I have about 50,000 data points in 3D on which I have run scipy.spatial.Delaunay from the new scipy (I'm using 0.10) which gives me a very useful triangulation.
Based on: http://en.wikipedia.org/wiki/Delaunay_triangulation (section "Relationship with the Voronoi diagram")
...I was wondering if there is an easy way to get to the "dual graph" of this triangulation, which is the Voronoi Tesselation.
Any clues? My searching around on this seems to show no pre-built in scipy functions, which I find almost strange!
Thanks,
Edward
The adjacency information can be found in the neighbors attribute of the Delaunay object. Unfortunately, the code does not expose the circumcenters to the user at the moment, so you'll have to recompute those yourself.
Also, the Voronoi edges that extend to infinity are not directly obtained in this way. It's still probably possible, but needs some more thinking.
import numpy as np
from scipy.spatial import Delaunay
points = np.random.rand(30, 2)
tri = Delaunay(points)
p = tri.points[tri.vertices]
# Triangle vertices
A = p[:,0,:].T
B = p[:,1,:].T
C = p[:,2,:].T
# See http://en.wikipedia.org/wiki/Circumscribed_circle#Circumscribed_circles_of_triangles
# The following is just a direct transcription of the formula there
a = A - C
b = B - C
def dot2(u, v):
return u[0]*v[0] + u[1]*v[1]
def cross2(u, v, w):
"""u x (v x w)"""
return dot2(u, w)*v - dot2(u, v)*w
def ncross2(u, v):
"""|| u x v ||^2"""
return sq2(u)*sq2(v) - dot2(u, v)**2
def sq2(u):
return dot2(u, u)
cc = cross2(sq2(a) * b - sq2(b) * a, a, b) / (2*ncross2(a, b)) + C
# Grab the Voronoi edges
vc = cc[:,tri.neighbors]
vc[:,tri.neighbors == -1] = np.nan # edges at infinity, plotting those would need more work...
lines = []
lines.extend(zip(cc.T, vc[:,:,0].T))
lines.extend(zip(cc.T, vc[:,:,1].T))
lines.extend(zip(cc.T, vc[:,:,2].T))
# Plot it
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
lines = LineCollection(lines, edgecolor='k')
plt.hold(1)
plt.plot(points[:,0], points[:,1], '.')
plt.plot(cc[0], cc[1], '*')
plt.gca().add_collection(lines)
plt.axis('equal')
plt.xlim(-0.1, 1.1)
plt.ylim(-0.1, 1.1)
plt.show()
As I spent a considerable amount of time on this, I'd like to share my solution on how to get the Voronoi polygons instead of just the edges.
The code is at https://gist.github.com/letmaik/8803860 and extends on the solution of tauran.
First, I changed the code to give me vertices and (pairs of) indices (=edges) separately, as many calculations can be simplified when working on indices instead of point coordinates.
Then, in the voronoi_cell_lines method I determine which edges belong to which cells. For that I use the proposed solution of Alink from a related question. That is, for each edge find the two nearest input points (=cells) and create a mapping from that.
The last step is to create the actual polygons (see voronoi_polygons method). First, the outer cells which have dangling edges need to be closed. This is as simple as looking through all edges and checking which ones have only one neighboring edge. There can be either zero or two such edges. In case of two, I then connect these by introducing an additional edge.
Finally, the unordered edges in each cell need to be put into the right order to derive a polygon from them.
The usage is:
P = np.random.random((100,2))
fig = plt.figure(figsize=(4.5,4.5))
axes = plt.subplot(1,1,1)
plt.axis([-0.05,1.05,-0.05,1.05])
vertices, lineIndices = voronoi(P)
cells = voronoi_cell_lines(P, vertices, lineIndices)
polys = voronoi_polygons(cells)
for pIdx, polyIndices in polys.items():
poly = vertices[np.asarray(polyIndices)]
p = matplotlib.patches.Polygon(poly, facecolor=np.random.rand(3,1))
axes.add_patch(p)
X,Y = P[:,0],P[:,1]
plt.scatter(X, Y, marker='.', zorder=2)
plt.axis([-0.05,1.05,-0.05,1.05])
plt.show()
which outputs:
The code is probably not suitable for large numbers of input points and can be improved in some areas. Nevertheless, it may be helpful to others who have similar problems.
