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I have a ply file that I am attempting to turn into a mesh for the purposes of ray tracing. It looks like this is the open3d visualizer and is supposed to represent a part of a city:
I used open3d to get make the following mesh as following(kdtree is just to get small number of points as file is huge):
input_file = "san.ply"
pcd = o3d.io.read_point_cloud(input_file)
point_cloud_in_numpy = np.asarray(pcd.points)
color = np.asarray(pcd.colors)
kd = scipy.spatial.cKDTree(point_cloud_in_numpy) #create kdtree for fast querying
near = kd.query_ball_point([0, 0, 0], 100)
items = point_cloud_in_numpy[near]
colors = color[near]
pcd2 = o3d.geometry.PointCloud()
pcd2.colors = o3d.utility.Vector3dVector(colors)
pcd2.points = o3d.utility.Vector3dVector(items)
pcd2.estimate_normals()
distances = pcd2.compute_nearest_neighbor_distance()
avg_dist = np.mean(distances)
radius = 2 * avg_dist
mesh = o3d.geometry.TriangleMesh.create_from_point_cloud_ball_pivoting(
pcd2,
o3d.utility.DoubleVector([radius, radius * 2]))
vertices = np.asarray(mesh.vertices)
faces = np.asarray(mesh.triangles)
o3d.visualization.draw_geometries([mesh])
However, when graphing the mesh, we get something that looks like this:
Many holes and just not at all optimal for ray tracing. I also tried using the create_from_point_cloud_poisson method instead however I kept on getting the following error:
[ERROR] /Users/yixing/repo/Open3D/build/poisson/src/ext_poisson/PoissonRecon/Src/FEMTree.IsoSurface.specialized.inl (Line 1463)
operator()
Failed to close loop [6: 87 64 18] | (113981): (2752,2560,2196)
which I found no way to fix online. I tried looking around but the best I found was pymeshfix which doesn't even work because "The input is assumed to represent a single closed solid object", which my point cloud is obviously not. I'm just looking for a good way to perform surface reconstruction that lets me keep the shape of the city while also fixing all the holes and making all surfaces created by points near eachother surfaces watertight.
Maybe you can close the holes with fill_holes() from the tensor-based TriangleMesh:
mesh = o3d.t.geometry.TriangleMesh.from_legacy(mesh).fill_holes().to_legacy()
fill_holes() takes a parameter for max. hole sizes to be closed
http://www.open3d.org/docs/latest/python_api/open3d.t.geometry.TriangleMesh.html#open3d.t.geometry.TriangleMesh.fill_holes
I'm currently working on a way to find rectangles/polygons in up to 15 given points (Image below).
Given Points
My goal is it to find polygons in that point array, like I marked in the image below. The polygons are rectangles in the real world but they are distorted a bit that's the reason why they can look like polygons or other shapes. I must find the best rectangle/polygon.
My idea was to check all connections between the points but the total amount of that is to big to run in and it took.
Does anyone has an idea how to solve that, I researched in the web and found the k-Nearest algorithm in sklearn for python but I don't have experience with that if this is the right way to solve it and how to do that. Maybe I'll also need a method to filter out some of the outliers to make it even easier for the algorithm to find the right corner points of the polygon.
The code snippet below splits the given point string into separate arrays, the array coordinatesOnly contains just the x and y values of the points.
Many thanks for you help.
