We Keep Coding

Extend part of an object

I'm attempting to extend the 'tail' of an arrow. So far I've been able to draw a line through the center of the arrow, but this line extends 'both' ways, rather than in just one direction. The script below shows my progress. Ideally I would be able to extend the tail of the arrow regardless of the orientation of the arrow image. Any suggestions on how to accomplish this. Image examples below, L:R start, progress, goal.
# import image and grayscale
image = cv2.imread("image path")
image = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
cv2.imshow("original",image)
# inverts black and white
gray = 255 - image
cv2.imshow("Inverted", gray)
# Extend the borders for the line
extended = cv2.copyMakeBorder(gray, 20, 20, 10, 10, cv2.BORDER_CONSTANT)
cv2.imshow("extended borders", extended)
# contour finding
contours, hierarchy = cv2.findContours(extended, 1, 2)
cont = contours[0]
rows,cols = extended.shape[:2]
[vx,vy,x,y] = cv2.fitLine(cont, cv2.DIST_L2,0,0.01,0.01)
leftish = int((-x*vy/vx) + y)
rightish = int(((cols-x)*vy/vx)+y)
line = cv2.line(extended,(cols-1,rightish),(0,leftish),(255,255,255), 6)
cv2.imshow("drawn line", line)

"Moments" can be strange things. They're building blocks and show up most often in statistics.
It helps to have a little background in statistics, and see the application of those calculations to image data, which can be considered a set of points. If you've ever calculated the weighted average or "centroid" of something, you'll recognize some of the sums that show up in "moments".
Higher order moments can be building blocks to higher statistical measures such as covariance and skewness.
Using covariance, you can calculate the major axis of your set of points, or your arrow in this case.
Using skewness, you can figure out which side of a distribution is heavier than the other... i.e. which side is the arrow's tip and which is its tail.
This should give you a very precise angle. The scale/radius however is best estimated using other ways. You'll notice that the radius estimated from the area of the arrow fluctuates a little. You could find the points belonging to the arrow that are furthest away from the center, and take that as a somewhat stable length.
Here's a longish program that implements the two ideas above and shows the direction of an arrow:
#!/usr/bin/env python3
import os
import sys
import numpy as np
import cv2 as cv
# utilities to convert between 2D vectors and complex numbers
# complex numbers are handy for rotating stuff
def to_complex(vec):
assert vec.shape[-1] == 2
if vec.dtype == np.float32:
return vec.view(np.complex64)
elif vec.dtype == np.float64:
return vec.view(np.complex128)
else:
assert False, vec.dtype
def from_complex(cplx):
if cplx.dtype == np.complex64:
return cplx.view(np.float32)
elif cplx.dtype == np.complex128:
return cplx.view(np.float64)
else:
assert False, cplx.dtype
# utilities for drawing with fractional bits of position
# just to make a pretty picture
def iround(val):
return int(round(val))
def ipt(vec, shift=0):
if isinstance(vec, (int, float)):
return iround(vec * 2**shift)
elif isinstance(vec, (tuple, list, np.ndarray)):
return tuple(iround(el * 2**shift) for el in vec)
else:
assert False, type(vec)
# utilities for affine transformation
# just to make a pretty picture
def rotate(degrees=0):
# we want positive rotation
# meaning move +x towards +y
# getRotationMatrix2D does it differently
result = np.eye(3).astype(np.float32)
result[0:2, 0:3] = cv.getRotationMatrix2D(center=(0,0), angle=-degrees, scale=1.0)
return result
def translate(dx=0, dy=0):
result = np.eye(3).astype(np.float32)
result[0:2,2] = [dx, dy]
return result
# main logic
def calculate_direction(im):
# using "nonzero" (default behavior) is a little noisy
mask = (im >= 128)
m = cv.moments(mask.astype(np.uint8), binaryImage=True)
# easier access... see below for details
m00 = m['m00']
m10 = m['m10']
m01 = m['m01']
mu00 = m00
mu20 = m['mu20']
mu11 = m['mu11']
mu02 = m['mu02']
nu30 = m['nu30']
nu03 = m['nu03']
# that's just the centroid
cx = m10 / m00
cy = m01 / m00
centroid = np.array([cx, cy]) # as a vector
# and that's the size in pixels:
size = m00
