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Counterclockwise sorting of x, y data

I have a set of points in a text file: random_shape.dat.
The initial order of points in the file is random. I would like to sort these points in a counter-clockwise order as follows (the red dots are the xy data):
I tried to achieve that by using the polar coordinates: I calculate the polar angle of each point (x,y) then sort by the ascending angles, as follows:
"""
Script: format_file.py
Description: This script will format the xy data file accordingly to be used with a program expecting CCW order of data points, By soting the points in Counterclockwise order
Example: python format_file.py random_shape.dat
"""
import sys
import numpy as np
# Read the file name
filename = sys.argv[1]
# Get the header name from the first line of the file (without the newline character)
with open(filename, 'r') as f:
header = f.readline().rstrip('\n')
angles = []
# Read the data from the file
x, y = np.loadtxt(filename, skiprows=1, unpack=True)
for xi, yi in zip(x, y):
angle = np.arctan2(yi, xi)
if angle < 0:
angle += 2*np.pi # map the angle to 0,2pi interval
angles.append(angle)
# create a numpy array
angles = np.array(angles)
# Get the arguments of sorted 'angles' array
angles_argsort = np.argsort(angles)
# Sort x and y
new_x = x[angles_argsort]
new_y = y[angles_argsort]
print("Length of new x:", len(new_x))
print("Length of new y:", len(new_y))
with open(filename.split('.')[0] + '_formatted.dat', 'w') as f:
print(header, file=f)
for xi, yi in zip(new_x, new_y):
print(xi, yi, file=f)
print("Done!")
By running the script:
python format_file.py random_shape.dat
Unfortunately I don't get the expected results in random_shape_formated.dat! The points are not sorted in the desired order.
Any help is appreciated.
EDIT: The expected resutls:
Create a new file named: filename_formatted.dat that contains the sorted data according to the image above (The first line contains the starting point, the next lines contain the points as shown by the blue arrows in counterclockwise direction in the image).
EDIT 2: The xy data added here instead of using github gist:
random_shape
0.4919261070361315 0.0861956168831175
0.4860816807027076 -0.06601587301587264
0.5023029456281289 -0.18238249845392662
0.5194784026079869 0.24347943722943777
0.5395164357511545 -0.3140611471861465
0.5570497147514262 0.36010146103896146
0.6074231036252226 -0.4142604617604615
0.6397066014669927 0.48590810704447085
0.7048302091822873 -0.5173701298701294
0.7499157837544145 0.5698170011806378
0.8000108666123336 -0.6199254449254443
0.8601249660418364 0.6500974025974031
0.9002010323281716 -0.7196585989767801
0.9703341483292582 0.7299242424242429
1.0104102146155935 -0.7931355765446666
1.0805433306166803 0.8102046438410078
1.1206193969030154 -0.865251869342778
1.1907525129041021 0.8909386068476981
1.2308285791904374 -0.9360074773711129
1.300961695191524 0.971219008264463
1.3410377614778592 -1.0076702085792988
1.4111708774789458 1.051499409681228
1.451246943765281 -1.0788793781975592
1.5213800597663678 1.1317798110979933
1.561456126052703 -1.1509956709956706
1.6315892420537896 1.2120602125147582
1.671665308340125 -1.221751279024005
1.7417984243412115 1.2923406139315234
1.7818744906275468 -1.2943211334120424
1.8520076066286335 1.3726210153482883
1.8920836729149686 -1.3596340023612745
1.9622167889160553 1.4533549783549786
2.0022928552023904 -1.4086186540731989
2.072425971203477 1.5331818181818184
2.1125020374898122 -1.451707005116095
2.182635153490899 1.6134622195985833
2.2227112197772345 -1.4884454939000387
2.292844335778321 1.6937426210153486
2.3329204020646563 -1.5192876820149541
2.403053518065743 1.774476584022039
2.443129584352078 -1.5433264462809912
2.513262700353165 1.8547569854388037
2.5533387666395 -1.561015348288075
2.6234718826405867 1.9345838252656438
2.663547948926922 -1.5719008264462806
2.7336810649280086 1.9858362849271942
2.7737571312143436 -1.5750757575757568
2.8438902472154304 2.009421487603306
2.883966313501766 -1.5687258953168035
2.954099429502852 2.023481896890988
2.9941754957891877 -1.5564797323888229
3.0643086117902745 2.0243890200708385
3.1043846780766096 -1.536523022432113
3.1745177940776963 2.0085143644234558
3.2145938603640314 -1.5088557654466737
3.284726976365118 1.9749508067689887
3.324803042651453 -1.472570838252656
3.39493615865254 1.919162731208186
3.435012224938875 -1.4285753640299088
3.5051453409399618 1.8343467138921687
3.545221407226297 -1.3786835891381335
3.6053355066557997 1.7260966810966811
3.655430589513719 -1.3197205824478546
3.6854876392284703 1.6130086580086582
3.765639771801141 -1.2544077134986225
3.750611246943765 1.5024152236652237
3.805715838087476 1.3785173160173163
3.850244800627849 1.2787337662337666
3.875848954088563 -1.1827449822904361
3.919007794704616 1.1336638361638363
3.9860581363759846 -1.1074537583628485
3.9860581363759846 1.0004485329485333
4.058012891753723 0.876878197560016
4.096267318663407 -1.0303482880755608
4.15638141809291 0.7443374218374221
4.206476500950829 -0.9514285714285711
4.256571583808748 0.6491902794175526
4.3166856832382505 -0.8738695395513574
4.36678076609617 0.593855765446675
4.426894865525672 -0.7981247540338443
4.476989948383592 0.5802489177489183
4.537104047813094 -0.72918339236521
4.587199130671014 0.5902272727272733
4.647313230100516 -0.667045454545454
4.697408312958435 0.6246979535615904
4.757522412387939 -0.6148858717040526
4.807617495245857 0.6754968516332154
4.8677315946753605 -0.5754260133805582
4.917826677533279 0.7163173947264858
4.977940776962782 -0.5500265643447455
5.028035859820701 0.7448917748917752
5.088149959250204 -0.5373268398268394
5.138245042108123 0.7702912239275879
5.198359141537626 -0.5445838252656432
5.2484542243955445 0.7897943722943728
5.308568323825048 -0.5618191656828015
5.358663406682967 0.8052154663518301
5.41877750611247 -0.5844972451790631
5.468872588970389 0.8156473829201105
5.5289866883998915 -0.6067217630853987
5.579081771257811 0.8197294372294377
5.639195870687313 -0.6248642266824076
5.689290953545233 0.8197294372294377
5.749405052974735 -0.6398317591499403
5.799500135832655 0.8142866981503349
5.859614235262157 -0.6493565525383702
5.909709318120076 0.8006798504525783
5.969823417549579 -0.6570670995670991
6.019918500407498 0.7811767020857934
6.080032599837001 -0.6570670995670991
6.13012768269492 0.7562308146399057
6.190241782124423 -0.653438606847697
6.240336864982342 0.7217601338055886
6.300450964411845 -0.6420995670995664
6.350546047269764 0.6777646595828419
6.410660146699267 -0.6225964187327819
6.4607552295571855 0.6242443919716649
6.520869328986689 -0.5922077922077915
6.570964411844607 0.5548494687131056
6.631078511274111 -0.5495730027548205
6.681173594132029 0.4686727666273125
6.7412876935615325 -0.4860743801652889
6.781363759847868 0.3679316979316982
6.84147785927737 -0.39541245791245716
6.861515892420538 0.25880333951762546
6.926639500135833 -0.28237987012986965
6.917336127605076 0.14262677798392165
6.946677533279001 0.05098957832291173
6.967431210462995 -0.13605442176870675
6.965045730326905 -0.03674603174603108