Polygon in Given Points
import math
import numpy as np
import matplotlib.pyplot as plt
import time
from sklearn.neighbors import NearestNeighbors
millis = round(int(time.time())) / 1000
####input String
print("2D to 3D convert")
resultString = "0,487.50,399.46,176.84,99.99;1,485.93,423.43,-4.01,95.43;2,380.53,433.28,1.52,94.90;3,454.47,397.68,177.07,90.63;4,490.20,404.10,-6.17,89.90;5,623.56,430.52,-176.09,89.00;6,394.66,385.44,90.22,87.74;7,625.61,416.77,-177.95,87.02;8,597.21,591.66,-91.04,86.49;9,374.03,540.89,-11.20,85.77;10,600.51,552.91,178.29,85.52;11,605.29,530.78,-179.89,85.34;12,583.73,653.92,-82.39,84.42;13,483.56,449.58,-91.12,83.37;14,379.01,451.62,-6.21,81.51"
resultString = resultString.split(";")
resultStringSplitted = list()
coordinatesOnly = list()
for i in range(len(resultString)):
resultStringSplitted .append(resultString[i].split(","))
newList = ((float(resultString[i].split(",")[1]),float(resultString[i].split(",")[2])))
coordinatesOnly.append(newList)
for j in range(len(resultStringSplitted[i])):
resultStringSplitted[i][j] = float(resultStringSplitted[i][j])
#Check if score is valid
validScoreList = list()
for i in range(len(resultStringSplitted)):
if resultStringSplitted[i][len(resultStringSplitted[i])-1] != 0:
validScoreList.append(resultStringSplitted[i])
resultStringSplitted = validScoreList
#Result String array contains all 2D results
# [Point Number, X Coordinate, Y Coordinate, Angle, Point Score]
for i in range(len(resultStringSplitted)):
plt.scatter(resultStringSplitted[i][1],resultStringSplitted[i][2])
plt.show(block=True)
Since you mentioned that you can have a maximum of 15 points, I suggest to check all possible combinations of 4 points and keep all rectangles that are close enough to perfect rectangles. For 15 points, it is "only" 15*14*13*12=32760 potential rectangles.
import math
import itertools
import numpy as np
coordinatesOnly = ((0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(1,3)) # Test data
rectangles = []
# Returns True if l0 and l1 are within 10% deviation
def isValid(l0, l1):
if l0 == 0 or l1 == 0:
return False
return abs(max(l0,l1)/min(l0,l1) - 1) < 0.1
for p in itertools.combinations(np.array(coordinatesOnly),4):
for r in itertools.permutations(p,4):
l01 = np.linalg.norm(r[1]-r[0]) # Side
l12 = np.linalg.norm(r[2]-r[1]) # Side
l23 = np.linalg.norm(r[3]-r[2]) # Side
l30 = np.linalg.norm(r[0]-r[3]) # Side
l02 = np.linalg.norm(r[2]-r[0]) # Diagonal
l13 = np.linalg.norm(r[2]-r[0]) # Diagonal
areSidesEqual = isValid(l01,l23) and isValid(l12,l30)
isDiag1Valid = isValid(math.sqrt(l01*l01+l30*l30),l13) # Pythagore
isDiag2Valid = isValid(math.sqrt(l01*l01+l12*l12),l02) # Pythagore
if areSidesEqual and isDiag1Valid and isDiag2Valid:
rectangles.append(r)
break
print(rectangles)
It takes about 1 second to run on 15 points on my computer. It really depends on what are your requirements for computation time, i.e., real time, interactive time, "I just don't want to spend days waiting for the answer" time.
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:
I am trying to follow the tutorial from scikit-image regarding Template Matching (check it here).
Using just this example, I would like to find all matching coins (maxima) in the image, not only this one which gave the highest score. I was thinking about using:
maxima = argrelextrema(result, np.greater)
but the problem is that it finds also very small local maxima, which are just a noise. Is there any way to screen numpy array and find the strongest maxima? Thanks!
To find all the coins the documentation suggests "...you should use a proper peak-finding function." The easiest of these is probably peak_local_max (as suggested in the comments) which is also from skimage, and has a manual page here. Using some reasonable numbers in the *args gets the peaks out of the response image.
The second comment about the peaks being displaced is also discussed in the documentation
"Note that the peaks in the output of match_template correspond to the origin (i.e. top-left corner) of the template."
One can manually correct for this (by translating the peaks by the side lengths of the template), or you can set the pad_input bool to True (source), which as a by-product means that the peaks in the response function line up with the center of the template at the point of maximal overlap.
