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I'm new to python so the code may not be the best. I'm trying to find the minimum Total Cost (TotalC) and the corresponding m,k and xM values that go with this minimum cost. I'm not sure how to do this. I have tried using min(TotalC) however this gives an error within the loop or outside the loop only returns the value of TotalC and not the corresponding m, k, and xM values. Any help would be appreciated. This section is at the end of the code, I have included my entire code.
I have tried using
minIndex = TotalC.argmin()
but I'm not sure how to use it and it only returns 0 each time.
import numpy as np
import matplotlib.pyplot as plt
def Load(x):
Fpeak = (1000 + (9*(x**2) - (183*x))) *1000 #Fpeak in N
td = (20 - ((0.12)*(x**2)) + (4.2*(x))) / 1000 #td in s
return Fpeak, td
#####################################################################################################
####################### Part 2 ########################
def displacement(m,k,x,dt): #Displacement function
Fpeak, td = Load(x) #Load Function from step 1
w = np.sqrt(k/m) # Natural circular frequency
T = 2 * np.pi /w #Natural period of blast (s)
time = np.arange(0,2*T,0.001) #Time array with range (0 - 2*T) with steps of 2*T/100
zt = [] #Create a lsit to store displacement values
for t in time:
if (t <= td):
zt.append((Fpeak/k) * (1 - np.cos(w*t)) + (Fpeak/(k*td)) * ((np.sin(w*t)/w) - t))
else:
zt.append((Fpeak/(k*w*td)) * (np.sin(w*t) - np.sin(w*(t-td))) - ((Fpeak/k) * np.cos(w*t)))
zmax=max(zt) #Find the max displacement from the list of zt values
return zmax #Return max displacement
k = 1E6
m = 200
dt = 0.0001
x = 0
z = displacement(m,k,x,dt)
###################################################################################
############### Part 3 #######################
# k = 1E6 , m = 200kg , Deflection = 0.1m
k_values = np.arange(1E6, 7E6, ((7E6-1E6)/10)) #List of k values between min and max (1E6 and 7E6).
m_values = np.arange(200,1200,((1200-200)/10)) #List of m values between min and max 200kg and 1200kg
xM = []
for k in k_values: # values of k
for m in m_values: # values of m within k for loop
def bisector(m,k,dpoint,dt): #dpoint = decimal point accuracy
xL = 0
xR = 10
xM = (xL + xR)/2
zmax = 99
while round(zmax, dpoint) !=0.1:
zmax = displacement(m,k,xM,dt)
if zmax > 0.1:
xL = xM
xM = (xL + xR)/2
else:
xR = xM
xM = (xL + xR)/2
return xM
xM = bisector(m, k, 4, 0.001)
print('xM value =',xM)
#####################################################
#######Step 4
def cost (m,k,xM):
Ck = 900 + 825*((k/1E6)**2) - (1725*(k/1E6))
Cm = 10*m - 2000
Cx = 2400*((xM**2)/4)
TotalC = Ck + Cm + Cx
minIndex = TotalC.argmin(0)
print(minIndex)
return TotalC
TotalC = cost(m, k, xM)
minIndex = TotalC.argmin()
print(minIndex)
print([xM, m, k, TotalC])
argmin() returns the index of a minimum value. If you are looking for the minimum itself, try using .min(). There is also a possibility that 0 is the lowest value in Your array so bear that in mind
In my following code i m running a lorentz chaotic equation from which i will get random numbers in terms of xs , ys and zs
import numpy as np
def lorenz(x, y, z, a=10,b=8/3,c=28 ):
x_dot = a*(y -x)
y_dot = - y +c*x - x*z
z_dot = -b*z + x*y
return x_dot, y_dot, z_dot
dt = 0.01
num_steps = 10000
# Need one more for the initial values
xs = np.empty(num_steps + 1)
ys = np.empty(num_steps + 1)
zs = np.empty(num_steps + 1)
# Set initial values
xs[0], ys[0], zs[0]= (1,1,1)
# Step through "time", calculating the partial derivatives at the current point
# and using them to estimate the next point
for i in range(num_steps):
x_dot, y_dot, z_dot= lorenz(xs[i], ys[i], zs[i])
xs[i + 1] = xs[i] + (x_dot * dt)