I came across the same problem and built a solution out of pv.'s answer and other code snippets I found across the web. The solution returns a complete Voronoi diagram, including the outer lines where no triangle neighbours are present.
#!/usr/bin/env python
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
from scipy.spatial import Delaunay
def voronoi(P):
delauny = Delaunay(P)
triangles = delauny.points[delauny.vertices]
lines = []
# Triangle vertices
A = triangles[:, 0]
B = triangles[:, 1]
C = triangles[:, 2]
lines.extend(zip(A, B))
lines.extend(zip(B, C))
lines.extend(zip(C, A))
lines = matplotlib.collections.LineCollection(lines, color='r')
plt.gca().add_collection(lines)
circum_centers = np.array([triangle_csc(tri) for tri in triangles])
segments = []
for i, triangle in enumerate(triangles):
circum_center = circum_centers[i]
for j, neighbor in enumerate(delauny.neighbors[i]):
if neighbor != -1:
segments.append((circum_center, circum_centers[neighbor]))
else:
ps = triangle[(j+1)%3] - triangle[(j-1)%3]
ps = np.array((ps[1], -ps[0]))
middle = (triangle[(j+1)%3] + triangle[(j-1)%3]) * 0.5
di = middle - triangle[j]
ps /= np.linalg.norm(ps)
di /= np.linalg.norm(di)
if np.dot(di, ps) < 0.0:
ps *= -1000.0
else:
ps *= 1000.0
segments.append((circum_center, circum_center + ps))
return segments
def triangle_csc(pts):
rows, cols = pts.shape
A = np.bmat([[2 * np.dot(pts, pts.T), np.ones((rows, 1))],
[np.ones((1, rows)), np.zeros((1, 1))]])
b = np.hstack((np.sum(pts * pts, axis=1), np.ones((1))))
x = np.linalg.solve(A,b)
bary_coords = x[:-1]
return np.sum(pts * np.tile(bary_coords.reshape((pts.shape[0], 1)), (1, pts.shape[1])), axis=0)
if __name__ == '__main__':
P = np.random.random((300,2))
X,Y = P[:,0],P[:,1]
fig = plt.figure(figsize=(4.5,4.5))
axes = plt.subplot(1,1,1)
plt.scatter(X, Y, marker='.')
plt.axis([-0.05,1.05,-0.05,1.05])
segments = voronoi(P)
lines = matplotlib.collections.LineCollection(segments, color='k')
axes.add_collection(lines)
plt.axis([-0.05,1.05,-0.05,1.05])
plt.show()
Black lines = Voronoi diagram, Red lines = Delauny triangles
I do not know of a function to do this, but it does not seem like an overly complicated task.
The Voronoi graph is the junction of the circumcircles, as described in the wikipedia article.
So you could start with a function that finds the center of the circumcircles of a triangle, which is basic mathematics (http://en.wikipedia.org/wiki/Circumscribed_circle).
Then, just join centers of adjacent triangles.
Related
Short version:
Is it possible to create a new scipy.spatial.Delaunay object with a subset of the triangles (2D data) from an existing object?
The goal would be to use the find_simplex method on the new object with filtered out simplices.
Similar but not quite the same
matplotlib contour/contourf of **concave** non-gridded data
How to deal with the (undesired) triangles that form between the edges of my geometry when using Triangulation in matplotlib
Long version:
I am looking at lat-lon data that I regrid with scipy.interpolate.griddata like in the pseudo-code below:
import numpy as np
from scipy.interpolate import griddata
from scipy.spatial import Delaunay
from scipy.interpolate.interpnd import _ndim_coords_from_arrays
#lat shape (a,b): 2D array of latitude values
#lon shape (a,b): 2D array of longitude values
#x shape (a,b): 2D array of variable of interest at lat and lon
# lat-lon data
nonan = ~np.isnan(lat)
flat_lat = lat[nonan]
flat_lon = lon[nonan]
flat_x = x[nonan]
# regular lat-lon grid for regridding
lon_ar = np.arange(loni,lonf,resolution)
lat_ar = np.arange(lati,latf,resolution)
lon_grid, lat_grid = np.meshgrid(lon_ar,lat_ar)
# regrid
x_grid = griddata((flat_lon,flat_lat),flat_x,(lon_grid,lat_grid), method='nearest')
# filter out extrapolated values
cloud_points = _ndim_coords_from_arrays((flat_lon,flat_lat))
regrid_points = _ndim_coords_from_arrays((lon_grid.ravel(),lat_grid.ravel()))
tri = Delaunay(cloud_points)