Combining these two bits into a script we get something like:
import numpy as np
import matplotlib.pyplot as plt
from skimage import data
from skimage.feature import match_template
from skimage.feature import peak_local_max # new import!
image = data.coins()
coin = image[170:220, 75:130]
result = match_template(image, coin,pad_input=True) #added the pad_input bool
peaks = peak_local_max(result,min_distance=10,threshold_rel=0.5) # find our peaks
# produce a plot equivalent to the one in the docs
plt.imshow(result)
# highlight matched regions (plural)
plt.plot(peaks[:,1], peaks[:,0], 'o', markeredgecolor='r', markerfacecolor='none', markersize=10)
I have been digging and found some solution but unfortunately I am not sure if I know what exactly is done in the script. I have slightly modified script found here:
neighborhood_size = 20 #how many pixels
threshold = 0.01 #threshold of maxima?
data_max = filters.maximum_filter(result, neighborhood_size)
maxima = (result == data_max)
data_min = filters.minimum_filter(result, neighborhood_size)
diff = ((data_max - data_min) > threshold)
maxima[diff == 0] = 0
x_image,y_image = [], []
temp_size = coin.shape[0]
labeled, num_objects = ndimage.label(maxima)
slices = ndimage.find_objects(labeled)
x, y = [], []
for dy,dx in slices:
x_center = (dx.start + dx.stop - 1)/2
x.append(x_center)
y_center = (dy.start + dy.stop - 1)/2
y.append(y_center)
fig, (raw,found) = plt.subplots(1,2)
raw.imshow(image,cmap=plt.cm.gray)
raw.set_axis_off()
found.imshow(result)
found.autoscale(False)
found.set_axis_off()
plt.plot(x,y, 'ro')
plt.show()
and does this:
I also realized, that coordinates of found peaks are shifted in comparison to raw image. I think the difference comes from the template size. I will update when I will find out more.
EDIT: with slight code modification I was able also to find places on the input image:
x_image_center = (dx.start + dx.stop - 1 + temp_size) / 2
x_image.append(x_image_center)
y_image_center = (dy.start + dy.stop - 1 + temp_size) / 2
y_image.append(y_image_center)
I am trying to plot contour lines of pressure level. I am using a netCDF file which contain the higher resolution data (ranges from 3 km to 27 km). Due to higher resolution data set, I get lot of pressure values which are not required to be plotted (rather I don't mind omitting certain contour line of insignificant values). I have written some plotting script based on the examples given in this link http://matplotlib.org/basemap/users/examples.html.
After plotting the image looks like this
From the image I have encircled the contours which are small and not required to be plotted. Also, I would like to plot all the contour lines smoother as mentioned in the above image. Overall I would like to get the contour image like this:-
Possible solution I think of are
Find out the number of points required for plotting contour and mask/omit those lines if they are small in number.
or
Find the area of the contour (as I want to omit only circled contour) and omit/mask those are smaller.
or
Reduce the resolution (only contour) by increasing the distance to 50 km - 100 km.
I am able to successfully get the points using SO thread Python: find contour lines from matplotlib.pyplot.contour()
But I am not able to implement any of the suggested solution above using those points.
Any solution to implement the above suggested solution is really appreciated.
Edit:-
# Andras Deak
I used print 'diameter is ', diameter line just above del(level.get_paths()[kp]) line to check if the code filters out the required diameter. Here is the filterd messages when I set if diameter < 15000::
diameter is 9099.66295612
diameter is 13264.7838257
diameter is 445.574234531
diameter is 1618.74618114
diameter is 1512.58974168
However the resulting image does not have any effect. All look same as posed image above. I am pretty sure that I have saved the figure (after plotting the wind barbs).
Regarding the solution for reducing the resolution, plt.contour(x[::2,::2],y[::2,::2],mslp[::2,::2]) it works. I have to apply some filter to make the curve smooth.
