Question
Given a plotting window, how does one generate random points at the perimeter of a square (perimeter of the plotting window)?
Background and attempt
I found a similar question with regards to a rectangle in javascript.
I managed to write a program to generate random points within limits but the question is regarding how one could find random points with the condition that they are at the edge of the plot (either x is equal to 5 or -5 ,or y is equal to 5 or -5 in this case).
import numpy as np
import matplotlib.pyplot as plt
# Parameters
n = 6 # number of points
a = 5 # upper bound
b = -5 # lower bound
# Random coordinates [b,a) uniform distributed
coordy = (b - a) * np.random.random_sample((n,)) + a # generate random y
coordx = (b - a) * np.random.random_sample((n,)) + a # generate random x
# Create limits (x,y)=((-5,5),(-5,5))
plt.xlim((b,a))
plt.ylim((b,a))
# Plot points
for i in range(n):
plt.plot(coordx[i],coordy[i],'ro')
plt.show()
Summary
So to summarize, my question is how do I generate random coordinates given that they are at the edge of the plot/canvas. Any advice or help will be appreciated.
Here is what you can do:
from random import choice
import matplotlib.pyplot as plt
from numpy.random import random_sample
n = 6
a = 5
b = -5
plt.xlim((b,a))
plt.ylim((b,a))
for i in range(n):
r = (b - a) * random_sample() + a
random_point = choice([(choice([a,b]), r),(r, choice([a,b]))])
plt.scatter(random_point[0],random_point[1])
plt.show()
Output:
One possible approach (despite not very elegant) is the following: divide horizontal and vertical points Suppose you want to draw a point at the top or at the bottom of the window. Then,
Select randomly the y coordinate as b or -b
Select randomly (uniform distribution) the x coordinate
Similar approach for right and left edges of the window.
Hope that helps.
You could use this, but this is assuming you want to discard them when it is found they aren't on the edge.
for x in coordx:
if x != a:
coordx.pop(x)
else:
continue
And then do the same for y.
Geometrically speaking, being on the edge requires that a point satisfy certain conditions. Assuming that we are talking about a grid whose dimensions are defined by x ~ [0, a] and y ~ [0, b]:
The y-coordinate is either 0 or b, with the x-coordinate within [0, a], or
The x-coordinate is either 0 or a, with the y-coordinate within [0, b]
There are obviously more than one way to go about this, but here is a simple method to get you started.
def plot_edges(n_points, x_max, y_max, x_min=0, y_min=0):
# if x_max - x_min = y_max - y_min, plot a square
# otherwise, plot a rectangle
vertical_edge_x = np.random.uniform(x_min, x_max, n_points)
vertical_edige_y = np.asarray([y_min, y_max])[
np.random.randint(2, size=n_points)
]
horizontal_edge_x = np.asarray([x_min, x_max])[
np.random.randint(2, size=n_points)
]
horizontal_edge_y = np.random.uniform(x_min, x_max, n_points)
# plot generated points
plt.scatter(vertical_edge_x, vertical_edige_y)
plt.scatter(horizontal_edge_x, horizontal_edge_y)
plt.show()
Can you try this out?
import numpy as np
import matplotlib.pyplot as plt
# Parameters
n = 6 # number of points
a = 5 # upper bound
b = -5 # lower bound
import random
coordx,coordy=[],[]
for i in range(n):
xy = random.choice(['x','y'])
if xy=='x':
coordx.append(random.choice([b,a])) # generate random x
coordy.append(random.random()) # generate random y
if xy=='y':
coordx.append(random.random()) # generate random x
coordy.append(random.choice([b,a])) # generate random y
# Create limits (x,y)=((-5,5),(-5,5))
plt.xlim((b,a))
plt.ylim((b,a))
# Plot points
for i in range(n):
plt.plot(coordx[i],coordy[i],'ro')
plt.show()
Here is a sample output:
Here's a way to do that:
import numpy as np
import matplotlib.pyplot as plt
# Parameters
n = 6 # number of points
a = 5 # upper bound
b = -5 # lower bound
# Random coordinates [b,a) uniform distributed
coordy = (b - a) * np.random.random_sample((n,)) + a # generate random y
coordx = (b - a) * np.random.random_sample((n,)) + a # generate random x
# This is the new code
reset_axis = np.random.choice([True, False], n) # select which axis to reset
reset_direction = np.random.choice([a,b], n) # select to go up / right or down / left
coordx[reset_axis] = reset_direction[reset_axis]
coordy[~reset_axis] = reset_direction[~reset_axis]
