I have an image (saved as a variable called canny_image) and it looks like this after preprocessing.
I am basically trying to find the distance between the first two vertical lines. I tried using the hough_line function from skimage, but it's unable to find the first line, so I thought it might be easier to solve this manually.
I am basically trying to solve this by going through each row in the image until I get to the first pixel with a value of 255, (the lines have a value of 255, while everything else is zero), and then I store the location of that pixel in an array. And I take the mode of the values in the array as the x location of the first line. I'll do the same for the 2nd line by using the first x-value as a starting point.
def find_lines(canny_image):
threshold = 255
for y in range(canny_image.shape[0]):
for x in range(canny_image.shape[1]):
if canny_image[x, y] == threshold:
return x
This is the code I wrote to get the x-location of the first line, however, I'm not getting the desired output. Any help on how to solve this will be much appreciated. Thanks!
Perhaps try something like this
# Returns an array of lines x positions in an image
def find_line_x_positions(image, lines_to_detect: int, buffer_zone: int):
threshold = 255
(height, width) = image.shape
x_position_sums = np.zeros((lines_to_detect, 2), np.double) # For each line, store x_pos sum[i,0] and point count[i,1]
for y in range(height):
buffer = 0
line_index = 0
for x in range(width):
if buffer > 0:
buffer -= 1
if (image[y, x] >= threshold) and (buffer == 0):
buffer = buffer_zone
x_position_sums[line_index, 0] += x
x_position_sums[line_index, 1] += 1
line_index += 1
if ((line_index) == lines_to_detect):
break
# Divide the x position sums by the point counts to get the average x position for each line
results = x_position_sums[np.all(x_position_sums,axis=1)]
results = np.divide(results[:,0],results[:,1])
return results
You can also try OpenCV's HoughLines() function which is simpler to implement than scikit lib. When I tested OpenCV implementation out, it seems to have a hard time finding vertical lines(within 10 degrees from vertical) but you can solve this by rotating your image X degrees and look for lines within that range of rotation.
I was making some experimentations with the OpenCV function cv2.warpPerspective when I decided to code it from scratch to better understand its pipeline. Though I followed (hopefully) every theoretical step, it seems I am still missing something and I am struggling a lot to understand what. Could you please help me?
SRC image (left) and True DST Image (right)
Output of the cv2.warpPerspective overlapped on the True DST
# Invert the homography SRC->DST to DST->SRC
hinv = np.linalg.inv(h)
src = gray1
dst = np.zeros(gray2.shape)
h, w = src.shape
# Remap back and check the domain
for ox in range(h):
for oy in range(w):
# Backproject from DST to SRC
xw, yw, w = hinv.dot(np.array([ox, oy, 1]).T)
# cv2.INTER_NEAREST
x, y = int(xw/w), int(yw/w)
# Check if it falls in the src domain
c1 = x >= 0 and y < h
c2 = y >= 0 and y < w
if c1 and c2:
dst[x, y] = src[ox, oy]
cv2.imshow(dst + gray2//2)
Output of my code
PS: The output images are the overlapping of Estimated DST and the True DST to better highlight differences.
Your issue amounts to a typo. You mixed up the naming of your coordinates. The homography assumes (x,y,1) order, which would correspond to (j,i,1).
Just use (x, y, 1) in the calculation, and (xw, yw, w) in the result of that (then x,y = xw/w, yw/w). the w factor mirrors the math, when formulated properly.
Avoid indexing into .shape. The indices don't "speak". Just do (height, width) = src.shape[:2] and use those.
I'd recommend to fix the naming scheme, or define it up top in a comment. I'd recommend sticking with x,y instead of i,j,u,v, and then extend those with prefixes/suffixes for the space they're in ("src/dst/in/out"). Perhaps something like ox,oy for iterating, just xw,yw,w for the homography result, which turns into x,y via division, and ix,iy (integerized) for sampling in the input? Then you can use dst[oy, ox] = src[iy, ix]
The shape of a shapely polygon can easily be converted to an array of points by using
x,y = polygon.exterior.xy
However, this returns only the actual points.
How can I convert a shapely polygon to an array, with a 0 for each pixel outside of the shape, and a 1 for each pixel inside the shape?
Thus, for an example, a polygon with width=100, height=100 should result in a (100,100) array.
I could do this by getting the exterior points, and then looping through each pixel and seeing if it's inside/on the shape, or outside. But I think there should be an easier method?
I don't know about your exact requirement, but the fastest to get the general result should be this:
width = int(shape.bounds[2] - shape.bounds[0])
height = int(shape.bounds[3] - shape.bounds[1])
points = MultiPoint( [(x,y) for x in range(width) for y in range(height)] )
zeroes = points.difference( shape )
ones = points.intersection( shape )
This could answer my question, but I'm wondering if there is a faster way.
data = []
width = int(shape.bounds[2] - shape.bounds[0])
height = int(shape.bounds[3] - shape.bounds[1])
for y in range(0,height):
row = []
for x in range(0,width):
val = 1 if shape.convex_hull.contains(Point(x,y)) else 0
row.append(val)
data.append(row)
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 -
I'm plotting a point cloud and coloring by residual error. I'd like the colormap to remain centered on 0, so that 0 error is white.
I see answers for matplotlib. What about Mayavi?
from mayavi import mlab
mlab.points3d(x, y, z, e, colormap='RdBu')
you can set the vmin and vmax of the colormap explicitly with mlab.points3d. So, you could just make sure that vmin = -vmax. Something like this:
mylimit = 10
mlab.points3d(x, y, z, e, colormap='RdBu',vmin=-mylimit,vmax=mylimit)
Or, you could set the limit automatically with something like:
mylimit = max(abs(e.min()),abs(e.max()))
In case anybody wishes to do this but so that the full extent of the colorbar is used, here is a solution I made (with help from here) for mayavi that stretches the colorbar so that the centre of it is on zero:
#Mayavi surface
s = mlab.surf(data)
#Get the lut table of the data
lut = s.module_manager.scalar_lut_manager.lut.table.asarray()
maxd = np.max(data)
mind = np.min(data)
#Data range
dran = maxd - mind
#Proportion of the data range at which the centred value lies
zdp = abs(mind / dran)
#The +0.5's here are because floats are rounded down when converted to ints
#index equal portion of distance along colormap
cmzi = int(zdp * 255 + 0.5)
#linspace from zero to 128, with number of points matching portion to side of zero
topi = np.linspace(0, 127, cmzi) + 0.5
#and for other side
boti = np.linspace(128, 255, 255 - cmzi) + 0.5
#convert these linspaces to ints and map the new lut from these
shift_index = np.hstack([topi.astype(int), boti.astype(int)])
s.module_manager.scalar_lut_manager.lut.table = self.lut[shift_index]
#Force update of the figure now that we have changed the LUT
mlab.draw()
Note that if you wish to do this multiple times for the same surface (ie. if you're modifying the mayavi scalars rather than redrawing the plot) you need to make a record of the initial lut table and modify that each time.