Full working example code for removing lines:-
Here is the example code for your review
#!/usr/bin/env python
from netCDF4 import Dataset
import matplotlib
matplotlib.use('agg')
import matplotlib.pyplot as plt
import numpy as np
import scipy.ndimage
from mpl_toolkits.basemap import interp
from mpl_toolkits.basemap import Basemap
# Set default map
west_lon = 68
east_lon = 93
south_lat = 7
north_lat = 23
nc = Dataset('ncfile.nc')
# Get this variable for later calucation
temps = nc.variables['T2']
time = 0 # We will take only first interval for this example
# Draw basemap
m = Basemap(projection='merc', llcrnrlat=south_lat, urcrnrlat=north_lat,
llcrnrlon=west_lon, urcrnrlon=east_lon, resolution='l')
m.drawcoastlines()
m.drawcountries(linewidth=1.0)
# This sets the standard grid point structure at full resolution
x, y = m(nc.variables['XLONG'][0], nc.variables['XLAT'][0])
# Set figure margins
width = 10
height = 8
plt.figure(figsize=(width, height))
plt.rc("figure.subplot", left=.001)
plt.rc("figure.subplot", right=.999)
plt.rc("figure.subplot", bottom=.001)
plt.rc("figure.subplot", top=.999)
plt.figure(figsize=(width, height), frameon=False)
# Convert Surface Pressure to Mean Sea Level Pressure
stemps = temps[time] + 6.5 * nc.variables['HGT'][time] / 1000.
mslp = nc.variables['PSFC'][time] * np.exp(9.81 / (287.0 * stemps) * nc.variables['HGT'][time]) * 0.01 + (
6.7 * nc.variables['HGT'][time] / 1000)
# Contour only at 2 hpa interval
level = []
for i in range(mslp.min(), mslp.max(), 1):
if i % 2 == 0:
if i >= 1006 and i <= 1018:
level.append(i)
# Save mslp values to upload to SO thread
# np.savetxt('mslp.txt', mslp, fmt='%.14f', delimiter=',')
P = plt.contour(x, y, mslp, V=2, colors='b', linewidths=2, levels=level)
# Solution suggested by Andras Deak
for level in P.collections:
for kp,path in enumerate(level.get_paths()):
# include test for "smallness" of your choice here:
# I'm using a simple estimation for the diameter based on the
# x and y diameter...
verts = path.vertices # (N,2)-shape array of contour line coordinates
diameter = np.max(verts.max(axis=0) - verts.min(axis=0))
if diameter < 15000: # threshold to be refined for your actual dimensions!
#print 'diameter is ', diameter
del(level.get_paths()[kp]) # no remove() for Path objects:(
#level.remove() # This does not work. produces ValueError: list.remove(x): x not in list
plt.gcf().canvas.draw()
plt.savefig('dummy', bbox_inches='tight')
plt.close()
After the plot is saved I get the same image
You can see that the lines are not removed yet. Here is the link to mslp array which we are trying to play with http://www.mediafire.com/download/7vi0mxqoe0y6pm9/mslp.txt
If you want x and y data which are being used in the above code, I can upload for your review.
Smooth line
You code to remove the smaller circles working perfectly. However the other question I have asked in the original post (smooth line) does not seems to work. I have used your code to slice the array to get minimal values and contoured it. I have used the following code to reduce the array size:-
slice = 15
CS = plt.contour(x[::slice,::slice],y[::slice,::slice],mslp[::slice,::slice], colors='b', linewidths=1, levels=levels)
The result is below.
After searching for few hours I found this SO thread having simmilar issue:-
Regridding regular netcdf data
But none of the solution provided over there works.The questions similar to mine above does not have proper solutions. If this issue is solved then the code is perfect and complete.
General idea
Your question seems to have 2 very different halves: one about omitting small contours, and another one about smoothing the contour lines. The latter is simpler, since I can't really think of anything else other than decreasing the resolution of your contour() call, just like you said.