ys[i + 1] = ys[i] + (y_dot * dt)
zs[i + 1] = zs[i] + (z_dot * dt)
I am actually trying to test the xs, ys and zs value for random number generating test via NIST 800 by using the code below
from __future__ import print_function
import math
from fractions import Fraction
from scipy.special import gamma, gammainc, gammaincc
# from gamma_functions import *
import numpy
import cmath
import random
#ones_table = [bin(i)[2:].count('1') for i in range(256)]
def count_ones_zeroes(bits):
ones = 0
zeroes = 0
for bit in bits:
if (bit == 1):
ones += 1
else:
zeroes += 1
return (zeroes,ones)
def runs_test(bits):
n = len(bits)
zeroes,ones = count_ones_zeroes(bits)
prop = float(ones)/float(n)
print(" prop ",prop)
tau = 2.0/math.sqrt(n)
print(" tau ",tau)
if abs(prop-0.5) > tau:
return (False,0.0,None)
vobs = 1.0
for i in range(n-1):
if bits[i] != bits[i+1]:
vobs += 1.0
print(" vobs ",vobs)
p = math.erfc(abs(vobs - (2.0*n*prop*(1.0-prop)))/(2.0*math.sqrt(2.0*n)*prop*(1-prop) ))
success = (p >= 0.01)
return (success,p,None)
print(runs_test(xs))
#%%
from __future__ import print_function
import math
def count_ones_zeroes(bits):
ones = 0
zeroes = 0
for bit in bits:
if (bit == 1):
ones += 1
else:
zeroes += 1
return (zeroes,ones)
def monobit_test(bits):
n = len(bits)
zeroes,ones = count_ones_zeroes(bits)
s = abs(ones-zeroes)
print(" Ones count = %d" % ones)
print(" Zeroes count = %d" % zeroes)
p = math.erfc(float(s)/(math.sqrt(float(n)) * math.sqrt(2.0)))
success = (p >= 0.01)
return (success,p,None)
print(runs_test(xs))
the output which i m getting is false i.e
output:
prop 0.00019998000199980003
tau 0.01999900007499375
(False, 0.0, None)
what should i do now?
The Lorenz system is chaotic, not random. You implemented the differential equation solver well, but it seems that count_ones_zeroes doesn't do what its name implies, at least, not on the data you provide. on xs, it returns that (zeroes, ones) = (9999, 2), which is not what you want. The code checks the value within the xs array, i.e. an x value (e.g. 8.2) against 1, but x is a float between -20 and 20, so it will be usually non1, and will be counted as 0. Only x==1 will be counted as ones.
In python, int/int results in float, so there is no need to cast it to float, in contrast to e.g. C or C++, so instead of prop = float(ones)/float(n), you can write prop = ones/n Similar statements hold for +,- and *
I want to generate two linear chains of 20 monomers each at some distance to each other. The following code generates a single chain. Could someone help me with how to generate the second chain?
The two chains are fixed to a surface i.e the first monomer of the chain is fixed and the rest of the monomers move freely in x-y-z directions but the z component of the monomers should be positive.
Something like this:
import numpy as np
import numba as nb
#import pandas as pd
#nb.jit()
def gen_chain(N):
x = np.zeros(N)
y = np.zeros(N)
z = np.linspace(0, (N)*0.9, num=N)
return np.column_stack((x, y, z)), np.column_stack((x1, y1, z1))
#coordinates = np.loadtxt('2GN_50_T_10.txt', skiprows=199950)
#return coordinates
#nb.jit()
def lj(rij2):
sig_by_r6 = np.power(sigma**2 / rij2, 3)
sig_by_r12 = np.power(sigma**2 / rij2, 6)
lje = 4 * epsilon * (sig_by_r12 - sig_by_r6)
return lje
#nb.jit()
def fene(rij2):
return (-0.5 * K * np.power(R, 2) * np.log(1 - ((np.sqrt(rij2) - r0) / R)**2))
#nb.jit()
def total_energy(coord):
# Non-bonded energy.
e_nb = 0.0
for i in range(N):
for j in range(i - 1):
ri = coord[i]
rj = coord[j]
rij = ri - rj
rij2 = np.dot(rij, rij)
if (rij2 < rcutoff_sq):
e_nb += lj(rij2)
# Bonded FENE potential energy.
e_bond = 0.0
for i in range(1, N):
ri = coord[i]
rj = coord[i - 1] # Can be [i+1] ??
