Consider a simple code running inside a while loop:
while m_diff >= convergence_lim and paramC <= M_lim:
paramC = temp_M
try:
paramB, M, P = some_function(paramA, paramB, paramC)
except:
temp_M+= 0.001
continue
totalM0 = paramB['Mo']
totalM1 = sum(M) # Sum of all M
m_diff = abs(totalM0 - totalM1)
temp_M+= 0.001
This would run forever and iterate each 0.001 sample until m_diff is acceptable. I would like to use scipy.optimize to arrive at the numbers faster.
The issue is with the paramB as the function some_function regenerates a different value of paramB. The optimizer should look for best paramC that minimizes m_diff.
I tried:
import scipy.optimize
def mass_diff_func(paramC, paramB):
paramA = 10 #Constant
try:
paramB, M, P = some_function(paramA, paramB, paramC)
except:
return float('inf'), paramB
totalM0 = paramB['Mo']
totalM1 = sum(M)
m_diff = abs(totalM0 - totalM1)
return m_diff, paramB
result = scipy.optimize.minimize(mass_diff_func, paramC, args=(paramB,), bounds=[(convergence_lim, M_lim)])
optimal_paramC = result.x
But changes to paramB doesn't seem to be carried to minimize function. Or minimize function doesn't register changes to paramB, either way any idea how to fix this?
Open to other optimizers as well.
Related
I am learning some Python basics and we had this code about wrapper functions and how they can be used to add functionality to wrapped functions. So we calculate time that was used to compute two different functions:
import time
def timer(fn):
def wrapper():
t0 = time.time()
result = fn()
dt = time.time() - t0
return dt
return wrapper
def pow_2():
return 10000000 ** 2
def in_build_pow():
return pow(10000000, 2)
pow_2 = timer(pow_2)
in_build_pow = timer(in_build_pow)
a, b = 0, 0
N = 100
for i in range(N):
a += pow_2()
b += in_build_pow()
print("a = ", a)
print("b = ", b)
print(f"Average time pow_2 = {a / N:.10f}")
print(f"Average time in_build_pow = {b / N:.10f}")
But my result looks like this:
Unless I increase the iteration count N to more than 1000. The higher N, the more often results are not 0.0.
This is N = 1000:
And this is N = 10,000:
Does anyone knows why it is like this? Why I cannot count time for fewer iterations? The code works fine in online compilers, though. But VSCode, VS2022, and PyCharm all have same issues for me.
Some OS has problems calculating very short time intervals with those functions. So its better to use perf_counter_ns() method of time module. Me personally had problems on Win10, while MacOS worked perfectly fine with time() method.
I am struggling to understand how scipy.solve_ivp() handles errors in a system of ODE. Lets say I have the following, simple code for a single ODEs, and I think I might be doing things wrong in some way. Lets say my rhs looks something like:
from scipy.integrate import solve_ivp
def rhs_func(t, y):
z = 1.0/( x - y + 1j )
return z
Suppose we call solve_ivp with the following signature:
Z_solution = ivp_adaptive.solve_ivp( fun = rhs_func,
t_span = [100,0],
y0 = y0, #some initial value of 0 for example
method ='RK45',
t_eval = None,
args = some_additional_arguments_to_rhs_func,
dense_output = False,
rtol = 1e-8
atol = 1e-10
)
Now, the absolute and relative tolerances are supposed to fix the error of the caculation. The problem I am having has to do with the "t_eval=None" in this case. Apparently, this choice lets the integrator (in this case of type RK45) to choose the time step according to the specified tolerances above being or not being exceeded, i.e., the steps are not fixed, but somehow taking a larger step in t would mean a solution has been found that lies below the tolerances above (atol=1e-10 , rtol=1e-8). This is particularly useful in problems with large variations of the time scale, where a uniform discretization of t is very inefficient.