I find that an easy way to sort points with x,y-coordinates like that is to sort them dependent on the angle between the line from the points and the center of mass of the whole polygon and the horizontal line which is called alpha in the example. The coordinates of the center of mass (x0 and y0) can easily be calculated by averaging the x,y coordinates of all points. Then you calculate the angle using numpy.arccos for instance. When y-y0 is larger than 0 you take the angle directly, otherwise you subtract the angle from 360° (2𝜋). I have used numpy.where for the calculation of the angle and then numpy.argsort to produce a mask for indexing the initial x,y-values. The following function sort_xy sorts all x and y coordinates with respect to this angle. If you want to start from any other point you could add an offset angle for that. In your case that would be zero though.
def sort_xy(x, y):
x0 = np.mean(x)
y0 = np.mean(y)
r = np.sqrt((x-x0)**2 + (y-y0)**2)
angles = np.where((y-y0) > 0, np.arccos((x-x0)/r), 2*np.pi-np.arccos((x-x0)/r))
mask = np.argsort(angles)
x_sorted = x[mask]
y_sorted = y[mask]
return x_sorted, y_sorted
Plotting x, y before sorting using matplotlib.pyplot.plot (points are obvisously not sorted):
Plotting x, y using matplotlib.pyplot.plot after sorting with this method:

If it is certain that the curve does not cross the same X coordinate (i.e. any vertical line) more than twice, then you could visit the points in X-sorted order and append a point to one of two tracks you follow: to the one whose last end point is the closest to the new one. One of these tracks will represent the "upper" part of the curve, and the other, the "lower" one.
The logic would be as follows:
dist2 = lambda a,b: (a[0]-b[0])*(a[0]-b[0]) + (a[1]-b[1])*(a[1]-b[1])
z = list(zip(x, y)) # get the list of coordinate pairs
z.sort() # sort by x coordinate
cw = z[0:1] # first point in clockwise direction
ccw = z[1:2] # first point in counter clockwise direction
# reverse the above assignment depending on how first 2 points relate
if z[1][1] > z[0][1]:
cw = z[1:2]
ccw = z[0:1]
for p in z[2:]:
# append to the list to which the next point is closest
if dist2(cw[-1], p) < dist2(ccw[-1], p):
cw.append(p)
else:
ccw.append(p)
cw.reverse()
result = cw + ccw
This would also work for a curve with steep fluctuations in the Y-coordinate, for which an angle-look-around from some central point would fail, like here:
No assumption is made about the range of the X nor of the Y coordinate: like for instance, the curve does not necessarily have to cross the X axis (Y = 0) for this to work.

Counter-clock-wise order depends on the choice of a pivot point. From your question, one good choice of the pivot point is the center of mass.
Something like this:
# Find the Center of Mass: data is a numpy array of shape (Npoints, 2)
mean = np.mean(data, axis=0)
# Compute angles
angles = np.arctan2((data-mean)[:, 1], (data-mean)[:, 0])
# Transform angles from [-pi,pi] -> [0, 2*pi]
angles[angles < 0] = angles[angles < 0] + 2 * np.pi
# Sort
sorting_indices = np.argsort(angles)
sorted_data = data[sorting_indices]

Not really a python question I think, but still I think you could try sorting by - sign(y) * x doing something like:
def counter_clockwise_sort(points):
return sorted(points, key=lambda point: point['x'] * (-1 if point['y'] >= 0 else 1))
should work fine, assuming you read your points properly into a list of dicts of format {'x': 0.12312, 'y': 0.912}
EDIT: This will work as long as you cross the X axis only twice, like in your example.

If:
the shape is arbitrarily complex and
the point spacing is ~random
then I think this is a really hard problem.
For what it's worth, I have faced a similar problem in the past, and I used a traveling salesman solver. In particular, I used the LKH solver. I see there is a Python repo for solving the problem, LKH-TSP. Once you have an order to the points, I don't think it will be too hard to decide on a clockwise vs clockwise ordering.

If we want to answer your specific problem, we need to pick a pivot point.
Since you want to sort according to the starting point you picked, I would take a pivot in the middle (x=4,y=0 will do).
Since we're sorting counterclockwise, we'll take arctan2(-(y-pivot_y),-(x-center_x)) (we're flipping the x axis).
We get the following, with a gradient colored scatter to prove correctness (fyi I removed the first line of the dat file after downloading):
import numpy as np
import matplotlib.pyplot as plt
points = np.loadtxt('points.dat')
#oneliner for ordering points (transform, adjust for 0 to 2pi, argsort, index at points)
ordered_points = points[np.argsort(np.apply_along_axis(lambda x: np.arctan2(-x[1],-x[0]+4) + np.pi*2, axis=1,arr=points)),:]
#color coding 0-1 as str for gray colormap in matplotlib
plt.scatter(ordered_points[:,0], ordered_points[:,1],c=[str(x) for x in np.arange(len(ordered_points)) / len(ordered_points)],cmap='gray')
Result (in the colormap 1 is white and 0 is black), they're numbered in the 0-1 range by order:

For points with comparable distances between their neighbouring pts, we can use KDTree to get two closest pts for each pt. Then draw lines connecting those to give us a closed shape contour. Then, we will make use of OpenCV's findContours to get contour traced always in counter-clockwise manner. Now, since OpenCV works on images, we need to sample data from the provided float format to uint8 image format. Given, comparable distances between two pts, that should be pretty safe. Also, OpenCV handles it well to make sure it traces even sharp corners in curvatures, i.e. smooth or not-smooth data would work just fine. And, there's no pivot requirement, etc. As such all kinds of shapes would be good to work with.
Here'e the implementation -
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial.distance import pdist
from scipy.spatial import cKDTree
import cv2
from scipy.ndimage.morphology import binary_fill_holes
def counter_clockwise_order(a, DEBUG_PLOT=False):
b = a-a.min(0)
d = pdist(b).min()
c = np.round(2*b/d).astype(int)
img = np.zeros(c.max(0)[::-1]+1, dtype=np.uint8)
d1,d2 = cKDTree(c).query(c,k=3)
b = c[d2]
p1,p2,p3 = b[:,0],b[:,1],b[:,2]
for i in range(len(b)):
cv2.line(img,tuple(p1[i]),tuple(p2[i]),255,1)
cv2.line(img,tuple(p1[i]),tuple(p3[i]),255,1)
img = (binary_fill_holes(img==255)*255).astype(np.uint8)
if int(cv2.__version__.split('.')[0])>=3:
_,contours,hierarchy = cv2.findContours(img.copy(),cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
else:
contours,hierarchy = cv2.findContours(img.copy(),cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
cont = contours[0][:,0]
f1,f2 = cKDTree(cont).query(c,k=1)
ordered_points = a[f2.argsort()[::-1]]
if DEBUG_PLOT==1:
NPOINTS = len(ordered_points)
for i in range(NPOINTS):
plt.plot(ordered_points[i:i+2,0],ordered_points[i:i+2,1],alpha=float(i)/(NPOINTS-1),color='k')
plt.show()
return ordered_points
Sample run -
# Load data in a 2D array with 2 columns
a = np.loadtxt('random_shape.csv',delimiter=' ')
ordered_a = counter_clockwise_order(a, DEBUG_PLOT=1)
Output -

Create array mask from randomly ordered list

I have a dataset made of velocity data on an unstructured grid from a CFD simulation, in the structure:
data = [[x1, y1, u1, v1], ... , [xn, yn, un, vn]]
I need to have a regular grid inside the area covered by this data. However, I do not have information about the boundaries of the x, y domain other than the x, y values itself. The boundary is defined by a complex geometrical shape.
My solution would be to create a rectangular grid with numpy.mgrid and then construct an array mask to mask out areas with no data.
But I have no idea how to get a mask just from the randomly ordered coordinates. I tried using scipy's ConvexHull to find the boundaries but it is a concave problem. However, even if I had the boundary points, I am not sure how to create the mask from it, since the indices are not the same as in the regular grid.
How to determine this grid? Is there any other possibility? Maybe its useful to reorder the dataset?

I'm really unfamiliar with your use-case so I may be way off base here. It sounds like you're effectively looking for the min/max for each coordinate system to create a mask for? Conceptually (maybe not efficient for large datasets):
x_min = min([a[0] for a in data])
With that I think you'd be able to say something like "the domain of x is [x_min, x_max]"
Here's a full example that you can copy/past to see if it produces what you're looking for:
from random import randint
# Convenience
def r():
return randint(-100, 100)
# Generate 100 random coordinates
data = [[r(), r(), r(), r()] for _ in range(0, 100)]
x_min = min([a[0] for a in data])
x_max = max([a[0] for a in data])
y_min = min([a[1] for a in data])
y_max = max([a[1] for a in data])
u_min = min([a[2] for a in data])
u_max = max([a[2] for a in data])
v_min = min([a[3] for a in data])
v_max = max([a[3] for a in data])
print(f'X-Range: {x_min} to {x_max}')
print(f'Y-Range: {y_min} to {y_max}')
print(f'U-Range: {u_min} to {u_max}')
print(f'V-Range: {v_min} to {v_max}')
That produces this:
X-Range: -98 to 96
Y-Range: -100 to 96
U-Range: -95 to 100
V-Range: -100 to 100
While any single entry within data might be this:
print(data[randint(0, len(data)])
[70, -69, -59, -49]