rij = ri - rj
rij2 = np.dot(rij, rij)
e_bond += fene(rij2)
return e_nb + e_bond
#nb.jit()
def move(coord):
trial = np.ndarray.copy(coord)
for i in range(1, N):
while True:
delta = (2 * np.random.rand(3) - 1) * max_delta
trial[i] += delta
#while True:
if trial[i,2] > 0.0:
break
trial[i] -= delta
return trial
#nb.jit()
def accept(delta_e):
beta = 1.0 / T
if delta_e < 0.0:
return True
random_number = np.random.rand(1)
p_acc = np.exp(-beta * delta_e)
if random_number < p_acc:
return True
return False
if __name__ == "__main__":
# FENE potential parameters.
K = 40.0
R = 0.3
r0 = 0.7
# L-J potential parameters
sigma = 0.5716
epsilon = 1.0
# MC parameters
N = 20 # Numbers of monomers
rcutoff = 2.5 * sigma
rcutoff_sq = rcutoff * rcutoff
max_delta = 0.01
n_steps = 100000
T = 10
# MAIN PART OF THE CODE
coord = gen_chain(N)
energy_current = total_energy(coord)
traj = open('2GN_20_T_10.xyz', 'w')
traj_txt = open('2GN_20_T_10.txt', 'w')
for step in range(n_steps):
if step % 1000 == 0:
traj.write(str(N) + '\n\n')
for i in range(N):
traj.write("C %10.5f %10.5f %10.5f\n" % (coord[i][0], coord[i][1], coord[i][2]))
traj_txt.write("%10.5f %10.5f %10.5f\n" % (coord[i][0], coord[i][1], coord[i][2]))
print(step, energy_current)
coord_trial = move(coord)
energy_trial = total_energy(coord_trial)
delta_e = energy_trial - energy_current
if accept(delta_e):
coord = coord_trial
energy_current = energy_trial
traj.close()
I except the chain of particles to collapse into a globule.
There is some problem with the logic of the MC you are implementing.
To perform a MC you need to ATTEMPT a move, evaluate the energy of the new state and then accept/reject according to a random number.
In your code there is not the slightest sign of the attempt to move a particle.
You need to move one (or more of them), evaluate the energy, and then update your coordinates.
By the way, I suppose this is not your entire code. There are many parameters that are not defined like the "k" and the "R0" in your fene potential
The FENE potential models bond interactions. What your code is saying is that all particles within the cutoff are bonded by FENE springs, and that the bonds are not fixed but rather defined by the cutoff. With a r_cutoff = 3.0, larger than equilibrium distance of the LJ well, you are essentially considering that each particle is bonded to potentially many others. You are treating the FENE potential as a non-bonded one.
For the bond interactions you should ignore the cutoff and only evaluate the energy for the actual pairs that are bonded according to your topology, which means that first you need to define a topology. I suggest generating a linear molecule of N atoms in a box big enough to contain the whole stretched molecule, and consider the i-th atom as bonded to the (i-1)-th atom, with i = 2, ..., N. In this way the topology is well defined and persistent. Then consider both interactions separately, non-bonded and bond, and add them at the end.
Something like this, in pseudo-code:
e_nb = 0
for particle i = 1 to N:
for particle j = 1 to i-1:
if (dist(i, j) < rcutoff):
e_nb += lj(i, j)
e_bond = 0
for particle i = 2 to N:
e_bond += fene(i, i-1)
e_tot = e_nb + e_bond
Below you can find a modified version of your code. To make things simpler, in this version there is no box and no boundary conditions, just a chain in free space. The chain is initialized as a linear sequence of particles each distant 80% of R0 from the next, since R0 is the maximum length of the FENE bond. The code considers that particle i is bonded with i+1 and the bond is not broken. This code is just a proof of concept.