# and that's an approximate "radius", if it were a circle which it isn't
radius = (size / np.pi) ** 0.5
# (since the "size" in pixels can fluctuate due to resampling, so will the "radius")
# wikipedia helpfully mentions "image orientation" as an example:
# https://en.wikipedia.org/wiki/Image_moment#Examples_2
# we'll use that for the major axis
mup20 = mu20 / mu00
mup02 = mu02 / mu00
mup11 = mu11 / mu00
theta = 0.5 * np.arctan2(2 * mup11, mup20 - mup02)
#print(f"angle: {theta / np.pi * 180:+6.1f} degrees")
# we only have the axis, not yet the direction
# we will assess "skewness" now
# https://en.wikipedia.org/wiki/Skewness#Definition
# note how "positive" skewness appears in a distribution:
# it points away from the heavy side, towards the light side
# fortunately, cv.moments() also calculates those "standardized moments"
# https://en.wikipedia.org/wiki/Standardized_moment#Standard_normalization
skew = np.array([nu30, nu03])
#print("skew:", skew)
# we'll have to *rotate* that so it *roughly* lies along the x axis
# then assess which end is the heavy/light end
# then use that information to maybe flip the axis,
# so it points in the direction of the arrow
skew_complex = to_complex(skew) # reinterpret two reals as one complex number
rotated_skew_complex = skew_complex * np.exp(1j * -theta) # rotation
rotated_skew = from_complex(rotated_skew_complex)
#print("rotated skew:", rotated_skew)
if rotated_skew[0] > 0: # pointing towards tail
theta = (theta + np.pi) % (2*np.pi) # flip direction 180 degrees
else: # pointing towards head
pass
print(f"angle: {theta / np.pi * 180:+6.1f} degrees")
# construct a vector that points like the arrow in the picture
direction = np.exp([1j * theta])
direction = from_complex(direction)
return (radius, centroid, direction)
def draw_a_picture(im, radius, centroid, direction):
height, width = im.shape[:2]
# take the source at half brightness
canvas = cv.cvtColor(im // 2, cv.COLOR_GRAY2BGR)
shift = 4 # prettier drawing
cv.circle(canvas,
center=ipt(centroid, shift),
radius=ipt(radius, shift),
thickness=iround(radius * 0.1),
color=(0,0,255),
lineType=cv.LINE_AA,
shift=shift)
# (-direction) meaning point the *opposite* of the arrow's direction, i.e. towards tail
cv.line(canvas,
pt1=ipt(centroid + direction * radius * -3.0, shift),
pt2=ipt(centroid + direction * radius * +3.0, shift),
thickness=iround(radius * 0.05),
color=(0,255,255),
lineType=cv.LINE_AA,
shift=shift)
cv.line(canvas,
pt1=ipt(centroid + (-direction) * radius * 3.5, shift),
pt2=ipt(centroid + (-direction) * radius * 4.5, shift),
thickness=iround(radius * 0.15),
color=(0,255,255),
lineType=cv.LINE_AA,
shift=shift)
return canvas
if __name__ == '__main__':
imfile = sys.argv[1] if len(sys.argv) >= 2 else "p7cmR.png"
src = cv.imread(imfile, cv.IMREAD_GRAYSCALE)
src = 255 - src # invert (white arrow on black background)
height, width = src.shape[:2]
diagonal = np.hypot(height, width)
outsize = int(np.ceil(diagonal * 1.3)) # fudge factor
cv.namedWindow("arrow", cv.WINDOW_NORMAL)
cv.resizeWindow("arrow", 5*outsize, 5*outsize)
angle = 0 # degrees
increment = +1
do_spin = True
while True:
print(f"{angle:+.0f} degrees")
M = translate(dx=+outsize/2, dy=+outsize/2) # rotate(degrees=angle) # translate(dx=-width/2, dy=-height/2)
im = cv.warpAffine(src, M=M[:2], dsize=(outsize, outsize), flags=cv.INTER_CUBIC, borderMode=cv.BORDER_REPLICATE)
# resampling introduces blur... except when it's an even number like 0 degrees, 90 degrees, ...
# so at even rotations, things will jump a little.
# this rotation is only for demo purposes
(radius, centroid, direction) = calculate_direction(im)
canvas = draw_a_picture(im, radius, centroid, direction)
cv.imshow("arrow", canvas)
if do_spin:
angle = (angle + increment) % 360
print()
key = cv.waitKeyEx(30 if do_spin else -1)
if key == -1:
continue
elif key in (0x0D, 0x20): # ENTER (CR), SPACE
do_spin = not do_spin # toggle spinning
elif key == 27: # ESC
break # end program
elif key == 0x250000: # VK_LEFT
increment = -abs(increment)
angle += increment
elif key == 0x270000: # VK_RIGHT
increment = +abs(increment)
angle += increment
else:
print(f"key 0x{key:02x}")
cv.destroyAllWindows()