How to interpolate a line between two other lines in python

Note: I asked this question before but it was closed as a duplicate, however, I, along with several others believe it was unduely closed, I explain why in an edit in my original post. So I would like to re-ask this question here again.
Does anyone know of a python library that can interpolate between two lines. For example, given the two solid lines below, I would like to produce the dashed line in the middle. In other words, I'd like to get the centreline. The input is a just two numpy arrays of coordinates with size N x 2 and M x 2 respectively.
Furthermore, I'd like to know if someone has written a function for this in some optimized python library. Although optimization isn't exactly a necessary.
Here is an example of two lines that I might have, you can assume they do not overlap with each other and an x/y can have multiple y/x coordinates.
array([[ 1233.87375018, 1230.07095987],
[ 1237.63559365, 1253.90749041],
[ 1240.87500801, 1264.43925132],
[ 1245.30875975, 1274.63795396],
[ 1256.1449357 , 1294.48254424],
[ 1264.33600095, 1304.47893299],
[ 1273.38192911, 1313.71468591],
[ 1283.12411536, 1322.35942538],
[ 1293.2559388 , 1330.55873344],
[ 1309.4817002 , 1342.53074698],
[ 1325.7074616 , 1354.50276051],
[ 1341.93322301, 1366.47477405],
[ 1358.15898441, 1378.44678759],
[ 1394.38474581, 1390.41880113]])
array([[ 1152.27115094, 1281.52899302],
[ 1155.53345506, 1295.30515742],
[ 1163.56506781, 1318.41642169],
[ 1168.03497425, 1330.03181319],
[ 1173.26135672, 1341.30559949],
[ 1184.07110925, 1356.54121651],
[ 1194.88086178, 1371.77683353],
[ 1202.58908737, 1381.41765447],
[ 1210.72465255, 1390.65097106],
[ 1227.81309742, 1403.2904646 ],
[ 1244.90154229, 1415.92995815],
[ 1261.98998716, 1428.56945169],
[ 1275.89219696, 1438.21626352],
[ 1289.79440676, 1447.86307535],
[ 1303.69661656, 1457.50988719],
[ 1323.80994319, 1470.41028655],
[ 1343.92326983, 1488.31068591],
[ 1354.31738934, 1499.33260989],
[ 1374.48879779, 1516.93734053],
[ 1394.66020624, 1534.54207116]])
Visualizing this we have:
So my attempt at this has been using the skeletonize function in the skimage.morphology library by first rasterizing the coordinates into a filled in polygon. However, I get branching at the ends like this:

First of all, pardon the overkill; I had fun with your question. If the description is too long, feel free to skip to the bottom, I defined a function that does everything I describe.
Your problem would be relatively straightforward if your arrays were the same length. In that case, all you would have to do is find the average between the corresponding x values in each array, and the corresponding y values in each array.
So what we can do is create arrays of the same length, that are more or less good estimates of your original arrays. We can do this by fitting a polynomial to the arrays you have. As noted in comments and other answers, the midline of your original arrays is not specifically defined, so a good estimate should fulfill your needs.
Note: In all of these examples, I've gone ahead and named the two arrays that you posted a1 and a2.
Step one: Create new arrays that estimate your old lines
Looking at the data you posted:
These aren't particularly complicated functions, it looks like a 3rd degree polynomial would fit them pretty well. We can create those using numpy:
import numpy as np
# Find the range of x values in a1
min_a1_x, max_a1_x = min(a1[:,0]), max(a1[:,0])
# Create an evenly spaced array that ranges from the minimum to the maximum
# I used 100 elements, but you can use more or fewer.
# This will be used as your new x coordinates
new_a1_x = np.linspace(min_a1_x, max_a1_x, 100)
# Fit a 3rd degree polynomial to your data
a1_coefs = np.polyfit(a1[:,0],a1[:,1], 3)
# Get your new y coordinates from the coefficients of the above polynomial
new_a1_y = np.polyval(a1_coefs, new_a1_x)
# Repeat for array 2:
min_a2_x, max_a2_x = min(a2[:,0]), max(a2[:,0])
new_a2_x = np.linspace(min_a2_x, max_a2_x, 100)
a2_coefs = np.polyfit(a2[:,0],a2[:,1], 3)
new_a2_y = np.polyval(a2_coefs, new_a2_x)
The result:
That's not bad so bad! If you have more complicated functions, you'll have to fit a higher degree polynomial, or find some other adequate function to fit to your data.
Now, you've got two sets of arrays of the same length (I chose a length of 100, you can do more or less depending on how smooth you want your midpoint line to be). These sets represent the x and y coordinates of the estimates of your original arrays. In the example above, I named these new_a1_x, new_a1_y, new_a2_x and new_a2_y.
Step two: calculate the average between each x and each y in your new arrays
Then, we want to find the average x and average y value for each of our estimate arrays. Just use np.mean:
midx = [np.mean([new_a1_x[i], new_a2_x[i]]) for i in range(100)]
midy = [np.mean([new_a1_y[i], new_a2_y[i]]) for i in range(100)]
midx and midy now represent the midpoint between our 2 estimate arrays. Now, just plot your original (not estimate) arrays, alongside your midpoint array:
plt.plot(a1[:,0], a1[:,1],c='black')
plt.plot(a2[:,0], a2[:,1],c='black')
plt.plot(midx, midy, '--', c='black')
plt.show()
And voilà:
This method still works with more complex, noisy data (but you have to fit the function thoughtfully):
As a function:
I've put the above code in a function, so you can use it easily. It returns an array of your estimated midpoints, in the format you had your original arrays in.
The arguments: a1 and a2 are your 2 input arrays, poly_deg is the degree polynomial you want to fit, n_points is the number of points you want in your midpoint array, and plot is a boolean, whether you want to plot it or not.
import matplotlib.pyplot as plt
import numpy as np
def interpolate(a1, a2, poly_deg=3, n_points=100, plot=True):
min_a1_x, max_a1_x = min(a1[:,0]), max(a1[:,0])
new_a1_x = np.linspace(min_a1_x, max_a1_x, n_points)
a1_coefs = np.polyfit(a1[:,0],a1[:,1], poly_deg)
new_a1_y = np.polyval(a1_coefs, new_a1_x)
min_a2_x, max_a2_x = min(a2[:,0]), max(a2[:,0])
new_a2_x = np.linspace(min_a2_x, max_a2_x, n_points)
a2_coefs = np.polyfit(a2[:,0],a2[:,1], poly_deg)
new_a2_y = np.polyval(a2_coefs, new_a2_x)
midx = [np.mean([new_a1_x[i], new_a2_x[i]]) for i in range(n_points)]
midy = [np.mean([new_a1_y[i], new_a2_y[i]]) for i in range(n_points)]
if plot:
plt.plot(a1[:,0], a1[:,1],c='black')
plt.plot(a2[:,0], a2[:,1],c='black')
plt.plot(midx, midy, '--', c='black')
plt.show()
return np.array([[x, y] for x, y in zip(midx, midy)])
[EDIT]:
I was thinking back on this question, and I overlooked a simpler way to do this, by "densifying" both arrays to the same number of points using np.interp. This method follows the same basic idea as the line-fitting method above, but instead of approximating lines using polyfit / polyval, it just densifies:
min_a1_x, max_a1_x = min(a1[:,0]), max(a1[:,0])
min_a2_x, max_a2_x = min(a2[:,0]), max(a2[:,0])
new_a1_x = np.linspace(min_a1_x, max_a1_x, 100)
new_a2_x = np.linspace(min_a2_x, max_a2_x, 100)
new_a1_y = np.interp(new_a1_x, a1[:,0], a1[:,1])
new_a2_y = np.interp(new_a2_x, a2[:,0], a2[:,1])
midx = [np.mean([new_a1_x[i], new_a2_x[i]]) for i in range(100)]
midy = [np.mean([new_a1_y[i], new_a2_y[i]]) for i in range(100)]
plt.plot(a1[:,0], a1[:,1],c='black')
plt.plot(a2[:,0], a2[:,1],c='black')
plt.plot(midx, midy, '--', c='black')
plt.show()

The "line between two lines" is not so well defined. You can obtain a decent though simple solution by triangulating between the two curves (you can triangulate by progressing from vertex to vertex, choosing the diagonals that produce the less skewed triangle).
Then the interpolated curve joins the middles of the sides.