#!/usr/bin/python
import numpy as np
def gen_chain(N, R):
x = np.linspace(0, (N-1)*R*0.8, num=N)
y = np.zeros(N)
z = np.zeros(N)
return np.column_stack((x, y, z))
def lj(rij2):
sig_by_r6 = np.power(sigma/rij2, 3)
sig_by_r12 = np.power(sig_by_r6, 2)
lje = 4.0 * epsilon * (sig_by_r12 - sig_by_r6)
return lje
def fene(rij2):
return (-0.5 * K * R0**2 * np.log(1-(rij2/R0**2)))
def total_energy(coord):
# Non-bonded
e_nb = 0
for i in range(N):
for j in range(i-1):
ri = coord[i]
rj = coord[j]
rij = ri - rj
rij2 = np.dot(rij, rij)
if (rij2 < rcutoff):
e_nb += lj(rij2)
# Bonded
e_bond = 0
for i in range(1, N):
ri = coord[i]
rj = coord[i-1]
rij = ri - rj
rij2 = np.dot(rij, rij)
e_bond += fene(rij2)
return e_nb + e_bond
def move(coord):
trial = np.ndarray.copy(coord)
for i in range(N):
delta = (2.0 * np.random.rand(3) - 1) * max_delta
trial[i] += delta
return trial
def accept(delta_e):
beta = 1.0/T
if delta_e <= 0.0:
return True
random_number = np.random.rand(1)
p_acc = np.exp(-beta*delta_e)
if random_number < p_acc:
return True
return False
if __name__ == "__main__":
# FENE parameters
K = 40
R0 = 1.5
# LJ parameters
sigma = 1.0
epsilon = 1.0
# MC parameters
N = 50 # number of particles
rcutoff = 3.5
max_delta = 0.01
n_steps = 10000000
T = 1.5
coord = gen_chain(N, R0)
energy_current = total_energy(coord)
traj = open('traj.xyz', 'w')
for step in range(n_steps):
if step % 1000 == 0:
traj.write(str(N) + '\n\n')
for i in range(N):
traj.write("C %10.5f %10.5f %10.5f\n" % (coord[i][0], coord[i][1], coord[i][2]))
print(step, energy_current)
coord_trial = move(coord)
energy_trial = total_energy(coord_trial)
delta_e = energy_trial - energy_current
if accept(delta_e):
coord = coord_trial
energy_current = energy_trial
traj.close()
The code prints the current configuration at each step, you can just load it up on VMD and see how it behaves. The bonds will not show correctly at first on VMD, you must use a bead representation for the particles and define the bonds manually or with a script within VMD. In any case, you don't need to see the bonds to notice that the chain does not collapse.
Please bear in mind that if you want to simulate a chain at a certain density, you need to be careful to generate the correct topology. I recommend the EMC package to efficiently generate polymers at the desired thermodynamic conditions. It is by no means a trivial problem, especially for larger chains.
By the way, your code had an error in the FENE energy evaluation. rij2 is already squared, you squared it again.
Below you can see how the total energy as a function of the number of steps behaves for T = 1.0, N = 20, rcutoff = 3.5, and also the last current configuration after 10 thousand steps.
And below for N = 50, T = 1.5, max_delta = 0.01, K = 40, R = 1.5, rcutoff = 3.5, and 10 million steps. This is the last current configuration.
The full "trajectory", which isn't really a trajectory since this is MC, you can find here (it's under 6 MB).
I am new to programming, so I hope my stupid questions do not bug you.
I am now trying to calculate the poisson sphere distribution(a 3D version of the poisson disk) using python and then plug in the result to POV-RAY so that I can generate some random distributed packing rocks.
I am following these two links:
[https://github.com/CodingTrain/Rainbow-Code/blob/master/CodingChallenges/CC_33_poisson_disc/sketch.js#L13]
[https://www.cs.ubc.ca/~rbridson/docs/bridson-siggraph07-poissondisk.pdf]
tl;dr
0.Create an n-dimensional grid array and cell size = r/sqrt(n) where r is the minimum distance between each sphere. All arrays are set to be default -1 which stands for 'without point'
1.Create an initial sample. (it should be placed randomly but I choose to put it in the middle). Put it in the grid array. Also, intialize an active array. Put the initial sample in the active array.