As for removing a few contour lines, here's a solution which is based on directly removing contour lines individually. You have to loop over the collections of the object returned by contour(), and for each element check each Path, and delete the ones you don't need. Redrawing the figure's canvas will get rid of the unnecessary lines:
# dummy example based on matplotlib.pyplot.clabel example:
import matplotlib
import numpy as np
import matplotlib.cm as cm
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
delta = 0.025
x = np.arange(-3.0, 3.0, delta)
y = np.arange(-2.0, 2.0, delta)
X, Y = np.meshgrid(x, y)
Z1 = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
# difference of Gaussians
Z = 10.0 * (Z2 - Z1)
plt.figure()
CS = plt.contour(X, Y, Z)
for level in CS.collections:
for kp,path in reversed(list(enumerate(level.get_paths()))):
# go in reversed order due to deletions!
# include test for "smallness" of your choice here:
# I'm using a simple estimation for the diameter based on the
# x and y diameter...
verts = path.vertices # (N,2)-shape array of contour line coordinates
diameter = np.max(verts.max(axis=0) - verts.min(axis=0))
if diameter<1: # threshold to be refined for your actual dimensions!
del(level.get_paths()[kp]) # no remove() for Path objects:(
# this might be necessary on interactive sessions: redraw figure
plt.gcf().canvas.draw()
Here's the original(left) and the removed version(right) for a diameter threshold of 1 (note the little piece of the 0 level at the top):
Note that the top little line is removed while the huge cyan one in the middle doesn't, even though both correspond to the same collections element i.e. the same contour level. If we didn't want to allow this, we could've called CS.collections[k].remove(), which would probably be a much safer way of doing the same thing (but it wouldn't allow us to differentiate between multiple lines corresponding to the same contour level).
To show that fiddling around with the cut-off diameter works as expected, here's the result for a threshold of 2:
All in all it seems quite reasonable.
Your actual case
Since you've added your actual data, here's the application to your case. Note that you can directly generate the levels in a single line using np, which will almost give you the same result. The exact same can be achieved in 2 lines (generating an arange, then selecting those that fall between p1 and p2). Also, since you're setting levels in the call to contour, I believe the V=2 part of the function call has no effect.
import numpy as np
import matplotlib.pyplot as plt
# insert actual data here...
Z = np.loadtxt('mslp.txt',delimiter=',')
X,Y = np.meshgrid(np.linspace(0,300000,Z.shape[1]),np.linspace(0,200000,Z.shape[0]))
p1,p2 = 1006,1018
# this is almost the same as the original, although it will produce
# [p1, p1+2, ...] instead of `[Z.min()+n, Z.min()+n+2, ...]`
levels = np.arange(np.maximum(Z.min(),p1),np.minimum(Z.max(),p2),2)
#control
plt.figure()
CS = plt.contour(X, Y, Z, colors='b', linewidths=2, levels=levels)
#modified
plt.figure()
CS = plt.contour(X, Y, Z, colors='b', linewidths=2, levels=levels)
for level in CS.collections:
for kp,path in reversed(list(enumerate(level.get_paths()))):
# go in reversed order due to deletions!
# include test for "smallness" of your choice here:
# I'm using a simple estimation for the diameter based on the
# x and y diameter...
verts = path.vertices # (N,2)-shape array of contour line coordinates
diameter = np.max(verts.max(axis=0) - verts.min(axis=0))
if diameter<15000: # threshold to be refined for your actual dimensions!
del(level.get_paths()[kp]) # no remove() for Path objects:(
# this might be necessary on interactive sessions: redraw figure
plt.gcf().canvas.draw()
plt.show()
Results, original(left) vs new(right):
Smoothing by resampling
I've decided to tackle the smoothing problem as well. All I could come up with is downsampling your original data, then upsampling again using griddata (interpolation). The downsampling part could also be done with interpolation, although the small-scale variation in your input data might make this problem ill-posed. So here's the crude version:
import scipy.interpolate as interp #the new one
# assume you have X,Y,Z,levels defined as before
# start resampling stuff
dN = 10 # use every dN'th element of the gridded input data
my_slice = [slice(None,None,dN),slice(None,None,dN)]
# downsampled data
X2,Y2,Z2 = X[my_slice],Y[my_slice],Z[my_slice]
# same as X2 = X[::dN,::dN] etc.
# upsampling with griddata over original mesh
Zsmooth = interp.griddata(np.array([X2.ravel(),Y2.ravel()]).T,Z2.ravel(),(X,Y),method='cubic')
# plot
plt.figure()
CS = plt.contour(X, Y, Zsmooth, colors='b', linewidths=2, levels=levels)
You can freely play around with the grids used for interpolation, in this case I just used the original mesh, as it was at hand. You can also play around with different kinds of interpolation: the default 'linear' one will be faster, but less smooth.