My big problem has to do with the following piece of code in scipy.integrate._ivp.solve_ivp() around line 575, with the "t_eval == None" case:
while status is None:
message = solver.step()
if solver.status == 'finished':
status = 0
elif solver.status == 'failed':
status = -1
break
t_old = solver.t_old
t = solver.t
y = solver.y
if dense_output:
sol = solver.dense_output()
interpolants.append(sol)
else:
sol = None
if events is not None:
g_new = [event(t, y) for event in events]
active_events = find_active_events(g, g_new, event_dir)
if active_events.size > 0:
if sol is None:
sol = solver.dense_output()
root_indices, roots, terminate = handle_events(
sol, events, active_events, is_terminal, t_old, t)
for e, te in zip(root_indices, roots):
t_events[e].append(te)
y_events[e].append(sol(te))
if terminate:
status = 1
t = roots[-1]
y = sol(t)
g = g_new
# HERE I HAVE MODIFIED THE FILE BY CALLING AN INTERPOLATION FUNCTION FOR THE SOLUTION
if t_eval is None:
ts.append(t)
#ys.append(y)
# this calls to adapt the solution to a new set of values x over which y(x,t) is
# defined
interp_solution(t,y,solver,args)
y = solver.y
ys.append(y)
where I have defined a function:
def interp_solution( t, y, solver, args ):
import numpy as np
from scipy import interpolate
x_old = args.get_old_grid() # this call just returns an array of the style of
# x_new, and is where y is defined
x_new = np.linspace( -t, t, dim ) # the new array where components of y are
# defined
y_interp = interpolate.interp1d( x_old, y )
y_new = y_interp( x_new )
solver.y = y_new # update the solver y
# finally, we change the maximum allowed step of the integrator if t is below
# some threshold value
if ( t < args.get_threshold() ):
solver.max_step = #some number
return y_new
When I look at the results, it seems that this is very sensitive to the tolerances and the way the integration steps are performed, but somehow I fail to see where errors could come from in this approach -- can anyone explain if this approach is somehow affecting the solution and the associated errors ? How can one implement a similar approach in this fashion? Any help is greatly appreaciated.
I was doing a benchmark for myself that I encountered this interesting thing. I am trying to get the first 30 keys of a dictionary, and I have written three ways to get it as follows:
import time
dic = {str(i): i for i in range(10 ** 6)}
start_time = time.time()
x = list(dic.keys())[0:30]
print(time.time() - start_time)
start_time = time.time()
y = list(dic.keys())
x = y[0:30]
print(time.time() - start_time)
start_time = time.time()
z = dic.keys()
y = list(z)
x = y[0:30]
print(time.time() - start_time)
The results are:
0.015970945358276367
0.010970354080200195
0.01691460609436035
Surprisingly, the second method is much faster! Any thoughts on this?
Using Python's timeit module to measure various alternatives. I added mine which doesn't convert the keys to list:
from timeit import timeit
dic = {str(i): i for i in range(10 ** 6)}
def f1():
x = list(dic.keys())[0:30]
return x
def f2():
y = list(dic.keys())
x = y[0:30]
return x
def f3():
z = dic.keys()
y = list(z)
x = y[0:30]
return x
def f4():
x = [k for _, k in zip(range(30), dic.keys())]
return x
t1 = timeit(lambda: f1(), number=10)
t2 = timeit(lambda: f2(), number=10)
t3 = timeit(lambda: f3(), number=10)
t4 = timeit(lambda: f4(), number=10)
print(t1)
print(t2)
print(t3)
print(t4)
Prints:
0.1911074290110264
0.20418328599771485
0.18727918600779958
3.5186996683478355e-05
Maybe this is due to inaccuracies in your measure of time. You can use timeit for doing this kind of things:
import timeit
dic = {str(i): i for i in range(10 ** 6)}
# 27.5125/29.0836/26.8525
timeit.timeit("x = list(dic.keys())[0:30]", number=1000, globals={"dic": dic})
# 28.6648/26.4684/30.9534
timeit.timeit("y = list(dic.keys());x=y[0:30]", number=1000)
# 31.7345/29.5301/30.7541
timeit.timeit("z=dic.keys();y=list(z);x=y[0:30]", number=1000, globals={'dic': dic})
The comments show the times I got when running the same code 3 different times. As you can see, even by performing a large number of repetitions, it is possible to obtain quite large variations in time measured. This can be due to several different things:
An item can be in the cache of your processor or not.