2.While the active list is not empty, pick a random index. Generate points near it and make sure the points are not overlapping with nearby points(only test with the nearby arrays). If no sample can be created near the 'random index', kick the 'random index' out. Loop the process.
And here is my code:
import math
from random import uniform
import numpy
import random
radius = 1 #you can change the size of each sphere
mindis = 2 * radius
maxx = 10 #you can change the size of the container
maxy = 10
maxz = 10
k = 30
cellsize = mindis / math.sqrt(3)
nrofx = math.floor(maxx / cellsize)
nrofy = math.floor(maxy / cellsize)
nrofz = math.floor(maxz / cellsize)
grid = []
active = []
default = numpy.array((-1, -1, -1))
for fillindex in range(nrofx * nrofy * nrofz):
grid.append(default)
x = uniform(0, maxx)
y = uniform(0, maxy)
z = uniform(0, maxz)
firstpos = numpy.array((x, y, z))
firsti = maxx // 2
firstj = maxy // 2
firstk = maxz // 2
grid[firsti + nrofx * (firstj + nrofy * firstk)] = firstpos
active.append(firstpos)
while (len(active) > 0) :
randindex = math.floor(uniform(0,len(active)))
pos = active[randindex]
found = False
for attempt in range(k):
offsetx = uniform(mindis, 2 * mindis)
offsety = uniform(mindis, 2 * mindis)
offsetz = uniform(mindis, 2 * mindis)
samplex = offsetx * random.choice([1,-1])
sampley = offsety * random.choice([1,-1])
samplez = offsetz * random.choice([1,-1])
sample = numpy.array((samplex, sampley, samplez))
sample = numpy.add(sample, pos)
xcoor = math.floor(sample.item(0) / cellsize)
ycoor = math.floor(sample.item(1) / cellsize)
zcoor = math.floor(sample.item(2) / cellsize)
attemptindex = xcoor + nrofx * (ycoor + nrofy * zcoor)
if attemptindex >= 0 and attemptindex < nrofx * nrofy * nrofz and numpy.all([sample, default]) == True and xcoor > 0 and ycoor > 0 and zcoor > 0 :
test = True
for testx in range(-1,2):
for testy in range(-1, 2):
for testz in range(-1, 2):
testindex = (xcoor + testx) + nrofx * ((ycoor + testy) + nrofy * (zcoor + testz))
if testindex >=0 and testindex < nrofx * nrofy * nrofz :
neighbour = grid[testindex]
if numpy.all([neighbour, sample]) == False:
if numpy.all([neighbour, default]) == False:
distance = numpy.linalg.norm(sample - neighbour)
if distance > mindis:
test = False
if test == True and len(active)<len(grid):
found = True
grid[attemptindex] = sample
active.append(sample)
if found == False:
del active[randindex]
for printout in range(len(grid)):
print("<" + str(active[printout][0]) + "," + str(active[printout][1]) + "," + str(active[printout][2]) + ">")
print(len(grid))
My code seems to run forever.
Therefore I tried to add a print(len(active)) in the last of the while loop.
Surprisingly, I think I discovered the bug as the length of the active list just keep increasing! (It is supposed to be the same length as the grid) I think the problem is caused by the active.append(), but I can't figure out where is the problem as the code is literally the 90% the same as the one made by Mr.Shiffman.
I don't want to free ride this but I have already checked again and again while correcting again and again for this code :(. Still, I don't know where the bug is. (why do the active[] keep appending!?)
Thank you for the precious time.
I want to find pattern in some spectra.
Spectrum image
Pattern should look like 2 in gray circles on picture, all data looks similarly. Light blue line is the original data, dotted dark blue line - average over 6 points. I was trying to do window with some size and scan data and check whether the y-flux value drops/rise below 60 ish % but that seems to find other regions and the one that I want, or only this I don't want.
The width of pattern is not always the same in spectra that I have. There is a picture of spectrum with pattern black dashed line but my program didn't found it.
not found picture
I tried changing size of window but it doesn't help. Can I use some pattern recognition algorithm to find this patterns? Could somebody point me in some direction? Or explain in easy way since I'm kinda lost in this, please?