outside_hull = tri.find_simplex(regrid_points) < 0
x_grid[outside_hull.reshape(x_grid.shape)] = np.nan
# filter out large triangles ??
# it would be easy if I could "subset" tri into a new scipy.spatial.Delaunay object
# new_tri = ??
# outside_hull = new_tri.find_simplex(regrid_points) < 0
The problem is that the convex hull has low quality (very large, shown in blue in example below) triangles that I would like to filter out as they don't represent the data well. I know how to filter them out in input points, but not in the regridded output. Here is the filter function:
def filter_large_triangles(
points: np.ndarray, tri: Optional[Delaunay] = None, coeff: float = 2.0
):
"""
Filter out triangles that have an edge > coeff * median(edge)
Inputs:
tri: scipy.spatial.Delaunay object
coeff: triangles with an edge > coeff * median(edge) will be filtered out
Outputs:
valid_slice: boolean array that selects "normal" triangles
"""
if tri is None:
tri = Delaunay(points)
edge_lengths = np.zeros(tri.vertices.shape)
seen = {}
# loop over triangles
for i, vertex in enumerate(tri.vertices):
# loop over edges
for j in range(3):
id0 = vertex[j]
id1 = vertex[(j + 1) % 3]
# avoid calculating twice for non-border edges
if (id0,id1) in seen:
edge_lengths[i, j] = seen[(id0,id1)]
else:
edge_lengths[i, j] = np.linalg.norm(points[id1] - points[id0])
seen[(id0,id1)] = edge_lengths[i, j]
median_edge = np.median(edge_lengths.flatten())
valid_slice = np.all(edge_lengths < coeff * median_edge, axis=1)
return valid_slice
The bad triangles are shown in blue below:
import matplotlib.pyplot as plt
no_large_triangles = filter_large_triangles(cloud_points,tri)
fig,ax = plt.subplot()
ax.triplot(points[:,0],points[:,1],tri.simplices,c='blue')
ax.triplot(points[:,0],points[:,1],tri.simplices[no_large_triangles],c='green')
plt.show()
Is it possible to create a new scipy.spatial.Delaunay object with only the no_large_triangles simplices? The goal would be to use the find_simplex method on that new object to easily filter out points.
As an alternative how could I find the indices of points in regrid_points that fall inside the blue triangles? (tri.simplices[~no_large_triangles])
So it is possible to modify the Delaunay object for the purpose of using find_simplex on a subset of simplices, but it seems only with the bruteforce algorithm.