I work with rivers, so this is a common problem. One of my solutions is exactly like the one you showed in your question--i.e. skeletonize the blob. You see that the boundaries have problems, so what I've done that seems to work well is to simply mirror the boundaries. For this approach to work, the blob must not intersect the corners of the image.
You can find my implementation in RivGraph; this particular algorithm is in rivers/river_utils.py called "mask_to_centerline".
Here's an example output showing how the ends of the centerline extend to the desired edge of the object:

sacuL's solution almost worked for me, but I needed to aggregate more than just two curves.
Here is my generalization for sacuL's solution:
def interp(*axis_list):
min_max_xs = [(min(axis[:,0]), max(axis[:,0])) for axis in axis_list]
new_axis_xs = [np.linspace(min_x, max_x, 100) for min_x, max_x in min_max_xs]
new_axis_ys = [np.interp(new_x_axis, axis[:,0], axis[:,1]) for axis, new_x_axis in zip(axis_list, new_axis_xs)]
midx = [np.mean([new_axis_xs[axis_idx][i] for axis_idx in range(len(axis_list))]) for i in range(100)]
midy = [np.mean([new_axis_ys[axis_idx][i] for axis_idx in range(len(axis_list))]) for i in range(100)]
for axis in axis_list:
plt.plot(axis[:,0], axis[:,1],c='black')
plt.plot(midx, midy, '--', c='black')
plt.show()
If we now run an example:
a1 = np.array([[x, x**2+5*(x%4)] for x in range(10)])
a2 = np.array([[x-0.5, x**2+6*(x%3)] for x in range(10)])
a3 = np.array([[x+0.2, x**2+7*(x%2)] for x in range(10)])
interp(a1, a2, a3)
we get the plot:

python (sympy) implicit function: get values instead of plot?

I am new to sympy but I already get a nice output when I plot the implicit function (actually the formula for Cassini's ovals) using sympy:
from sympy import plot_implicit, symbols, Eq, solve
x, y = symbols('x y')
k=2.7
a=3
eq = Eq((x**2 + y**2)**2-2*a**2*(x**2-y**2), k**4-a**4)
plot_implicit(eq)
Now is it actually possible to somehow get the x and y values corresponding to the plot? or alternatively solve the implicit equation without plotting at all?
thanks! :-)

This is an answer addressing your
is it actually possible to somehow get the x and y values corresponding to the plot?
and I say "addressing" because it's not possible to get the x and y values used to draw the curves — because the curves are not drawn using a sequenc of 2D points… more on this later,
TL;DR
pli = plot_implicit(...)
series = pli[0]
data, action = series.get_points()
data = np.array([(x_int.mid, y_int.mid) for x_int, y_int in data])
Let's start with your code
from sympy import plot_implicit, symbols, Eq, solve
x, y = symbols('x y')
k=2.7
a=3
eq = Eq((x**2 + y**2)**2-2*a**2*(x**2-y**2), k**4-a**4)
and plot it, with a twist: we save the Plot object and print it
pli = plot_implicit(eq)
print(pli)
to get
Plot object containing:
[0]: Implicit equation: Eq(-18*x**2 + 18*y**2 + (x**2 + y**2)**2, -27.8559000000000) for x over (-5.0, 5.0) and y over (-5.0, 5.0)
We are interested in this object indexed by 0,
ob = pli[0]
print(dir(ob))
that gives (ellipsis are mine)
['__class__', …, get_points, …, 'var_y']
The name get_points sounds full of promise, doesn't it?
print(ob.get_points())
that gives (edited for clarity and with a big cut)
([
[interval(-3.759774, -3.750008), interval(-0.791016, -0.781250)],
[interval(-3.876961, -3.867195), interval(-0.634768, -0.625003)],
[interval(-3.837898, -3.828133), interval(-0.693361, -0.683596)],
[interval(-3.847664, -3.837898), interval(-0.673830, -0.664065)],
...
[interval(3.837895, 3.847661), interval(0.664064, 0.673830)],
[interval(3.828130, 3.837895), interval(0.683596, 0.693362)],
[interval(3.867192, 3.876958), interval(0.625001, 0.634766)],
[interval(3.750005, 3.759770), interval(0.781255, 0.791021)]
], 'fill')
What is this? the documentation of plot_implicit has
plot_implicit, by default, uses interval arithmetic to plot functions.
Following the source code of plot_implicit.py and plot,py one realizes that, in this case, the actual plotting (speaking of the matpolotlib backend) is just a line of code
self.ax.fill(x, y, facecolor=s.line_color, edgecolor='None')
where x and y are constructed from the list of intervals, as returned from .get_points(), as follows
x, y = [], []
for intervals in interval_list:
intervalx = intervals[0]
intervaly = intervals[1]
x.extend([intervalx.start, intervalx.start,
intervalx.end, intervalx.end, None])
y.extend([intervaly.start, intervaly.end,
intervaly.end, intervaly.start, None])
so that for each couple of intervals matplotlib is directed to draw a filled rectangle, small enough that the eye sees a continuous line (note the use of None to have disjoint rectangles).
We can conclude that the list of couples of intervals
l_xy_intervals = ((pli[0]).get_points())[0]
represents rectangular areas where the implicit expression you are plotting is
"true enough"