Active Brownian Motion

I am attempting to write a Python code to simulate many particles in a confined box. These particles behave in such a way that they move in the box in straight lines with a slight angular noise (small changes in the direction of the particle path). They should interact by acknowledging the other particle and 'shuffle/squeeze' past each other and continue on their intended path, much like humans on a busy street. Eventually, the particles should cluster together when the density of particles (or packing fraction) reaches a certain value, but I haven't got to this stage yet.
The code currently has particle interactions and I have attempted the angular noise but without success so far. The problems I would like some help with are the angular noise and the particle interactions. My idea for the angular noise was to multiply the data by a number between roughly 0.9 and 1.1 to change the direction of the particle slightly but after adding the lines in, nothing changed. The particles do interact but they seem to move in a fast semi-circle around the other interacting which is not what I want.
I don't think any knowledge or understanding of ABM is needed to write the angular noise code but some may be needed for the forces, I am not 100% sure though.
If anyone has any improvements for the code speed or ideas which may help with the interactions and/or angular noise that would be much appreciated. I will also leave an example of an animation which is my aim: https://warwick.ac.uk/fac/sci/physics/staff/research/cwhitfield/abpsimulations
The above link shows the animation I am looking for, although I don't need the sliders, just the box, and moving particles. The whole code is shown below:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
def set_initial_coordinates():
x_co = [np.random.uniform(0, 2) for i in range(n_particles)]
y_co = [np.random.uniform(0, 2) for i in range(n_particles)]
return x_co, y_co
def set_initial_velocities():
x_vel = np.array([np.random.uniform(-1, 1) for i in range(n_particles)])
y_vel = np.array([np.random.uniform(-1, 1) for i in range(n_particles)])
return x_vel, y_vel
def init():
ax.set_xlim(-0.05, 2.05)
ax.set_ylim(-0.07, 2.07)
return ln,
def update(dt):
xdata = initialx + vx * dt
ydata = initialy + vy * dt
fx = np.abs((xdata + 2) % 4 - 2)
fy = np.abs((ydata + 2) % 4 - 2)
for i in range(n_particles):
for j in range(n_particles):
if i == j:
continue
dx = fx[j] - fx[i] # distance in x direction
dy = fy[j] - fy[i] # distance in y direction
dr = np.sqrt((dx ** 2) + (dy ** 2)) # distance between x
if dr <= r:
force = k * ((2 * r) - dr) # size of the force if distance is less than or equal to radius
# Imagine a unit vector going from i to j
x_comp = dx / dr # x component of force
y_comp = dy / dr # y component of force
fx[i] += -x_comp * force # x force
fy[i] += -y_comp * force # y force
ln.set_data(fx, fy)
return ln,
# theta = np.random.uniform(0, 2) for i in range(n_particles)
n_particles = 10
initialx, initialy = set_initial_coordinates()
vx, vy = set_initial_velocities()
fig, ax = plt.subplots()
x_co, y_co = [], []
ln, = plt.plot([], [], 'bo', markersize=15) # radius 0.05
plt.xlim(0, 2)
plt.ylim(0, 2)
k = 1
r = 0.1
t = np.linspace(0, 10, 1000)
ani = FuncAnimation(fig, update, t, init_func=init, blit=True, repeat=False)
plt.show()

How to map rectangle image to quadrilateral with PIL?

Python PIL library allows me to map any quadrilateral in an image to rectangle using
im.transform(size, QUAD, data)
What I need is a function that does the opposite, i.e. map a rectangular image to specified quadrilateral.
I figured this might be achieved with the above mentioned function like this:
I.e. I would find such quad (the red one in the image) that would, using the function im.transform(size, QUAD, data) transform the image to quad I want. The problem is I don't know how to find the red quad.
I would appreciate any idea on how to find the red quad or any other way to map a rect image to quad, only with PIL if possible.