That's my code:
import numpy as np
import matplotlib.pyplot as plt
from astropy.io import ascii
import glob
def reading(file_name):
data = ascii.read(file_name)
lam = data['col0'][1:-1]
#data offset *10**17 + 5
flux = data['col1'][1:-1]*10**17 + 5
return lam, flux
def percentChange(startPoint,currentPoint):
return abs(((currentPoint-startPoint)/startPoint))*100.00
def window(data, size):
n = len(data)
out = []
wind = data[0 : size]
i = size
while i + size/2 < n:
wind = data[i - size/2 : i + size/2]
tmp = percentChange(wind[0], wind[-1])
if tmp > 50.:
out.append([tmp, i - size/2, i + size/2])
i = i + size
return out
def window2(data, size):
n = len(data)
out = []
wind = data[0 : size]
i = size
while i + size/2 < n:
wind = data[i - size/2 : i + size/2]
tmp = percentChange(wind[0], wind[len(wind)/2])
if tmp > 50.:
out.append([tmp, i - size/2, i + size/2])
i = i + size
return out
def plotting(lamb, flux):
plt.rcParams['font.family'] = 'freeserif'
plt.rcParams['font.size'] = 12
plt.rcParams['axes.labelsize'] = 15
plt.rcParams['xtick.labelsize'] = 12
plt.rcParams['ytick.labelsize'] = 12
plt.rcParams['legend.fontsize'] = 12
plt.rcParams['figure.titlesize'] = 12
plt.rcParams['xtick.minor.visible'] = True
plt.rcParams['ytick.minor.visible'] = True
plt.plot(lamb, flux)
plt.xlabel("wavelenght [A]")
plt.ylabel("flux [erg/cm^2/s/A]")
def averaging(lamb, flux, param):
end = 1480
bin_flux_1 = [np.mean(flux[i : i + param]) for i in range(0, end, param)]
bin_lam_1 = [np.mean(lamb[i : i + param]) for i in range(0, end, param)]
return bin_lam_1, bin_flux_1
def main():
param = 6
stack = 6
for name in glob.glob('TRAIN/*.dat'):
print name
lamb, flux = reading(name)
lamb_a, flux_a = averaging(lamb, flux, param)
plotting(lamb, flux)
plotting(lamb_a, flux_a)
change = window(flux_a, stack)
change2 = window2(flux_a, stack)
minim = flux_a.index(min(flux_a))
for i in range(len(change)):
plt.axvline(lamb_a[change[i][1]], color='r', linestyle='--',linewidth=1)
plt.axvline(lamb_a[change[i][2]], color='r', linestyle='--',linewidth=1)
for i in range(len(change2)):
plt.axvline(lamb_a[change2[i][1]], color='y', linestyle='-',linewidth=1)
plt.axvline(lamb_a[change2[i][2]], color='y', linestyle='-',linewidth=1)
plt.axvline(lamb_a[minim], color='k', linestyle='--',linewidth=1)
plt.show()
if __name__ == "__main__":
main()
You can do it by using Knuth–Morris–Pratt algorithm in linear O(n + m) time complexity where n and m are the lengths of text and pattern.
KMP algorithm is basically a pattern matching algorithm (finding the starting position of a needle in haystack) which works on character string.
def kmp_matcher(t, d):
n=len(t)
m=len(d)
pi = compute_prefix_function(d)
q = 0
i = 0
while i < n:
if d[q]==t[i]:
q=q+1
i = i + 1
else:
if q != 0:
q = pi[q-1]
else:
i = i + 1
if q == m:
print "pattern occurs with shift "+str(i-q)
q = pi[q-1]
def compute_prefix_function(p):
m=len(p)
pi =range(m)
k=1
l = 0
while k < m:
if p[k] <= p[l]:
l = l + 1
pi[k] = l
k = k + 1
else:
if l != 0:
l = pi[l-1]
else:
pi[k] = 0
k = k + 1
return pi
t = 'brownfoxlazydog'
p = 'lazy'
kmp_matcher(t, p)