You can do this, even with interval math, if you try getting the mid point of each interval. Starting from your code, and slightly change it, by saving the plot_implicit object in a variable called g we have:
from sympy import plot_implicit, symbols, Eq, solve
x, y = symbols('x y')
k=2.7
a=3
eq = Eq((x**2 + y**2)**2-2*a**2*(x**2-y**2), k**4-a**4)
g = plot_implicit(eq)
Now let's save in a variable named ptos the intervals that were used to draw the plot.
ptos = g[0].get_points()[0]
This way ptos[0][0] will be the first interval in the x axis and ptos[0][1] will be its pair in the y axis. The intervals have a property called mid which gives the middle point of the interval. So you can suppose that ptos[0][0].mid, ptos[0][1].mid will be a pair x,y "true enough" to be one of our numerical solutions.
This way, a data frame composed of this middle point pairs can be generated with:
intervs = np.array(dtype='object')
meio = lambda x0:x0.mid
px = list(map(meio, intervs[:,0]))
py = list(map(meio, intervs[:,1]))
import pandas as pd
dados = pd.DataFrame({'x':px, 'y':px})
dados.head()
Which in this example would give us:
x y
0 -1.177733 0.598826
1 -1.175389 0.596483
2 -1.175389 0.598826
3 -1.173045 0.596483
4 -1.173045 0.598826
This idea of getting the intervals middle points can be used whenever one needs to move from "interval math" to "standard" point level math. Hope this helps. Regards.

Return the value of a 2D PDF given x and y in Python?

I have some data that I plotted the PDF using matplotlib's hist2D function.
The result looks like this:
The hist2d function returns a triple of arrays: H,xedges,yedges. H being the 2D histogram value.
Now I'd like to turn this discrete H matrix and turn it into a function, that returns the value of H for any given (x,y) input.
In other words I'd like to turn my 2D histogram into a 2D step function. Is there a specific function that would be computationally cheap that I could use on that purpose?
This looks like a pretty simple operation (usually done for image processing but with pixel indices instead of real numbers) but I'm unable to find anything about it, can you please help me?

You can construct an interpolator from the counts like this:
from numpy import random, histogram2d, diff
import matplotlib.pyplot as plt
from scipy.interpolate import interp2d
# Generate sample data
n = 10000
x = random.randn(n)
y = -x + random.randn(n)
# bin
nbins = 100
H, xedges, yedges = histogram2d(x, y, bins=nbins)
# Figure out centers of bins
def centers(edges):
return edges[:-1] + diff(edges[:2])/2
xcenters = centers(xedges)
ycenters = centers(yedges)
# Construct interpolator
pdf = interp2d(xcenters, ycenters, H)
# test
plt.pcolor(xedges, yedges, pdf(xedges, yedges))
Result:
Note that this will be linearly interpolated rather than step-wise. For a quicker version which assumes a regular grid, this will also work:
from numpy import meshgrid, vectorize
def position(edges, value):
return int((value - edges[0])/diff(edges[:2]))
#vectorize
def pdf2(x, y):
return H[position(yedges, y), position(xedges, x)]
# test - note we need the meshgrid here to get the right shapes
xx, yy = meshgrid(xcenters, ycenters)
plt.pcolor(xedges, yedges, pdf2(xx, yy))