# end of new code.
# Create limits (x,y)=((-5,5),(-5,5))
plt.xlim((b,a))
plt.ylim((b,a))
# Plot points
for i in range(n):
plt.plot(coordx[i],coordy[i],'ro')
plt.show()
The result is:
Related
I want to generate a toy example to illustrate a convex piecewise linear function in python, but I couldn't figure out the best way to do this. What I want to do is to indicate the number of lines and generate the function randomly.
A convex piecewise-linear function is defined as:
For instance, if I want to have four linear lines, then I want to generate something as shown below.
Since there are four lines. I need to generate four increasing random integers to determine the intervals in x-axis.
import random
import numpy as np
random.seed(1)
x_points = np.array(random.sample(range(1, 20), 4))
x_points.sort()
x_points = np.append(0, x_points)
x_points
[0 3 4 5 9]
I can now use the first two points and create a random linear function, but I don't know how I should continue from there to preserve the convexity. Note that a function is called convex if the line segment between any two points on the graph of the function does not lie below the graph between the two points.
The slope increases monotonously by a random value from the range [0,1), starting from 0. The first y value is also zero, see the comments.
import numpy as np
np.random.seed(0)
x_points = np.random.randint(low=1, high=20, size=4)
x_points.sort()
x_points = np.append(0, x_points) # the first 0 point is 0
slopes = np.add.accumulate(np.random.random(size=3))
slopes = np.append(0,slopes) # the first slope is 0
y_incr = np.ediff1d(x_points)*slopes
y_points = np.add.accumulate(y_incr)
y_points = np.append(0,y_points) # the first y values is 0
A possible output looks like this:
print(x_points)
print(y_points)
# [ 0 1 4 13 16]
# [ 0. 0. 2.57383685 17.92061306 24.90689622]
To print this figure:
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
ax.plot(x_points,y_points, '-o', label="convex piecewise-linear function")
ax.legend()
fig.patch.set_facecolor('white')
plt.show()
make sure the gradient (=dx/dy) is increasing.
Pseudocode:
s = 1;
x = 0;
y = 0;
n = 4;
while(--n>0)
{
//increase x randomly
dx = rand(3);
dy = dx * s;
x += dx;
y += dy;
//increase gradient randomly
s += rand(3);
print x + "/" +y;
}
I am trying to label x and y points based on their being in a specific section of a meshgrid in python. The points are stored in a pandas dataframe.
Here I have a scatter plot of the coordinates and above them I am plotting the grid.
The entire grid is way bigger, from the bottom left point (500,1250) to upper right point (2750, 3250), which means the whole grid is 225x200 sections.
I want to iterate through the sections of the grid and check if a point is inside. If a point is inside the section I want to add a label to the point. The label should be the same of the section name.
I want to add a column to the dataframe called 'section' that stores the section a point belongs to.
In the example (picture above) I would like to label all the points with
770 <= x <= 780 and 1795 <= y <= 1805 with the section name 'A3'.
my code currently looks like this:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
df = pd.read_csv('./file.csv', sep=';')
x_min = df['X[mm]'].min()
x_max = df['X[mm]'].max()
y_min = df['Y[mm]'].min()
y_max = df['Y[mm]'].max()
#side of the square in mm:
square_side = 10
xs = np.arange(x_min, x_max+square_side, square_side)
ys = np.arange(y_min, y_max+square_side, square_side)
x_2, y_2 = np.meshgrid(xs, ys, indexing = 'ij')
fig, ax = plt.subplots(figsize=(9,9))
ax.plot(df['X[mm]'], df['Y[mm]'], linewidth=0.2, c='black')
#plot meshgrid as grid instead of points:
segs1 = np.stack((x_2[:,[0,-1]],y_2[:,[0,-1]]), axis=2)
segs2 = np.stack((x_2[[0,-1],:].T,y_2[[0,-1],:].T), axis=2)
plt.gca().add_collection(LineCollection(np.concatenate((segs1, segs2))))
ax.set_aspect('equal', 'box')
plt.show()
I have also a function that determines if the points are inside of a rectangle (this does not use meshgrid):
def is_inside_rect(M, A, B, D):
'''Check if a point M is inside a rectangle with corners A, B, C, D'''
# 0 <= dot(BC,BM) <= dot(BC,BC)
#print(np.dot(B - A, D - A))
return 0 <= np.dot(B - A, M - A) <= np.dot(B - A, B - A) and 0 <= np.dot(D - B, M - B) <= np.dot(D - B, D - B)
I thought of using it in a while loop like this:
x = x_min
y = y_min
while (x <= x_max + square_side) and (y <= y_max + square_side):
A = np.array([x, y])
B = np.array([x + square_side, y])
D = np.array([x + square_side, y + square_side])
print(A, B, D)
df['c'] = df[['X[mm]', 'Y[mm]']].apply(lambda coord: 'red' if is_inside_rect(np.array(coord), A, B, D) else 'black', axis=1)
x += square_side
y += square_side
but this very slow and it changes the colors of all the points in every iteration.