Result after downsampling(left) and upsampling(right):
Of course you should still apply the small-line-removal algorithm after this resampling business, and keep in mind that this heavily distorts your input data (since if it wasn't distorted, then it wouldn't be smooth). Also, note that due to the crude method used in the downsampling step, we introduce some missing values near the top/right edges of the region under consideraton. If this is a problem, you should consider doing the downsampling based on griddata as I've noted earlier.
This is a pretty bad solution, but it's the only one that I've come up with. Use the get_contour_verts function in this solution you linked to, possibly with the matplotlib._cntr module so that nothing gets plotted initially. That gives you a list of contour lines, sections, vertices, etc. Then you have to go through that list and pop the contours you don't want. You could do this by calculating a minimum diameter, for example; if the max distance between points is less than some cutoff, throw it out.
That leaves you with a list of LineCollection objects. Now if you make a Figure and Axes instance, you can use Axes.add_collection to add all of the LineCollections in the list.
I checked this out really quick, but it seemed to work. I'll come back with a minimum working example if I get a chance. Hope it helps!
Edit: Here's an MWE of the basic idea. I wasn't familiar with plt._cntr.Cntr, so I ended up using plt.contour to get the initial contour object. As a result, you end up making two figures; you just have to close the first one. You can replace checkDiameter with whatever function works. I think you could turn the line segments into a Polygon and calculate areas, but you'd have to figure that out on your own. Let me know if you run into problems with this code, but it at least works for me.
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
def checkDiameter(seg, tol=.3):
# Function for screening line segments. NB: Not actually a proper diameter.
diam = (seg[:,0].max() - seg[:,0].min(),
seg[:,1].max() - seg[:,1].min())
return not (diam[0] < tol or diam[1] < tol)
# Create testing data
x = np.linspace(-1,1, 21)
xx, yy = np.meshgrid(x,x)
z = np.exp(-(xx**2 + .5*yy**2))
# Original plot with plt.contour
fig0, ax0 = plt.subplots()
# Make sure this contour object actually has a tiny contour to remove
cntrObj = ax0.contour(xx,yy,z, levels=[.2,.4,.6,.8,.9,.95,.99,.999])
# Primary loop: Copy contours into a new LineCollection
lineNew = list()
for lineOriginal in cntrObj.collections:
# Get properties of the original LineCollection
segments = lineOriginal.get_segments()
propDict = lineOriginal.properties()
propDict = {key: value for (key,value) in propDict.items()
if key in ['linewidth','color','linestyle']} # Whatever parameters you want to carry over
# Filter out the lines with small diameters
segments = [seg for seg in segments if checkDiameter(seg)]
# Create new LineCollection out of the OK segments
if len(segments) > 0:
lineNew.append(mpl.collections.LineCollection(segments, **propDict))
# Make new plot with only these line collections; display results
fig1, ax1 = plt.subplots()
ax1.set_xlim(ax0.get_xlim())
ax1.set_ylim(ax0.get_ylim())
for line in lineNew:
ax1.add_collection(line)
plt.show()
FYI: The bit with propDict is just to automate bringing over some of the line properties from the original plot. You can't use the whole dictionary at once, though. First, it contains the old plot's line segments, but you can just swap those for the new ones. But second, it appears to contain a number of parameters that are in conflict with each other: multiple linewidths, facecolors, etc. The {key for key in propDict if I want key} workaround is my way to bypass that, but I'm sure someone else can do it more cleanly.