, without having to shift the coordinates and data manually? That is, is there a way to have the axis on the left of the figure above inherit the polar transformation, but not the projection?
I would prefer to do this without defining my own transformation or projection objects. I thought there should be a way to inherit this small piece of the polar transformation, without doing the "full" transformation to Cartesian x,y.
TL/DR: How to use Wedge() in polar coordinates?
I'm generating a 2D histogram plot in polar coordinates (r, theta). At various values of r there can be different numbers of theta values (to preserve equal area sized bins). To draw the color coded bins I'm currently using pcolormesh() calls for each radial ring. This works ok, but near the center of the plot where there may be only 3 bins (each 120 degrees "wide" in theta space), pcolormesh() draws triangles that don't "sweep" out full arc (just connecting the two outer arc points with a straight line).
I've found a workaround using ax.bar() call, one for each radial ring and passing in arrays of theta values (each bin rendering as an individual bar). But when doing 90 rings with 3 to 360 theta bins in each, it's incredibly slow (minutes).
I tried using Wedge() patches, but can't get them to render correctly in the polar projection. Here is sample code showing both approaches:
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.patches import Wedge
from matplotlib.collections import PatchCollection
# Theta coordinates in degrees
theta1=45
theta2=80
# Radius coordinates
r1 = 0.4
r2 = 0.5
# Plot using bar()
fig, ax = plt.subplots(figsize=[6,6], subplot_kw={'projection': 'polar'})
theta_mid = np.deg2rad((theta1 + theta2)/2)
theta_width = np.deg2rad(theta2 - theta1)
height = r2 - r1
ax.bar(x=theta_mid, height = height, width=theta_width, bottom=r1)
ax.set_rlim(0, 1)
plt.savefig('bar.png')
# Plot using Wedge()
fig, ax = plt.subplots(figsize=[6,6], subplot_kw={'projection': 'polar'})
patches = []
patches.append( Wedge(center=(0, 0), r = r1, theta1=theta1, theta2=theta2, width = r2-r1, color='blue'))
p = PatchCollection(patches)
ax.add_collection(p)
ax.set_rlim(0, 1)
plt.savefig('wedge.png')
The outputs of each are:
Bar
Wedge
I've tried using radians for the wedge (because polar plots usually want their angle values in radians). That didn't help.
Am I missing something in how I'm using the Wedge? If I add thousands of Wedges to my Patch collection should I have any expectation it will be faster than bar()?
Thinking this was an actual bug, I opened this issue https://github.com/matplotlib/matplotlib/issues/22717 on matplotlib where one of the maintainers nicely pointed out that I should be using Rectangle() instead of Wedge().
The solution they provided is
from matplotlib.patches import Rectangle
fig, ax = plt.subplots(figsize=[6,6], subplot_kw={'projection': 'polar'})
p = PatchCollection([Rectangle((np.deg2rad(theta1), r1), theta_width, height, color='blue')])
ax.add_collection(p)
ax.set_rlim(0, 1)
plt.savefig('wedge.png')
How to use matplotlib or pyqtgraph draw plot like this:
Line AB is a two-directions street, green part represents the direction from point A to point B, red part represents B to A, width of each part represents the traffic volume. Widths are measured in point, will not changed at different zoom levels or dpi settings.
This is only an example, in fact I have hunderds of streets. This kind of plot is very common in many traffic softwares. I tried to use matplotlib's patheffect but result is frustrated:
from matplotlib import pyplot as plt
import matplotlib.patheffects as path_effects
x=[0,1,2,3]
y=[1,0,0,-1]
ab_width=20
ba_width=30
fig, axes= plt.subplots(1,1)
center_line, = axes.plot(x,y,color='k',linewidth=2)
center_line.set_path_effects(
[path_effects.SimpleLineShadow(offset=(0, -ab_width/2),shadow_color='g', alpha=1, linewidth=ab_width),
path_effects.SimpleLineShadow(offset=(0, ba_width/2), shadow_color='r', alpha=1, linewidth=ba_width),
path_effects.SimpleLineShadow(offset=(0, -ab_width), shadow_color='k', alpha=1, linewidth=2),
path_effects.SimpleLineShadow(offset=(0, ba_width), shadow_color='k', alpha=1, linewidth=2),
path_effects.Normal()])
axes.set_xlim(-1,4)
axes.set_ylim(-1.5,1.5)
One idea came to me is to take each part of the line as a standalone line, and recalculate it's position when changing zoom level, but it's too complicated and slow.
If there any easy way to use matplotlib or pyqtgraph draw what I want? Any suggestion will be appreciated!