# filter out extrapolated values
cloud_points = _ndim_coords_from_arrays((flat_lon,flat_lat))
regrid_points = _ndim_coords_from_arrays((lon_grid.ravel(),lat_grid.ravel()))
tri = Delaunay(cloud_points)
outside_hull = tri.find_simplex(regrid_points) < 0
# filter out large triangles
large_triangles = ~filter_large_triangles(cloud_points,tri)
large_triangle_ids = np.where(large_triangles)[0]
subset_tri = tri # this doesn't preserve tri, effectively just a renaming
# the _find_simplex_bruteforce method only needs the simplices and neighbors
subset_tri.nsimplex = large_triangle_ids.size
subset_tri.simplices = tri.simplices[large_triangles]
subset_tri.neighbors = tri.neighbors[large_triangles]
# update neighbors
for i,triangle in enumerate(subset_tri.neighbors):
for j,neighbor_id in enumerate(triangle):
if neighbor_id in large_triangle_ids:
# reindex the neighbors to match the size of the subset
subset_tri.neighbors[i,j] = np.where(large_triangle_ids==neighbor_id)[0]
elif neighbor_id>=0 and (neighbor_id not in large_triangle_ids):
# that neighbor was a "normal" triangle that should not exist in the subset
subset_tri.neighbors[i,j] = -1
inside_large_triangles = subset_tri.find_simplex(regrid_points,bruteforce=True) >= 0
invalid_slice = np.logical_or(outside_hull,inside_large_triangles)
x_grid[invalid_slice.reshape(x_grid.shape)] = np.nan
Showing that the new Delaunay object has only the subset of large triangles
import matplotlib.pyplot as plt
fig,ax = plt.subplot()
ax.triplot(cloud_points[:,0],cloud_points[:,1],subset_tri.simplices,color='red')
plt.show()
Plotting x_grid with pcolormesh before the filtering for large triangles (zoomed in the blue circle above):
After the filtering:
I want to fill a bunch of polygons with line hatch. The lines must have a specific angle with respect to x-axis. I found that matplotlib already suppots some hatch classes and one can define a custom class (like How to fill a polygon with a custom hatch in matplotlib?). I tried to generate a custom hatch but when I append it to the list of hatches the init function doesn't know the angle. I tried with the following class:
class AngularHatch(HatchPatternBase):
def __init__(self, hatch, density, angle):
self.num_lines = int((hatch.count('{'))*density*3)
self.num_vertices = self.num_lines * 2
self.R = np.array([[np.cos(angle), -np.sin(angle)],
[np.sin(angle), np.cos(angle)]])
def set_vertices_and_codes(self, vertices, codes):
steps, stepsize = np.linspace(0.0, 1.0, self.num_lines, False,
retstep=True)
steps += stepsize / 2.
vertices[0::2, 0] = 0
vertices[0::2, 1] = steps
vertices[1::2, 0] = 1
vertices[1::2, 1] = steps
for i, v in enumerate(vertices):
vertices[i] = self.R.dot(v)
codes[0::2] = Path.MOVETO
codes[1::2] = Path.LINETO
Then I add this class to the list of available classes for hatching. However this will not generate the correct lines since the code is modified from the HorizontalHatch source code here and I think this generates lines in the unit square. Moreover I need to generate this patch for a specific angle for each polygon I want to render. ¿Any ideas on how to give the correct angle to this class per polygon?
The following does not solve this issue. It just solves part of the problem and shows at which point the approach fails. I am currently convinced that hatching with arbitrary angles is not possible with matplotlib, because the size of the unit cell is fixed.
To overcome the problem of setting the angle, one may define a custom format from which to take the angle information. E.g. "{angle}{factor}", such that "{45}{2}" would produce a hatching with an angle of 45° and a density factor of 2.
I then do not completely understand the attempt of calculating the vertices. To replicate the behaviour of the hatches which are built-in, one may rotate them directly.
The problem is that this way the line hatches work only for angles of 45°. This is because the lines at the edges of the unit cell do not align well. See the following:
import numpy as np
import matplotlib.hatch
import matplotlib.path
import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse, Rectangle
class AngularHatch(matplotlib.hatch.HatchPatternBase):
def __init__(self, hatch, density):
self.num_lines=0
self.num_vertices=0
if hatch[0] == "{":
h = hatch.strip("{}").split("}{")
angle = np.deg2rad(float(h[0])-45)
d = float(h[1])
self.num_lines = int(density*d)
self.num_vertices = (self.num_lines + 1) * 2
self.R = np.array([[np.cos(angle), -np.sin(angle)],
[np.sin(angle), np.cos(angle)]])
def set_vertices_and_codes(self, vertices, codes):
steps = np.linspace(-0.5, 0.5, self.num_lines + 1, True)
vertices[0::2, 0] = 0.0 + steps
vertices[0::2, 1] = 0.0 - steps
vertices[1::2, 0] = 1.0 + steps
vertices[1::2, 1] = 1.0 - steps
codes[0::2] = matplotlib.path.Path.MOVETO
codes[1::2] = matplotlib.path.Path.LINETO
vertices[:,:] = np.dot((vertices-0.5),self.R)+0.5
matplotlib.hatch._hatch_types.append(AngularHatch)
fig = plt.figure()
ax = fig.add_subplot(111)
ellipse = ax.add_patch(Rectangle((0.1, 0.1), 0.4, 0.8, fill=False))
ellipse.set_hatch('{45}{1}')
ellipse.set_color('red')
ellipse = ax.add_patch(Rectangle((0.55, 0.1), 0.4, 0.8, fill=False))
ellipse.set_hatch('{22}{1}')
ellipse.set_color('blue')
plt.show()
I would like to plot parallel lines with different colors. E.g. rather than a single red line of thickness 6, I would like to have two parallel lines of thickness 3, with one red and one blue.