So I solved the issue with a simple forward mapping, rather than inverse mapping, which is usually better, but in my application I only ever map the rectangle to a quad that is smaller than the rectangle, so there are usually no holes in the transformed image. The code is as follows:
def reverse_quad_transform(image, quad_to_map_to, alpha):
# forward mapping, for simplicity
result = Image.new("RGBA",image.size)
result_pixels = result.load()
width, height = result.size
for y in range(height):
for x in range(width):
result_pixels[x,y] = (0,0,0,0)
p1 = (quad_to_map_to[0],quad_to_map_to[1])
p2 = (quad_to_map_to[2],quad_to_map_to[3])
p3 = (quad_to_map_to[4],quad_to_map_to[5])
p4 = (quad_to_map_to[6],quad_to_map_to[7])
p1_p2_vec = (p2[0] - p1[0],p2[1] - p1[1])
p4_p3_vec = (p3[0] - p4[0],p3[1] - p4[1])
for y in range(height):
for x in range(width):
pixel = image.getpixel((x,y))
y_percentage = y / float(height)
x_percentage = x / float(width)
# interpolate vertically
pa = (p1[0] + p1_p2_vec[0] * y_percentage, p1[1] + p1_p2_vec[1] * y_percentage)
pb = (p4[0] + p4_p3_vec[0] * y_percentage, p4[1] + p4_p3_vec[1] * y_percentage)
pa_to_pb_vec = (pb[0] - pa[0],pb[1] - pa[1])
# interpolate horizontally
p = (pa[0] + pa_to_pb_vec[0] * x_percentage, pa[1] + pa_to_pb_vec[1] * x_percentage)
try:
result_pixels[p[0],p[1]] = (pixel[0],pixel[1],pixel[2],min(int(alpha * 255),pixel[3]))
except Exception:
pass
return result

Elliptic orbit with polar method tracing phases, axes and Earth direction

I would like to represent the elliptical orbit of a binary system of two stars. What I aim to, is something like this:
Where I have a grid of the sizes along the axes, an in-scale star at the focus, and the orbit of the secondary star. The decimal numbers along the orbit are the orbital phases. The arrow at the bottom is the Earth direction, and the thick part at the orbit is related to the observation for that specific case - I don't need it. What I want to change from this plot is:
Orbital phase: instead of numbers along the orbit, I would like "dashed rays" from the focus to the orbit, and the orbital phase above them:
I don't want the cross along (0, 0);
I would like to re-orient the orbit, in order that the 0.0 phase is around the top left part of the plot, and the Earth direction is an upward pointing straight arrow (the parameters of my system are different from the one plotted here).
I tried to look for python examples, but the only thing I came out with (from here), is a polar plot:
which is not really representative of what I want, but still is a beginning:
import numpy as np
import matplotlib.pyplot as plt
cos = np.cos
pi = np.pi
a = 10
e = 0.1
theta = np.linspace(0,2*pi, 360)
r = (a*(1-e**2))/(1+e*cos(theta))
fig = plt.figure()
ax = fig.add_subplot(111, polar=True)
ax.set_yticklabels([])
ax.plot(theta,r)
print(np.c_[r,theta])
plt.show()