If you can have each independent line, this can be done easily with the fill_between function.
from matplotlib import pyplot as plt
import numpy as np
x=np.array([0,1,2,3])
y=np.array([1,0,0,-1])
y1width=-1
y2width=3
y1 = y + y1width
y2 = y + y2width
fig = plt.figure()
ax = fig.add_subplot(111)
plt.plot(x,y, 'k', x,y1, 'k',x,y2, 'k',linewidth=2)
ax.fill_between(x, y1, y, color='g')
ax.fill_between(x, y2, y, color='r')
plt.xlim(-1,4)
plt.ylim(-3,6)
plt.show()
Here I considered the center line as the reference (thus the negative y1width), but could be done differently. The result is then:
If the lines are 'complicated', eventually intersecting at some point, then the keyword argument interpolate=True must be used to fill the crossover regions properly. Another interesting argument probably useful for your use case is where, to condition the region, for instance, where=y1 < 0. For more information you can check out the documentation.
One way of solving your issue is using filled polygons, some linear algebra and some calculus. The main idea is to draw a polygon along your x and y coordinates and along shifted coordinates to close and fill the polygon.
These are my results:
And here is the code:
from __future__ import division
import numpy
from matplotlib import pyplot, patches
def road(x, y, w, scale=0.005, **kwargs):
# Makes sure input coordinates are arrays.
x, y = numpy.asarray(x, dtype=float), numpy.asarray(y, dtype=float)
# Calculate derivative.
dx = x[2:] - x[:-2]
dy = y[2:] - y[:-2]
dy_dx = numpy.concatenate([
[(y[1] - y[0]) / (x[1] - x[0])],
dy / dx,
[(y[-1] - y[-2]) / (x[-1] - x[-2])]
])
# Offsets the input coordinates according to the local derivative.
offset = -dy_dx + 1j
offset = w * scale * offset / abs(offset)
y_offset = y + w * scale
#
AB = zip(
numpy.concatenate([x + offset.real, x[::-1]]),
numpy.concatenate([y + offset.imag, y[::-1]]),
)
p = patches.Polygon(AB, **kwargs)
# Returns polygon.
return p
if __name__ == '__main__':
# Some plot initializations
pyplot.close('all')
pyplot.ion()
# This is the list of coordinates of each point
x = [0, 1, 2, 3, 4]
y = [1, 0, 0, -1, 0]
# Creates figure and axes.
fig, ax = pyplot.subplots(1,1)
ax.axis('equal')
center_line, = ax.plot(x, y, color='k', linewidth=2)
AB = road(x, y, 20, color='g')
BA = road(x, y, -30, color='r')
ax.add_patch(AB)
ax.add_patch(BA)
The first step in calculating how to offset each data point is by calculating the discrete derivative dy / dx. I like to use complex notation to handle vectors in Python, i.e. A = 1 - 1j. This makes life easier for some mathematical operations.
The next step is to remember that the derivative gives the tangent to the curve and from linear algebra that the normal to the tangent is n=-dy_dx + 1j, using complex notation.
The final step in determining the offset coordinates is to ensure that the normal vector has unity size n_norm = n / abs(n) and multiply by the desired width of the polygon.
Now that we have all the coordinates for the points in the polygon, the rest is quite straightforward. Use patches.Polygon and add them to the plot.
This code allows you also to define if you want the patch on top of your route or below it. Just give a positive or negative value for the width. If you want to change the width of the polygon depending on your zoom level and/or resolution, you adjust the scale parameter. It also gives you freedom to add additional parameters to the patches such as fill patterns, transparency, etc.
I have a complicated curve defined as a set of points in a table like so (the full table is here):
# x y
1.0577 12.0914
1.0501 11.9946
1.0465 11.9338
...
If I plot this table with the commands:
plt.plot(x_data, y_data, c='b',lw=1.)
plt.scatter(x_data, y_data, marker='o', color='k', s=10, lw=0.2)
I get the following:
where I've added the red points and segments manually. What I need is a way to calculate those segments for each of those points, that is: a way to find the minimum distance from a given point in this 2D space to the interpolated curve.
I can't use the distance to the data points themselves (the black dots that generate the blue curve) since they are not located at equal intervals, sometimes they are close and sometimes they are far apart and this deeply affects my results further down the line.
