Consider the following simple code
using PyPlot
x = [2,5,3,4]
y = [1,2,3,4]
plot(x,y,".-")
As you note the lines that connect the points on the graph are displayed according to the order of the data into the arrays. I mean, the first "point x-y" (2,1) is connected to the point (5,2), that is connected to (3,3) and so on.
How do I get a graph that reflects this piece of code:
using PyPlot
x = [2,3,4,5]
y = [1,3,4,2]
plot(x,y,".-")
?
Or in other words, how can I sort the x-array and preserve the x-y correspondence?
Related
I have two numpy array : X shape is (68,44,13) and X_toadd shape is: (68,44,7)I want to add them together in a way that I will have X_new shape as (68,44, 20). So, I need to keep the first two dimensions of X and add the 7 columns from X_toadd's third dimension to the 13 columns.
how should I do that?
add, append, and concatenate are tried but the result is not what I want which should have the shape (68,44,20)!
You need to specify which axis to use to glue things together.
Here, -1 denotes the first axis from the the back.
import numpy as np
a,b = np.zeros((68,44,13)), np.zeros((68,44,7))
c = np.concatenate([a,b], axis=-1)
c.shape
(68, 44, 20)
I received this problem as one of my homework problems, and I'm catching my ears with it. I would like some help with it:
The problem of the cuts: A rectangular metal sheet plate, of Lp x Hp size, has n
holes, in integer coordinates points. Cut a piece of plate of maximum surface, without holes, knowing that only parallel with the sides of the plate cuts are allowed.
Note: The coordinates of the holes are retained in two vectors vx[i] for the abscises holes
and vy[i] for ordinates (these vectors aren't certainly sorted, wholes being able to be stored in chronologycal order, for example). The initial rectangle, and then
rectangles that appear in the cutting process, are stored through the coordinates of the bottom left corner
(x, y) by length L and height h (each rectangle is identified through a set of 4 variables: x, y, L, h, by with the help of which the 4 corners coordinates are formed). We apply "Divide and conquer" technique.
What I have tried:
### [1,3] :vx (holes)
def cuts(x,y,L,h):
for i in range(nrg):
if vx[i] > x and vx[i] < x+L and vy[i] > y and vy[i] <y+h:
found = True
break
if found:
1: couts(x,y,L-vx[i],h)
2: cuts(vx[i],y,L-vx[i]+x,h)
3: cuts(L-vx[i],y,L-vx[i],h)
4: cuts(x,h-y,L,h-y)
else:
ar = L*h
if ar > arMax:
arMax = ar
nrg = 2
vx[1] = [1,3]
vx[2] = [2,3]
[x,y,L,h]
sol[0] = x
sol[1] = y
sol[2] = L
sol[3] = h
I am doing a clustering task and I have a distance matrix. I wish to visualize this distance matrix as a 2D graph. Please let me know if there is any way to do it online or in programming languages like R or python.
My distance matrix is as follows,
I used the classical Multidimensional scaling functionality (in R) and obtained a 2D plot that looks like:
But What I am looking for is a graph with nodes and weighted edges running between them.
Possibility 1
I assume, that you want a 2dimensional graph, where distances between nodes positions are the same as provided by your table.
In python, you can use networkx for such applications. In general there are manymethods of doing so, remember, that all of them are just approximations (as in general it is not possible to create a 2 dimensional representataion of points given their pairwise distances) They are some kind of stress-minimizatin (or energy-minimization) approximations, trying to find the "reasonable" representation with similar distances as those provided.
As an example you can consider a four point example (with correct, discrete metric applied):
p1 p2 p3 p4
---------------
p1 0 1 1 1
p2 1 0 1 1
p3 1 1 0 1
p4 1 1 1 0
In general, drawing actual "graph" is redundant, as you have fully connected one (each pair of nodes is connected) so it should be sufficient to draw just points.