Any thoughts would be appreciated.
Merci
Even with the smart offsetting (s. below), there is still an issue in a view that has sharp angles between consecutive points.
Zoomed view of smart offsetting:
Overlaying lines of varying thickness:
Plotting parallel lines is not an easy task. Using a simple uniform offset will of course not show the desired result. This is shown in the left picture below.
Such a simple offset can be produced in matplotlib as shown in the transformation tutorial.
Method1
A better solution may be to use the idea sketched on the right side. To calculate the offset of the nth point we can use the normal vector to the line between the n-1st and the n+1st point and use the same distance along this normal vector to calculate the offset point.
The advantage of this method is that we have the same number of points in the original line as in the offset line. The disadvantage is that it is not completely accurate, as can be see in the picture.
This method is implemented in the function offset in the code below.
In order to make this useful for a matplotlib plot, we need to consider that the linewidth should be independent of the data units. Linewidth is usually given in units of points, and the offset would best be given in the same unit, such that e.g. the requirement from the question ("two parallel lines of width 3") can be met.
The idea is therefore to transform the coordinates from data to display coordinates, using ax.transData.transform. Also the offset in points o can be transformed to the same units: Using the dpi and the standard of ppi=72, the offset in display coordinates is o*dpi/ppi. After the offset in display coordinates has been applied, the inverse transform (ax.transData.inverted().transform) allows a backtransformation.
Now there is another dimension of the problem: How to assure that the offset remains the same independent of the zoom and size of the figure?
This last point can be addressed by recalculating the offset each time a zooming of resizing event has taken place.
Here is how a rainbow curve would look like produced by this method.
And here is the code to produce the image.
import numpy as np
import matplotlib.pyplot as plt
dpi = 100
def offset(x,y, o):
""" Offset coordinates given by array x,y by o """
X = np.c_[x,y].T
m = np.array([[0,-1],[1,0]])
R = np.zeros_like(X)
S = X[:,2:]-X[:,:-2]
R[:,1:-1] = np.dot(m, S)
R[:,0] = np.dot(m, X[:,1]-X[:,0])
R[:,-1] = np.dot(m, X[:,-1]-X[:,-2])
On = R/np.sqrt(R[0,:]**2+R[1,:]**2)*o
Out = On+X
return Out[0,:], Out[1,:]
def offset_curve(ax, x,y, o):
""" Offset array x,y in data coordinates
by o in points """
trans = ax.transData.transform
inv = ax.transData.inverted().transform
X = np.c_[x,y]
Xt = trans(X)
xto, yto = offset(Xt[:,0],Xt[:,1],o*dpi/72. )
Xto = np.c_[xto, yto]
Xo = inv(Xto)
return Xo[:,0], Xo[:,1]
# some single points
y = np.array([1,2,2,3,3,0])
x = np.arange(len(y))
#or try a sinus
x = np.linspace(0,9)
y=np.sin(x)*x/3.
fig, ax=plt.subplots(figsize=(4,2.5), dpi=dpi)
cols = ["#fff40b", "#00e103", "#ff9921", "#3a00ef", "#ff2121", "#af00e7"]
lw = 2.
lines = []
for i in range(len(cols)):
l, = plt.plot(x,y, lw=lw, color=cols[i])
lines.append(l)
def plot_rainbow(event=None):
xr = range(6); yr = range(6);
xr[0],yr[0] = offset_curve(ax, x,y, lw/2.)
xr[1],yr[1] = offset_curve(ax, x,y, -lw/2.)
xr[2],yr[2] = offset_curve(ax, xr[0],yr[0], lw)
xr[3],yr[3] = offset_curve(ax, xr[1],yr[1], -lw)
xr[4],yr[4] = offset_curve(ax, xr[2],yr[2], lw)
xr[5],yr[5] = offset_curve(ax, xr[3],yr[3], -lw)
for i in range(6):
lines[i].set_data(xr[i], yr[i])
plot_rainbow()
fig.canvas.mpl_connect("resize_event", plot_rainbow)
fig.canvas.mpl_connect("button_release_event", plot_rainbow)
plt.savefig(__file__+".png", dpi=dpi)
plt.show()
Method2
To avoid overlapping lines, one has to use a more complicated solution.