Since this is not a well behaved curve I'm not really sure what I could do. I've tried interpolating it with a UnivariateSpline but it returns a very poor fit:
# Sort data according to x.
temp_data = zip(x_data, y_data)
temp_data.sort()
# Unpack sorted data.
x_sorted, y_sorted = zip(*temp_data)
# Generate univariate spline.
s = UnivariateSpline(x_sorted, y_sorted, k=5)
xspl = np.linspace(0.8, 1.1, 100)
yspl = s(xspl)
# Plot.
plt.scatter(xspl, yspl, marker='o', color='r', s=10, lw=0.2)
I also tried increasing the number of interpolating points but got a mess:
# Sort data according to x.
temp_data = zip(x_data, y_data)
temp_data.sort()
# Unpack sorted data.
x_sorted, y_sorted = zip(*temp_data)
t = np.linspace(0, 1, len(x_sorted))
t2 = np.linspace(0, 1, 100)
# One-dimensional linear interpolation.
x2 = np.interp(t2, t, x_sorted)
y2 = np.interp(t2, t, y_sorted)
plt.scatter(x2, y2, marker='o', color='r', s=10, lw=0.2)
Any ideas/pointers will be greatly appreciated.
If you're open to using a library for this, have a look at shapely: https://github.com/Toblerity/Shapely
As a quick example (points.txt contains the data you linked to in your question):
import shapely.geometry as geom
import numpy as np
coords = np.loadtxt('points.txt')
line = geom.LineString(coords)
point = geom.Point(0.8, 10.5)
# Note that "line.distance(point)" would be identical
print(point.distance(line))
As an interactive example (this also draws the line segments you wanted):
import numpy as np
import shapely.geometry as geom
import matplotlib.pyplot as plt
class NearestPoint(object):
def __init__(self, line, ax):
self.line = line
self.ax = ax
ax.figure.canvas.mpl_connect('button_press_event', self)
def __call__(self, event):
x, y = event.xdata, event.ydata
point = geom.Point(x, y)
distance = self.line.distance(point)
self.draw_segment(point)
print 'Distance to line:', distance
def draw_segment(self, point):
point_on_line = line.interpolate(line.project(point))
self.ax.plot([point.x, point_on_line.x], [point.y, point_on_line.y],
color='red', marker='o', scalex=False, scaley=False)
fig.canvas.draw()
if __name__ == '__main__':
coords = np.loadtxt('points.txt')
line = geom.LineString(coords)
fig, ax = plt.subplots()
ax.plot(*coords.T)
ax.axis('equal')
NearestPoint(line, ax)
plt.show()
Note that I've added ax.axis('equal'). shapely operates in the coordinate system that the data is in. Without the equal axis plot, the view will be distorted, and while shapely will still find the nearest point, it won't look quite right in the display:
The curve is by nature parametric, i.e. for each x there isn't necessary a unique y and vice versa. So you shouldn't interpolate a function of the form y(x) or x(y). Instead, you should do two interpolations, x(t) and y(t) where t is, say, the index of the corresponding point.
Then you use scipy.optimize.fminbound to find the optimal t such that (x(t) - x0)^2 + (y(t) - y0)^2 is the smallest, where (x0, y0) are the red dots in your first figure. For fminsearch, you could specify the min/max bound for t to be 1 and len(x_data)
You could try implementing a calculation of distance from point to line on incremental pairs of points on the curve and finding that minimum. This will introduce a small bit of error from the curve as drawn, but it should be very small, as the points are relatively close together.
http://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
You can easily use the package trjtrypy in PyPI: https://pypi.org/project/trjtrypy/
All needed computations and visualizations are available in this package. You can get your answer within a line of code like:
to get the minimum distance use: trjtrypy.basedists.distance(points, curve)
to visualize the curve and points use: trjtrypy.visualizations.draw_landmarks_trajectory(points, curve)
I'd like to create an Argand Diagram from a set of complex numbers using matplotlib.
Are there any pre-built functions to help me do this?
Can anyone recommend an approach?
Image by LeonardoG, CC-SA-3.0
I'm not sure exactly what you're after here...you have a set of complex numbers, and want to map them to the plane by using their real part as the x coordinate and the imaginary part as y?
If so you can get the real part of any python imaginary number with number.real and the imaginary part with number.imag. If you're using numpy, it also provides a set of helper functions numpy.real and numpy.imag etc. which work on numpy arrays.
So for instance if you had an array of complex numbers stored something like this:
In [13]: a = n.arange(5) + 1j*n.arange(6,11)
In [14]: a
Out[14]: array([ 0. +6.j, 1. +7.j, 2. +8.j, 3. +9.j, 4.+10.j])
...you can just do
In [15]: fig,ax = subplots()
In [16]: ax.scatter(a.real,a.imag)
This plots dots on an argand diagram for each point.
edit: For the plotting part, you must of course have imported matplotlib.pyplot via from matplotlib.pyplot import * or (as I did) use the ipython shell in pylab mode.