Python example
import networkx as nx
import numpy as np
import string
dt = [('len', float)]
A = np.array([(0, 0.3, 0.4, 0.7),
(0.3, 0, 0.9, 0.2),
(0.4, 0.9, 0, 0.1),
(0.7, 0.2, 0.1, 0)
])*10
A = A.view(dt)
G = nx.from_numpy_matrix(A)
G = nx.relabel_nodes(G, dict(zip(range(len(G.nodes())),string.ascii_uppercase)))
G = nx.to_agraph(G)
G.node_attr.update(color="red", style="filled")
G.edge_attr.update(color="blue", width="2.0")
G.draw('distances.png', format='png', prog='neato')
In R you can try multidimensional scaling
# Classical MDS
# N rows (objects) x p columns (variables)
# each row identified by a unique row name
d <- dist(mydata) # euclidean distances between the rows
fit <- cmdscale(d,eig=TRUE, k=2) # k is the number of dim
fit # view results
# plot solution
x <- fit$points[,1]
y <- fit$points[,2]
plot(x, y, xlab="Coordinate 1", ylab="Coordinate 2",
main="Metric MDS", type="n")
text(x, y, labels = row.names(mydata), cex=.7)
Possibility 2
You just want to draw a graph with labeled edges
Again, networkx can help:
import networkx as nx
# Create a graph
G = nx.Graph()
# distances
D = [ [0, 1], [1, 0] ]
labels = {}
for n in range(len(D)):
for m in range(len(D)-(n+1)):
G.add_edge(n,n+m+1)
labels[ (n,n+m+1) ] = str(D[n][n+m+1])
pos=nx.spring_layout(G)
nx.draw(G, pos)
nx.draw_networkx_edge_labels(G,pos,edge_labels=labels,font_size=30)
import pylab as plt
plt.show()
Multidimensional scaling (MDS) is exactly what you want. See here and here for more.
You did not mentioned if you want a 2 dimensional graph or not. I suppose that you want to build a graph on 2 dimensions due to the fact that you need that for visualization. Considering that you have to be aware that for the most of the graphs this is simply not possible.
What can be probably done is to approximate somehow the values from distance matrix, something like small values to have relative small edges and big values to have a relative big length.
With all previous considerations one option would be graphviz. See neato function.
In general what you are interested in is force-directed drawing. See wikipedia for further reference.
You can use d3js Force Directed Graph and configure distance between nodes. d3js force layout has some clustering capability to separate nodes with similar distances. Here's an example with values as distance between nodes:
http://vida.io/documents/SyT7DREdQmGSpsBkK
Another way to visualize is to use same distance between nodes but different line thickness. In that case, you'd want to calculate stroke-width based on values:
.style("stroke-width", function(d) { return Math.sqrt(d.value / 50); });
I'm trying to find the shortest path between two points in a grid with no obstacles and move in all directions (N NE E ES S SW W WN).
It seems to be a common task... Is this not implemented already in Matlab? When Matlab plots two points joined by a line ( plot(X,Y,'-') ) seems to internally do this calculation as I guess that the generated image is a grid too.
Example: From [1,1] to [3,6] one solution is [1,1; 2,2; 2,3; 2,4; 3,5; 3,6]
I have tried:
dist_x = length(linspace(p1(1),p2(1), dist(p1(1),p2(1))+1));
dist_y = length(linspace(p1(2),p2(2), dist(p1(2),p2(2))+1));
num_points = max(dist_x, dist_y);
x = round(linspace(p1(1),p2(1),num_points));
y = round(linspace(p1(2),p2(2),num_points));
But I think that it returns more points than it should and maybe there is an implemented routine.
Thanks a lot
The solution (given by J.F. Sebastian) is the Bresenham Line Algorithm.
I'm using procedural techniques to generate graphics for a game I am writing.
To generate some woods I would like to scatter trees randomly within a regular hexagonal area centred at <0,0>.
What is the best way to generate these points in a uniform way?
If you can find a good rectangular bounding box for your hexagon, the easiest way to generate uniformly random points is by rejection sampling (http://en.wikipedia.org/wiki/Rejection_sampling)
That is, find a rectangle that entirely contains your hexagon, and then generate uniformly random points within the rectangle (this is easy, just independently generate random values for each coordinate in the right range). Check if the random point falls within the hexagon. If yes, keep it. If no, draw another point.
So long as you can find a good bounding box (the area of the rectangle should not be more than a constant factor larger than the area of the hexagon it encloses), this will be extremely fast.
A possibly simple way is the following:
F ____ B
/\ /\
A /__\/__\ E
\ /\ /
\/__\/
D C
Consider the parallelograms ADCO (center is O) and AOBF.
Any point in this can be written as a linear combination of two vectors AO and AF.
An point P in those two parallelograms satisfies
P = x* AO + y * AF or xAO + yAD.
where 0 <= x < 1 and 0 <= y <= 1 (we discount the edges shared with BECO).
Similarly any point Q in the parallelogram BECO can be written as the linear combination of vectors BO and BE such that
Q = xBO + yBE where 0 <=x <=1 and 0 <=y <= 1.
Thus to select a random point
we select
A with probability 2/3 and B with probability 1/3.