Since all your points are equally sized, there is no need to define all of your squares beforehand and then determine which squares have which points. I would use the coordinates of each point to directly determine which square it will land in.
Let's take the 1-dimensional case, for the sake of simplicity. You want to group points on the number line into "squares" (really 1-d line segments). If your first square starts at x=0, your second at x=10, your third at x=20, and so on, how do you find the square for an arbitrary point x? You know that your squares are spaced by 10 (and you know they start at 0, which makes things easier), so you can simply divide by 10 and round down to get the square index.
You can just as easily do the same thing in 3-dimensions (or n-dimensions).
square_side = 10
x_min = df['X[mm]'].min()
y_min = df['Y[mm]'].min()
def label_point(x, y):
# Double forward slash is integer (round down) division
# Add 1 here if you really want 1-based indexing
x_label = (x - x_min) // square_side
y_label = chr(ord('A') + (y - y_min) // square_side)
return f'{y_label}{x_label}'
df['label'] = df[['X[mm]', 'Y[mm]']].apply(lambda coord: label_point(*coord), axis=1)
As for the efficiency, this solution looks at each point only once, and does a constant amount of work with each point, so it is O(n) in the number of points. Your solution looks at each square once, and for each square looks at each point this is O(n × m) where n is the number of points and m is the number of squares.
Your solution is more general, in that your is_inside_rect function works when your grid of rectangles has an arbitrary rotation. In this case, I would recommend rotating all your points about the origin, and then running my solution.
Also, your loop is adding 10 to x and y every loop, so you are traversing your space diagonally. I don't think you meant to do that.
For the following code whose job is to perform Monte Carlo integration for a function f, I was wondering what would happen if I define f as y = sqrt(1-x^2), which is the equation for a unit quarter circle, and specify an endpoint that is greater than 1, since we know that f is only defined for 0<x<1.
import numpy as np
import matplotlib.pyplot as plt
def definite_integral_show(f, x0, x1, N):
"""Approximate the definite integral of f(x)dx between x0 and x1 using
N random points
Arguments:
f -- a function of one real variable, must be nonnegative on [x0, x1]
N -- the number of random points to use
"""
#First, let's compute fmax. We do that by evaluating f(x) on a grid
#of points between x0 and x1
#This assumes that f is generally smooth. If it's not, we're in trouble!
x = np.arange(x0, x1, 0.01)
y = f(x)
print(y)
f_max = max(y)
#Now, let's generate the random points. The x's should be between
#x0 and x1, so we first create points beterrm 0 and (x1-x0), and
#then add x0
#The y's should be between 0 and fmax
#
# 0...(x1-x0)
x_rand = x0 + np.random.random(N)*(x1-x0)
print(x_rand)
y_rand = 0 + np.random.random(N)*f_max
#Now, let's find the indices of the poitns above and below
#the curve. That is, for points below the curve, let's find
# i s.t. y_rand[i] < f(x_rand)[i]
#And for points above the curve, find
# i s.t. y_rand[i] >= f(x_rand)[i]
ind_below = np.where(y_rand < f(x_rand))
ind_above = np.where(y_rand >= f(x_rand))
#Finally, let's display the results
plt.plot(x, y, color = "red")
pts_below = plt.scatter(x_rand[ind_below[0]], y_rand[ind_below[0]], color = "green")
pts_above = plt.scatter(x_rand[ind_above[0]], y_rand[ind_above[0]], color = "blue")
plt.legend((pts_below, pts_above),
('Pts below the curve', 'Pts above the curve'),
loc='lower left',
ncol=3,
fontsize=8)
def f1(x):
return np.sqrt(1-x**2)
definite_integral_show(f1, 0, 6, 200)
To my surprise, the program still works and gives me the following picture.