Here's something that gets you very close. You do not need polar coordinates to plot a decent ellipse. There is a so-called artist you can readily utilize.
You probably have to customize the axis labels and maybe insert an arrow or two if you want:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse
# initializing the figure:
fig = plt.figure()
# the (carthesian) axis:
ax = fig.add_subplot(111,aspect='equal')
ax.grid(True)
# parameters of the ellipse:
a = 5.0
e = 4.0
b = np.sqrt(a**2.0 - e**2.0)
# the center of the ellipse:
x = 6.0
y = 6.0
# the angle by which the ellipse is rotated:
angle = -45.0
#angle = 0.0
# plotting the ellipse, using an artist:
ax.add_artist(Ellipse(xy=[x,y], width=2.0*a, height=2.0*b, \
angle=angle, facecolor='none'))
ax.set_xlim(0,2.0*x)
ax.set_ylim(0,2.0*y)
# marking the focus (actually, both)
# and accounting for the rotation of the ellipse by angle
xf = [x - e*np.cos(angle * np.pi/180.0),
x + e*np.cos(angle * np.pi/180.0)]
yf = [y - e*np.sin(angle * np.pi/180.0),
y + e*np.sin(angle * np.pi/180.0)]
ax.plot(xf,yf,'xr')
# plotting lines from the focus to the ellipse:
# these should be your "rays"
t = np.arange(np.pi,3.0*np.pi,np.pi/5.0)
p = b**2.0 / a
E = e / a
r = [p/(1-E*np.cos(ti)) for ti in t]
# converting the radius based on the focus
# into x,y coordinates on the ellipse:
xr = [ri*np.cos(ti) for ri,ti in zip(r,t)]
yr = [ri*np.sin(ti) for ri,ti in zip(r,t)]
# accounting for the rotation by anlge:
xrp = [xi*np.cos(angle * np.pi/180.0) - \
yi*np.sin(angle * np.pi/180.0) for xi,yi in zip(xr,yr)]
yrp = [xi*np.sin(angle * np.pi/180.0) + \
yi*np.cos(angle * np.pi/180.0) for xi,yi in zip(xr,yr)]
for q in range(0,len(t)):
ax.plot([xf[0], xf[0]+xrp[q]],[yf[0], yf[0]+yrp[q]],'--b')
# put labels outside the "rays"
offset = 0.75
rLabel = [ri+offset for ri in r]
xrl = [ri*np.cos(ti) for ri,ti in zip(rLabel,t)]
yrl = [ri*np.sin(ti) for ri,ti in zip(rLabel,t)]
xrpl = [xi*np.cos(angle * np.pi/180.0) - \
yi*np.sin(angle * np.pi/180.0) for xi,yi in zip(xrl,yrl)]
yrpl = [xi*np.sin(angle * np.pi/180.0) + \
yi*np.cos(angle * np.pi/180.0) for xi,yi in zip(xrl,yrl)]
# for fancy label rotation reduce the range of the angle t:
tlabel = [(ti -np.pi)*180.0/np.pi for ti in t]
for q in range(0,len(tlabel)):
if tlabel[q] >= 180.0:
tlabel[q] -= 180.0
# convert the angle t from radians into degrees:
tl = [(ti-np.pi)*180.0/np.pi for ti in t]
for q in range(0,len(t)):
rotate_label = angle + tlabel[q]
label_text = '%.1f' % tl[q]
ax.text(xf[0]+xrpl[q],yf[0]+yrpl[q],label_text,\
va='center', ha='center',rotation=rotate_label)
plt.show()
The example above will result in this figure:
Explanations:
You can use an artist to plot the ellipse, instead of using polar coordinates
The nomenclature is based on the definitions available on Wikipedia
The angle in the artist setup rotates the ellipse. This angle is later used to rotate coordinates for the rays and labels (this is just math)
The rays are derived from the polar form of the ellipse relative to a focus.
The angle t runs from pi to 3.0*pi because I assumed that this would correspond to your idea of where the rays should start. You get the same rays for 0 to 2.0*pi. I used np.arange instead of linspace because I wanted a defined increment (pi/5.0, or 36 degrees) in this example.
The labels at the end of the rays are placed as text, the variable offset controls the distance between the ellipse and the labels. Adjust this as needed.
For the alignment of the label text orientation with the rays, I reduced the angle, t, to the range 0 to 180 degrees. This makes for better readability compared to the full range of 0 to 360 degrees.
For the label text I just used the angle, t, for simplicity. Replace this with whatever information better suits your purpose.
The angle, t, was converted from radians to degrees before the loop that places the label. Inside the loop, each element of tl is converted to a string. This allows for more formatting control (e.g. %.3f if you needed 3 decimals)