If you selected A, select x in [0,1) (note, half-open interval [0,1)) and y in [-1,1] and choose point P = xAO+yAF if y > 0 else choose P = x*AO + |y|*AD.
If you selected B, select x in [0,1] and y in [0,1] and choose point Q = xBO + yBE.
So it will take three random number calls to select one point, which might be good enough, depending on your situation.
If it's a regular hexagon, the simplest method that comes to mind is to divide it into three rhombuses. That way (a) they have the same area, and (b) you can pick a random point in any one rhombus with two random variables from 0 to 1. Here is a Python code that works.
from math import sqrt
from random import randrange, random
from matplotlib import pyplot
vectors = [(-1.,0),(.5,sqrt(3.)/2.),(.5,-sqrt(3.)/2.)]
def randinunithex():
x = randrange(3);
(v1,v2) = (vectors[x], vectors[(x+1)%3])
(x,y) = (random(),random())
return (x*v1[0]+y*v2[0],x*v1[1]+y*v2[1])
for n in xrange(500):
v = randinunithex()
pyplot.plot([v[0]],[v[1]],'ro')
pyplot.show()
A couple of people in the discussion raised the question of uniformly sampling a discrete version of the hexagon. The most natural discretization is with a triangular lattice, and there is a version of the above solution that still works. You can trim the rhombuses a little bit so that they each contain the same number of points. They only miss the origin, which has to be allowed separately as a special case. Here is a code for that:
from math import sqrt
from random import randrange, random
from matplotlib import pyplot
size = 10
vectors = [(-1.,0),(.5,sqrt(3.)/2.),(.5,-sqrt(3.)/2.)]
def randinunithex():
if not randrange(3*size*size+1): return (0,0)
t = randrange(3);
(v1,v2) = (vectors[t], vectors[(t+1)%3])
(x,y) = (randrange(0,size),randrange(1,size))
return (x*v1[0]+y*v2[0],x*v1[1]+y*v2[1])
# Plot 500 random points in the hexagon
for n in xrange(500):
v = randinunithex()
pyplot.plot([v[0]],[v[1]],'ro')
# Show the trimmed rhombuses
for t in xrange(3):
(v1,v2) = (vectors[t], vectors[(t+1)%3])
corners = [(0,1),(0,size-1),(size-1,size-1),(size-1,1),(0,1)]
corners = [(x*v1[0]+y*v2[0],x*v1[1]+y*v2[1]) for (x,y) in corners]
pyplot.plot([x for (x,y) in corners],[y for (x,y) in corners],'b')
pyplot.show()
And here is a picture.
alt text http://www.freeimagehosting.net/uploads/0f80ad5d9a.png
The traditional approach (applicable to regions of any polygonal shape) is to perform trapezoidal decomposition of your original hexagon. Once that is done, you can select your random points through the following two-step process:
1) Select a random trapezoid from the decomposition. Each trapezoid is selected with probability proportional to its area.
2) Select a random point uniformly in the trapezoid chosen on step 1.
You can use triangulation instead of trapezoidal decomposition, if you prefer to do so.
Chop it up into six triangles (hence this applies to any regular polygon), randomly choose one triangle, and randomly choose a point in the selected triangle.
Choosing random points in a triangle is a well-documented problem.
And of course, this is quite fast and you'll only have to generate 3 random numbers per point --- no rejection, etc.
Update:
Since you will have to generate two random numbers, this is how you do it:
R = random(); //Generate a random number called R between 0-1
S = random(); //Generate a random number called S between 0-1
if(R + S >=1)
{
R = 1 – R;
S = 1 – S;
}
You may check my 2009 paper, where I derived an "exact" approach to generate "random points" inside different lattice shapes: "hexagonal", "rhombus", and "triangular". As far as I know it is the "most optimized approach" because for every 2D position you only need two random samples. Other works derived earlier require 3 samples for each 2D position!
Hope this answers the question!
http://arxiv.org/abs/1306.0162
1) make biection from points to numbers (just enumerate them), get random number -> get point.
Another solution.
2) if N - length of hexagon's side, get 3 random numbers from [1..N], start from some corner and move 3 times with this numbers for 3 directions.
The rejection sampling solution above is intuitive and simple, but uses a rectangle, and (presumably) euclidean, X/Y coordinates. You could make this slightly more efficient (though still suboptimal) by using a circle with radius r, and generate random points using polar coordinates from the center instead, where distance would be rand()*r, and theta (in radians) would be rand()*2*PI.