I suspect that it works because in NumPy, nan's in an array are just ignored when performing operations on the array. However, I don't understand why the picture only contains points whose x and y coordinates are both between 0 to 1. Where are the points that aren't within this range, but whose values are computed by
x_rand = x0 + np.random.random(N)*(x1-x0)
y_rand = 0 + np.random.random(N)*f_max
You can just print out the arrays (for example by generating only one random point) and see that they go into neither ind_below nor ind_above...
That's because all comparisons that involves nan returns False. (See also: What is the rationale for all comparisons returning false for IEEE754 NaN values?). (so y_rand < nan and y_rand >= nan both evaluates to False)
The easiest way to change the code is
ind_below = np.where(y_rand < f(x_rand))
ind_above = np.where(~(y_rand < f(x_rand)))
(optionally only compute the array once)
linspace generates a linear space. How can I generate a grid using an arbitrary density function?
Say, I would like to have a grid from 0 to 1, with 100 grid points, and where the density of points is given by (x - 0.5)**2 - how would I create such a grid in Python?
That is, I want many grid-points where the function (x - 0.5)**2) is large, and few points where the function is small. I do not want a grid that has values according to this function.
For example like this:
x = (np.linspace(0.5,1.5,100)-0.5)**2
The start and end values have to be chosen so that f(start) = 0 and f(end)=1.
In that case the following solution should work. Be sure that func is positive throughout the range...
import numpy as np
from matplotlib import pyplot as plt
def func(x):
return (x-0.5)**2
start = 0
end = 1
npoints = 100
x = np.linspace(start,end,npoints)
fx = func(x)
# take density (or intervals) as inverse of fx
# g in [0,1] controls how much warping you want.
# g = 0: fully warped
# g = 1: linearly spaced
g = 0
density = (1+g*(fx-1))/fx
# sum the intervals to get new grid
x_density = np.cumsum(density)
# rescale to match old range
x_density -= x_density.min()
x_density/= x_density.max()
x_density *= (end-start)
x_density += start
fx_density = func(x_density)
plt.plot(x,fx,'ok',ms = 10,label = 'linear')
plt.plot(x_density,fx_density,'or',ms = 10,label = 'warped')
plt.legend(loc = 'upper center')
plt.show()
I have an application that requires a disk populated with 'n' points in a quasi-random fashion. I want the points to be somewhat random, but still have a more or less regular density over the disk.
My current method is to place a point, check if it's inside the disk, and then check if it is also far enough away from all other points already kept. My code is below:
import os
import random
import math
# ------------------------------------------------ #
# geometric constants
center_x = -1188.2
center_y = -576.9
center_z = -3638.3
disk_distance = 2.0*5465.6
disk_diam = 5465.6
# ------------------------------------------------ #
pts_per_disk = 256
closeness_criteria = 200.0
min_closeness_criteria = disk_diam/closeness_criteria
disk_center = [(center_x-disk_distance),center_y,center_z]
pts_in_disk = []
while len(pts_in_disk) < (pts_per_disk):
potential_pt_x = disk_center[0]
potential_pt_dy = random.uniform(-disk_diam/2.0, disk_diam/2.0)
potential_pt_y = disk_center[1]+potential_pt_dy
potential_pt_dz = random.uniform(-disk_diam/2.0, disk_diam/2.0)
potential_pt_z = disk_center[2]+potential_pt_dz
potential_pt_rad = math.sqrt((potential_pt_dy)**2+(potential_pt_dz)**2)
if potential_pt_rad < (disk_diam/2.0):
far_enough_away = True
for pt in pts_in_disk:
if math.sqrt((potential_pt_x - pt[0])**2+(potential_pt_y - pt[1])**2+(potential_pt_z - pt[2])**2) > min_closeness_criteria:
pass
else:
far_enough_away = False
break
if far_enough_away:
pts_in_disk.append([potential_pt_x,potential_pt_y,potential_pt_z])
outfile_name = "pt_locs_x_lo_"+str(pts_per_disk)+"_pts.txt"
outfile = open(outfile_name,'w')
for pt in pts_in_disk:
outfile.write(" ".join([("%.5f" % (pt[0]/1000.0)),("%.5f" % (pt[1]/1000.0)),("%.5f" % (pt[2]/1000.0))])+'\n')
outfile.close()
In order to get the most even point density, what I do is basically iteratively run this script using another script, with the 'closeness' criteria reduced for each successive iteration. At some point, the script can not finish, and I just use the points of the last successful iteration.