One could first offset every point normal to the two line segments it is part of (green points in the picture below). Then calculate the line through those offset points and find their intersection.
A particular case would be when the slopes of two subsequent line segments equal. This has to be taken care of (eps in the code below).
from __future__ import division
import numpy as np
import matplotlib.pyplot as plt
dpi = 100
def intersect(p1, p2, q1, q2, eps=1.e-10):
""" given two lines, first through points pn, second through qn,
find the intersection """
x1 = p1[0]; y1 = p1[1]; x2 = p2[0]; y2 = p2[1]
x3 = q1[0]; y3 = q1[1]; x4 = q2[0]; y4 = q2[1]
nomX = ((x1*y2-y1*x2)*(x3-x4)- (x1-x2)*(x3*y4-y3*x4))
denom = float( (x1-x2)*(y3-y4) - (y1-y2)*(x3-x4) )
nomY = (x1*y2-y1*x2)*(y3-y4) - (y1-y2)*(x3*y4-y3*x4)
if np.abs(denom) < eps:
#print "intersection undefined", p1
return np.array( p1 )
else:
return np.array( [ nomX/denom , nomY/denom ])
def offset(x,y, o, eps=1.e-10):
""" Offset coordinates given by array x,y by o """
X = np.c_[x,y].T
m = np.array([[0,-1],[1,0]])
S = X[:,1:]-X[:,:-1]
R = np.dot(m, S)
norm = np.sqrt(R[0,:]**2+R[1,:]**2) / o
On = R/norm
Outa = On+X[:,1:]
Outb = On+X[:,:-1]
G = np.zeros_like(X)
for i in xrange(0, len(X[0,:])-2):
p = intersect(Outa[:,i], Outb[:,i], Outa[:,i+1], Outb[:,i+1], eps=eps)
G[:,i+1] = p
G[:,0] = Outb[:,0]
G[:,-1] = Outa[:,-1]
return G[0,:], G[1,:]
def offset_curve(ax, x,y, o, eps=1.e-10):
""" Offset array x,y in data coordinates
by o in points """
trans = ax.transData.transform
inv = ax.transData.inverted().transform
X = np.c_[x,y]
Xt = trans(X)
xto, yto = offset(Xt[:,0],Xt[:,1],o*dpi/72., eps=eps )
Xto = np.c_[xto, yto]
Xo = inv(Xto)
return Xo[:,0], Xo[:,1]
# some single points
y = np.array([1,1,2,0,3,2,1.,4,3]) *1.e9
x = np.arange(len(y))
x[3]=x[4]
#or try a sinus
#x = np.linspace(0,9)
#y=np.sin(x)*x/3.
fig, ax=plt.subplots(figsize=(4,2.5), dpi=dpi)
cols = ["r", "b"]
lw = 11.
lines = []
for i in range(len(cols)):
l, = plt.plot(x,y, lw=lw, color=cols[i], solid_joinstyle="miter")
lines.append(l)
def plot_rainbow(event=None):
xr = range(2); yr = range(2);
xr[0],yr[0] = offset_curve(ax, x,y, lw/2.)
xr[1],yr[1] = offset_curve(ax, x,y, -lw/2.)
for i in range(2):
lines[i].set_data(xr[i], yr[i])
plot_rainbow()
fig.canvas.mpl_connect("resize_event", plot_rainbow)
fig.canvas.mpl_connect("button_release_event", plot_rainbow)
plt.show()
Note that this method should work well as long as the offset between the lines is smaller then the distance between subsequent points on the line. Otherwise method 1 may be better suited.
The best that I can think of is to take your data, generate a series of small offsets, and use fill_between to make bands of whatever color you like.