Stop pyplot.contour from drawing a contour along a discontinuity

I have a 2d map of a coordinate transform. The data at each point is the aximuthal angle in the original coordinate system, which goes from 0 to 360. I'm trying to use pyplot.contour to plot lines of constant angle, e.g. 45 degrees. The contour appears along the 45 degree line between the two poles, but there's an additional part to the contour that connects the two poles along the 0/360 discontinuity. This makes a very jagged ugly line as it basically just traces the pixels with a number close to 0 on one side and another close to 360 on the other.
Examples:
Here is an image using full colour map:
You can see the discontinuity along the blue/red curve on the left side. One side is 360 degrees, the other is 0 degrees. When plotting contours, I get:
Note that all contours connect the two poles, but even though I have NOT plotted the 0 degree contour, all the other contours follow along the 0 degree discontinuity (because pyplot thinks if it's 0 on one side and 360 on the other, there must be all other angles in between).
Code to produce this data:
import numpy as np
import matplotlib.pyplot as plt
jgal = np.array(
[
[-0.054875539726, -0.873437108010, -0.483834985808],
[0.494109453312, -0.444829589425, 0.746982251810],
[-0.867666135858, -0.198076386122, 0.455983795705],
]
)
def s2v3(rra, rdec, r):
pos0 = r * np.cos(rra) * np.cos(rdec)
pos1 = r * np.sin(rra) * np.cos(rdec)
pos2 = r * np.sin(rdec)
return np.array([pos0, pos1, pos2])
def v2s3(pos):
x = pos[0]
y = pos[1]
z = pos[2]
if np.isscalar(x):
x, y, z = np.array([x]), np.array([y]), np.array([z])
rra = np.arctan2(y, x)
low = np.where(rra < 0.0)
high = np.where(rra > 2.0 * np.pi)
if len(low[0]):
rra[low] = rra[low] + (2.0 * np.pi)
if len(high[0]):
rra[high] = rra[high] - (2.0 * np.pi)
rxy = np.sqrt(x ** 2 + y ** 2)
rdec = np.arctan2(z, rxy)
r = np.sqrt(x ** 2 + y ** 2 + z ** 2)
if x.size == 1:
rra = rra[0]
rdec = rdec[0]
r = r[0]
return rra, rdec, r
def gal2fk5(gl, gb):
rgl = np.deg2rad(gl)
rgb = np.deg2rad(gb)
r = 1.0
pos = s2v3(rgl, rgb, r)
pos1 = np.dot(pos.transpose(), jgal).transpose()
rra, rdec, r = v2s3(pos1)
dra = np.rad2deg(rra)
ddec = np.rad2deg(rdec)
return dra, ddec
def make_coords(resolution=50):
width = 9
height = 6
px = width * resolution
py = height * resolution
coords = np.zeros((px, py, 4))
for ix in range(0, px):
for iy in range(0, py):
l = 360.0 / px * ix - 180.0
b = 180.0 / py * iy - 90.0
dra, ddec = gal2fk5(l, b)
coords[ix, iy, 0] = dra
coords[ix, iy, 1] = ddec
coords[ix, iy, 2] = l
coords[ix, iy, 3] = b
return coords
coords = make_coords()
# now do one of these
# plt.imshow(coords[:,:,0],origin='lower') # color plot
plt.contour(
coords[:, :, 0], levels=[45, 90, 135, 180, 225, 270, 315]
) # contour plot with jagged ugliness
plt.show()
How can I either:
stop pyplot.contour from drawing a contour along the discontinuity
make pyplot.contour recognize that the 0/360 discontinuity in angle is not a real discontinuity at all.
I can just increase the resolution of the underlying data, but before I get a nice smooth line it starts to take a very long time and a lot of memory to plot.
I will also want to plot a contour along 0 degrees, but if I can figure out how to hide the discontinuity I can just shift it to somewhere else not near a contour. Or, if I can make #2 happen, it won't be an issue.

This is definitely still a hack, but you can get nice smooth contours with a two-fold approach:
Plot contours of the absolute value of the phase (going from -180˚ to 180˚) so that there is no discontinuity.
Plot two sets of contours in a finite region so that numerical defects close to the tops and bottoms of the extrema do not creep in.
Here is the complete code to append to your example:
Z = np.exp(1j*np.pi*coords[:,:,0]/180.0)
Z *= np.exp(0.25j*np.pi/2.0) # Shift to get same contours as in your example
X = np.arange(300)
Y = np.arange(450)
N = 2
levels = 90*(0.5 + (np.arange(N) + 0.5)/N)
c1 = plt.contour(X, Y, abs(np.angle(Z)*180/np.pi), levels=levels)
c2 = plt.contour(X, Y, abs(np.angle(Z*np.exp(0.5j*np.pi))*180/np.pi), levels=levels)
One can generalize this code to get smooth contours for any "periodic" function. What is left to be done is to generate a new set of contours with the correct values so that colormaps apply correctly, labels will be applied correctly etc. However, there does not seem to be a simple way of doing this with matplotlib: the relevant QuadContourSet class does everything and I do not see a simple way of constructing an appropriate contour object from the contours c1 and c2.

I was interested in the exact same problem. One solution is to NaN out the contours along the branch cut; see here; another is to use the max_jump argument in matplotx's contour().
I molded the solution into a Python package, cplot.
import cplot
import numpy as np
def f(z):
return np.exp(1 / z)
cplot.show(f, (-1.0, +1.0, 400), (-1.0, +1.0, 400))