So my question is rather broad: is there a better way to do this? My method is ok for now, but my gut says that there is a better way to generate such a field of points.
An illustration of the output is graphed below, one with a high closeness criteria, and another with a 'lowest found' closeness criteria (what I want).
A simple solution based on Disk Point Picking from MathWorld:
import numpy as np
import matplotlib.pyplot as plt
n = 1000
r = np.random.uniform(low=0, high=1, size=n) # radius
theta = np.random.uniform(low=0, high=2*np.pi, size=n) # angle
x = np.sqrt(r) * np.cos(theta)
y = np.sqrt(r) * np.sin(theta)
# for plotting circle line:
a = np.linspace(0, 2*np.pi, 500)
cx,cy = np.cos(a), np.sin(a)
fg, ax = plt.subplots(1, 1)
ax.plot(cx, cy,'-', alpha=.5) # draw unit circle line
ax.plot(x, y, '.') # plot random points
ax.axis('equal')
ax.grid(True)
fg.canvas.draw()
plt.show()
It gives.
Alternatively, you also could create a regular grid and distort it randomly:
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.tri as tri
n = 20
tt = np.linspace(-1, 1, n)
xx, yy = np.meshgrid(tt, tt) # create unit square grid
s_x, s_y = xx.ravel(), yy.ravel()
ii = np.argwhere(s_x**2 + s_y**2 <= 1).ravel() # mask off unwanted points
x, y = s_x[ii], s_y[ii]
triang = tri.Triangulation(x, y) # create triangluar grid
# distort the grid
g = .5 # distortion factor
rx = x + np.random.uniform(low=-g/n, high=g/n, size=x.shape)
ry = y + np.random.uniform(low=-g/n, high=g/n, size=y.shape)
rtri = tri.Triangulation(rx, ry, triang.triangles) # distorted grid
# for circle:
a = np.linspace(0, 2*np.pi, 500)
cx,cy = np.cos(a), np.sin(a)
fg, ax = plt.subplots(1, 1)
ax.plot(cx, cy,'k-', alpha=.2) # circle line
ax.triplot(triang, "g-", alpha=.4)
ax.triplot(rtri, 'b-', alpha=.5)
ax.axis('equal')
ax.grid(True)
fg.canvas.draw()
plt.show()
It gives
The triangles are just there for visualization. The obvious disadvantage is that depending on your choice of grid, either in the middle or on the borders (as shown here), there will be more or less large "holes" due to the grid discretization.
If you have a defined area like a disc (circle) that you wish to generate random points within you are better off using an equation for a circle and limiting on the radius:
x^2 + y^2 = r^2 (0 < r < R)
or parametrized to two variables
cos(a) = x/r
sin(a) = y/r
sin^2(a) + cos^2(a) = 1
To generate something like the pseudo-random distribution with low density you should take the following approach:
For randomly distributed ranges of r and a choose n points.
This allows you to generate your distribution to roughly meet your density criteria.
To understand why this works imagine your circle first divided into small rings of length dr, now imagine your circle divided into pie slices of angle da. Your randomness now has equal probability over the whole boxed area arou d the circle. If you divide the areas of allowed randomness throughout your circle you will get a more even distribution around the overall circle and small random variation for the individual areas giving you the psudo-random look and feel you are after.