I wrote a function to do this. I don't know what shape you're trying to plot, so this may or may not work for you. I tested it on a parabola and got decent results. You can also play around with the list of colors.
def rainbow_plot(x, y, spacing=0.1):
fig, ax = plt.subplots()
colors = ['red', 'yellow', 'green', 'cyan','blue']
top = max(y)
lines = []
for i in range(len(colors)+1):
newline_data = y - top*spacing*i
lines.append(newline_data)
for i, c in enumerate(colors):
ax.fill_between(x, lines[i], lines[i+1], facecolor=c)
return fig, ax
x = np.linspace(0,1,51)
y = 1-(x-0.5)**2
rainbow_plot(x,y)
I have figured out a method to cluster disperse point data into structured 2-d array(like rasterize function). And I hope there are some better ways to achieve that target.
My work
1. Intro
1000 point data has there dimensions of properties (lon, lat, emission) whicn represent one factory located at (x,y) emit certain amount of CO2 into atmosphere
grid network: predefine the 2-d array in the shape of 20x20
http://i4.tietuku.com/02fbaf32d2f09fff.png
The code reproduced here:
#### define the map area
xc1,xc2,yc1,yc2 = 113.49805889531724,115.5030664238035,37.39995194888143,38.789235929357105
map = Basemap(llcrnrlon=xc1,llcrnrlat=yc1,urcrnrlon=xc2,urcrnrlat=yc2)
#### reading the point data and scatter plot by their position
df = pd.read_csv("xxxxx.csv")
px,py = map(df.lon, df.lat)
map.scatter(px, py, color = "red", s= 5,zorder =3)
#### predefine the grid networks
lon_grid,lat_grid = np.linspace(xc1,xc2,21), np.linspace(yc1,yc2,21)
lon_x,lat_y = np.meshgrid(lon_grid,lat_grid)
grids = np.zeros(20*20).reshape(20,20)
plt.pcolormesh(lon_x,lat_y,grids,cmap = 'gray', facecolor = 'none',edgecolor = 'k',zorder=3)
2. My target
Finding the nearest grid point for each factory
Add the emission data into this grid number
3. Algorithm realization
3.1 Raster grid
note: 20x20 grid points are distributed in this area represented by blue dot.
http://i4.tietuku.com/8548554587b0cb3a.png
3.2 KD-tree
Find the nearest blue dot of each red point
sh = (20*20,2)
grids = np.zeros(20*20*2).reshape(*sh)
sh_emission = (20*20)
grids_em = np.zeros(20*20).reshape(sh_emission)
k = 0
for j in range(0,yy.shape[0],1):
for i in range(0,xx.shape[0],1):
grids[k] = np.array([lon_grid[i],lat_grid[j]])
k+=1
T = KDTree(grids)
x_delta = (lon_grid[2] - lon_grid[1])
y_delta = (lat_grid[2] - lat_grid[1])
R = np.sqrt(x_delta**2 + y_delta**2)
for i in range(0,len(df.lon),1):
idx = T.query_ball_point([df.lon.iloc[i],df.lat.iloc[i]], r=R)
# there are more than one blue dot which are founded sometimes,
# So I'll calculate the distances between the factory(red point)
# and all blue dots which are listed
if (idx > 1):
distance = []
for k in range(0,len(idx),1):
distance.append(np.sqrt((df.lon.iloc[i] - grids[k][0])**2 + (df.lat.iloc[i] - grids[k][1])**2))
pos_index = distance.index(min(distance))
pos = idx[pos_index]
# Only find 1 point
else:
pos = idx
grids_em[pos] += df.so2[i]
4. Result
co2 = grids_em.reshape(20,20)
plt.pcolormesh(lon_x,lat_y,co2,cmap =plt.cm.Spectral_r,zorder=3)
http://i4.tietuku.com/6ded65c4ac301294.png
5. My question
Can someone point out some drawbacks or error of this method?
Is there some algorithms more aligned with my target?
Thanks a lot!
There are many for-loop in your code, it's not the numpy way.