To follow up #inclement's answer; the following function produces an argand plot that is centred around 0,0 and scaled to the maximum absolute value in the set of complex numbers.
I used the plot function and specified solid lines from (0,0). These can be removed by replacing ro- with ro.
def argand(a):
import matplotlib.pyplot as plt
import numpy as np
for x in range(len(a)):
plt.plot([0,a[x].real],[0,a[x].imag],'ro-',label='python')
limit=np.max(np.ceil(np.absolute(a))) # set limits for axis
plt.xlim((-limit,limit))
plt.ylim((-limit,limit))
plt.ylabel('Imaginary')
plt.xlabel('Real')
plt.show()
For example:
>>> a = n.arange(5) + 1j*n.arange(6,11)
>>> from argand import argand
>>> argand(a)
produces:
EDIT:
I have just realised there is also a polar plot function:
for x in a:
plt.polar([0,angle(x)],[0,abs(x)],marker='o')
If you prefer a plot like the one below
one type of plot
or this one second type of plot
you can do this simply by these two lines (as an example for the plots above):
z=[20+10j,15,-10-10j,5+15j] # array of complex values
complex_plane2(z,1) # function to be called
by using a simple jupyter code from here
https://github.com/osnove/other/blob/master/complex_plane.py
I have written it for my own purposes. Even better it it helps to others.
To get that:
You can use:
cmath.polar to convert a complex number to polar rho-theta coordinates. In the code below this function is first vectorized in order to process an array of complex numbers instead of a single number, this is just to prevent the use an explicit loop.
A pyplot axis with its projection type set to polar. Plot can be done using pyplot.stem or pyplot.scatter.
In order to plot horizontal and vertical lines for Cartesian coordinates there are two possibilities:
Add a Cartesian axis and plot Cartesian coordinates. This solution is described in this question. I don't think it's an easy solution as the Cartesian axis won't be centered, nor it will have the correct scaling factor.
Use the polar axis, and translate Cartesian coordinates for projections into polar coordinates. This is the solution I used to plot the graph above. To not clutter the graph I've shown only one point with its projected Cartesian coordinates.
Code used for the plot above:
from cmath import pi, e, polar
from numpy import linspace, vectorize, sin, cos
from numpy.random import rand
from matplotlib import pyplot as plt
# Arrays of evenly spaced angles, and random lengths
angles = linspace(0, 2*pi, 12, endpoint=False)
lengths = 3*rand(*angles.shape)
# Create an array of complex numbers in Cartesian form
z = lengths * e ** (1j*angles)
# Convert back to polar form
vect_polar = vectorize(polar)
rho_theta = vect_polar(z)
# Plot numbers on polar projection
fig, ax = plt.subplots(subplot_kw={'projection': 'polar'})
ax.stem(rho_theta[1], rho_theta[0])
# Get a number, find projections on axes
n = 11
rho, theta = rho_theta[0][n], rho_theta[1][n]
a = cos(theta)
b = sin(theta)
rho_h, theta_h = abs(a)*rho, 0 if a >= 0 else -pi
rho_v, theta_v = abs(b)*rho, pi/2 if b >= 0 else -pi/2
# Plot h/v lines on polar projection
ax.plot((theta_h, theta), (rho_h, rho), c='r', ls='--')
ax.plot((theta, theta_v), (rho, rho_v), c='g', ls='--')
import matplotlib.pyplot as plt
from numpy import *
'''
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
This draws the axis for argand diagram
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
'''
r = 1
Y = [r*exp(1j*theta) for theta in linspace(0,2*pi, 200)]
Y = array(Y)
plt.plot(real(Y), imag(Y), 'r')
plt.ylabel('Imaginary')
plt.xlabel('Real')
plt.axhline(y=0,color='black')
plt.axvline(x=0, color='black')
def argand(complex_number):
'''
This function takes a complex number.
'''
y = complex_number
x1,y1 = [0,real(y)], [0, imag(y)]
x2,y2 = [real(y), real(y)], [0, imag(y)]
plt.plot(x1,y1, 'r') # Draw the hypotenuse
plt.plot(x2,y2, 'r') # Draw the projection on real-axis
plt.plot(real(y), imag(y), 'bo')
[argand(r*exp(1j*theta)) for theta in linspace(0,2*pi,100)]
plt.show()
https://github.com/QuantumNovice/Matplotlib-Argand-Diagram/blob/master/argand.py