Now your job is just to generate n points for each given area. You will want to have n be dependant on r as the area of each division changes as you move out of the circle. You can proportion this to the exact change in area each space brings:
for the n-th to n+1-th ring:
d(Area,n,n-1) = Area(n) - Area(n-1)
The area of any given ring is:
Area = pi*(dr*n)^2 - pi*(dr*(n-1))
So the difference becomes:
d(Area,n,n-1) = [pi*(dr*n)^2 - pi*(dr*(n-1))^2] - [pi*(dr*(n-1))^2 - pi*(dr*(n-2))^2]
d(Area,n,n-1) = pi*[(dr*n)^2 - 2*(dr*(n-1))^2 + (dr*(n-2))^2]
You could expound this to gain some insight on how much n should increase but it may be faster to just guess at some percentage increase (30%) or something.
The example I have provided is a small subset and decreasing da and dr will dramatically improve your results.
Here is some rough code for generating such points:
import random
import math
R = 10.
n_rings = 10.
n_angles = 10.
dr = 10./n_rings
da = 2*math.pi/n_angles
base_points_per_division = 3
increase_per_level = 1.1
points = []
ring = 0
while ring < n_rings:
angle = 0
while angle < n_angles:
for i in xrange(int(base_points_per_division)):
ra = angle*da + da*math.random()
rr = r*dr + dr*random.random()
x = rr*math.cos(ra)
y = rr*math.sin(ra)
points.append((x,y))
angle += 1
base_points_per_division = base_points_per_division*increase_per_level
ring += 1
I tested it with the parameters:
n_rings = 20
n_angles = 20
base_points = .9
increase_per_level = 1.1
And got the following results:
It looks more dense than your provided image, but I imagine further tweaking of those variables could be beneficial.
You can add an additional part to scale the density properly by calculating the number of points per ring.
points_per_ring = densitymath.pi(dr**2)*(2*n+1)
points_per_division = points_per_ring/n_angles
This will provide a an even better scaled distribution.
density = .03
points = []
ring = 0
while ring < n_rings:
angle = 0
base_points_per_division = density*math.pi*(dr**2)*(2*ring+1)/n_angles
while angle < n_angles:
for i in xrange(int(base_points_per_division)):
ra = angle*da + min(da,da*random.random())
rr = ring*dr + dr*random.random()
x = rr*math.cos(ra)
y = rr*math.sin(ra)
points.append((x,y))
angle += 1
ring += 1
Giving better results using the following parameters
R = 1.
n_rings = 10.
n_angles = 10.
density = 10/(dr*da) # ~ ten points per unit area
With a graph...
and for fun you can graph the divisions to see how well it is matching your distriubtion and adjust.
Depending on how random the points need to be, it may be simple enough to just make a grid of points within the disk, and then displace each point by some small but random amount.
It may be that you want more randomness, but if you just want to fill your disc with an even-looking distribution of points that aren't on an obvious grid, you could try a spiral with a random phase.
import math
import random
import pylab
n = 300
alpha = math.pi * (3 - math.sqrt(5)) # the "golden angle"
phase = random.random() * 2 * math.pi
points = []
for k in xrange(n):
theta = k * alpha + phase
r = math.sqrt(float(k)/n)
points.append((r * math.cos(theta), r * math.sin(theta)))
pylab.scatter(*zip(*points))
pylab.show()
Probability theory ensures that the rejection method is an appropriate method
to generate uniformly distributed points within the disk, D(0,r), centered at origin and of radius r. Namely, one generates points within the square [-r,r] x [-r,r], until a point falls within the disk:
do{
generate P in [-r,r]x[-r,r];
}while(P[0]**2+P[1]**2>r);
return P;
unif_rnd_disk is a generator function implementing this rejection method:
import matplotlib.pyplot as plt
import numpy as np
import itertools
def unif_rnd_disk(r=1.0):
pt=np.zeros(2)
while True:
yield pt
while True:
pt=-r+2*r*np.random.random(2)
if (pt[0]**2+pt[1]**2<=r):
break
G=unif_rnd_disk()# generator of points in disk D(0,r=1)
X,Y=zip(*[pt for pt in itertools.islice(G, 1, 1000)])
plt.scatter(X, Y, color='r', s=3)
plt.axis('equal')
If we want to generate points in a disk centered at C(a,b), we have to apply a translation to the points in the disk D(0,r):
C=[2.0, -3.5]
plt.scatter(C[0]+np.array(X), C[1]+np.array(Y), color='r', s=3)
plt.axis('equal')