Make some sample data first:
import numpy as np
import pandas as pd
from scipy.spatial import KDTree
import pylab as pl
xc1, xc2, yc1, yc2 = 113.49805889531724, 115.5030664238035, 37.39995194888143, 38.789235929357105
N = 1000
GSIZE = 20
x, y = np.random.multivariate_normal([(xc1 + xc2)*0.5, (yc1 + yc2)*0.5], [[0.1, 0.02], [0.02, 0.1]], size=N).T
value = np.ones(N)
df_points = pd.DataFrame({"x":x, "y":y, "v":value})
For equal space grids you can use hist2d():
pl.hist2d(df_points.x, df_points.y, weights=df_points.v, bins=20, cmap="viridis");
Here is the output:
Here is the code to use KdTree:
X, Y = np.mgrid[x.min():x.max():GSIZE*1j, y.min():y.max():GSIZE*1j]
grid = np.c_[X.ravel(), Y.ravel()]
points = np.c_[df_points.x, df_points.y]
tree = KDTree(grid)
dist, indices = tree.query(points)
grid_values = df_points.groupby(indices).v.sum()
df_grid = pd.DataFrame(grid, columns=["x", "y"])
df_grid["v"] = grid_values
fig, ax = pl.subplots(figsize=(10, 8))
ax.plot(df_points.x, df_points.y, "kx", alpha=0.2)
mapper = ax.scatter(df_grid.x, df_grid.y, c=df_grid.v,
cmap="viridis",
linewidths=0,
s=100, marker="o")
pl.colorbar(mapper, ax=ax);
the output is:
I'm trying to visualize a few graphs whose nodes represent different objects. I want to create an image that looks like the one here:
Basically, I need a 3D plot and the ability to draw edges between nodes on the same level or nodes on different levels.
This answer below may not be a complete solution, but is a working demo for rendering 3D graphs using networkx.
networkx as such cannot render 3D graphs. We will have to install mayavi for that to happen.
import networkx as nx
import matplotlib.pyplot as plt
import numpy as np
from mayavi import mlab
import random
def draw_graph3d(graph, graph_colormap='winter', bgcolor = (1, 1, 1),
node_size=0.03,
edge_color=(0.8, 0.8, 0.8), edge_size=0.002,
text_size=0.008, text_color=(0, 0, 0)):
H=nx.Graph()
# add edges
for node, edges in graph.items():
for edge, val in edges.items():
if val == 1:
H.add_edge(node, edge)
G=nx.convert_node_labels_to_integers(H)
graph_pos=nx.spring_layout(G, dim=3)
# numpy array of x,y,z positions in sorted node order
xyz=np.array([graph_pos[v] for v in sorted(G)])
# scalar colors
scalars=np.array(G.nodes())+5
mlab.figure(1, bgcolor=bgcolor)
mlab.clf()
#----------------------------------------------------------------------------
# the x,y, and z co-ordinates are here
# manipulate them to obtain the desired projection perspective
pts = mlab.points3d(xyz[:,0], xyz[:,1], xyz[:,2],
scalars,
scale_factor=node_size,
scale_mode='none',
colormap=graph_colormap,
resolution=20)
#----------------------------------------------------------------------------
for i, (x, y, z) in enumerate(xyz):
label = mlab.text(x, y, str(i), z=z,
width=text_size, name=str(i), color=text_color)
label.property.shadow = True
pts.mlab_source.dataset.lines = np.array(G.edges())
tube = mlab.pipeline.tube(pts, tube_radius=edge_size)
mlab.pipeline.surface(tube, color=edge_color)
mlab.show() # interactive window
# create tangled hypercube
def make_graph(nodes):
def make_link(graph, i1, i2):
graph[i1][i2] = 1
graph[i2][i1] = 1
n = len(nodes)
if n == 1: return {nodes[0]:{}}
nodes1 = nodes[0:n/2]
nodes2 = nodes[n/2:]
G1 = make_graph(nodes1)
G2 = make_graph(nodes2)
# merge G1 and G2 into a single graph
G = dict(G1.items() + G2.items())
# link G1 and G2
random.shuffle(nodes1)
random.shuffle(nodes2)
for i in range(len(nodes1)):
make_link(G, nodes1[i], nodes2[i])
return G
# graph example
nodes = range(10)
graph = make_graph(nodes)
draw_graph3d(graph)
This code was modified from one of the examples here.
Please post the code in this case, when you succeed in reaching the objective.