Your processor can be occupied doing several other things.
Etc...
As stated by #Andrej Kesely, your bottleneck is due to the fact that you cast your dictionary keys into a list. By doing so, Python goes through the entire dictionary keys, because that's how it converts something to a list generally. Hence, by avoiding this, you can get much better results.
I'm trying to write some code to compute mean, Variance, Standard Deviation, FWHM, and finally evaluate the Gaussian Integral. I've been running into a division by zero error that I can't get past and I would like to know the solution for this ?
Where it's throwing an error I've tried to throw an exception handler as follows
Average = (sum(yvalues)) / (len(yvalues)) try: return (sum(yvalues) / len(yvalues))
expect ZeroDivisionError:
return 0
xvalues = []
yvalues = []
def generate():
for i in range(0,300):
a = rand.uniform((float("-inf") , float("inf")))
b = rand.uniform((float("-inf") , float("inf")))
xvalues.append(i)
### Defining the variable 'y'
y = a * (b + i)
yvalues.append(y) + 1
def mean():
Average = (sum(yvalues))/(len(yvalues))
print("The average is", Average)
return Average
def varience():
# This calculates the SD and the varience
s = []
for i in yvalues:
z = i - mean()
z = (np.abs(i-z))**2
s.append(y)**2
t = mean()
v = numpy.sqrt(t)
print("Answer for Varience is:", v)
return v
Traceback (most recent call last):
File "Tuesday.py", line 42, in <module>
def make_gauss(sigma=varience(), mu=mean(), x = random.uniform((float("inf"))*-1, float("inf"))):
File "Tuesday.py", line 35, in varience
t = mean()
File "Tuesday.py", line 25, in mean
Average = (sum(yvalues))/(len(yvalues))
ZeroDivisionError: division by zero
There are a few things that are not quite right as people noted above.
import random
import numpy as np
def generate():
xvalues, yvalues = [], []
for i in range(0,300):
a = random.uniform(-1000, 1000)
b = random.uniform(-1000, 1000)
xvalues.append(i)
### Defining the variable 'y'
y = a * (b + i)
yvalues.append(y)
return xvalues, yvalues
def mean(yvalues):
return sum(yvalues)/len(yvalues)
def variance(yvalues):
# This calculates the SD and the varience
s = []
yvalues_mean = mean(yvalues)
for y in yvalues:
z = (y - yvalues_mean)**2
s.append(z)
t = mean(s)
return t
def variance2(yvalues):
yvalues_mean = mean(yvalues)
return sum( (y-yvalues_mean)**2 for y in yvalues) / len(yvalues)
# Generate the xvalues and yvalues
xvalues, yvalues = generate()
# Now do the calculation, based on the passed parameters
mean_yvalues = mean(yvalues)
variance_yvalues = variance(yvalues)
variance_yvalues2 = variance2(yvalues)
print('Mean {} variance {} {}'.format(mean_yvalues, variance_yvalues, variance_yvalues2))
# Using Numpy
np_mean = np.mean(yvalues)
np_var = np.var(yvalues)
print('Numpy: Mean {} variance {}'.format(np_mean, np_var))
The way variance was calculated isn't quite right, but given the comment of "SD and variance" you were probably going to calculate both.
The code above gives 2 (well, 3) ways to do what I understand you were trying to do but I changed a few of the methods to clean them up a bit. generate() returns two lists now. mean() returns the mean, etc. The function variance2() gives an alternative way to calculate the variance but using a list comprehension style.
The last couple of lines are an example using numpy which has all of it built in and, if available, is a great way to go.
The one part that wasn't clear was the random.uniform(float("-inf"), float("inf"))) which seems to be an error (?).
You are calling mean before you call generate.
This is obvious since yvalues.append(y) + 1 (in generate) would have caused another error (TypeError) since .append returns None and you can't add 1 to None.
Change yvalues.append(y) + 1 to yvalues.append(y + 1) and then make sure to call generate before you call mean.
Also notice that you have the same error in varience (which should be called variance, btw). s.append(y)**2 should be s.append(y ** 2).
Another error you have is that the stacktrace shows make_gauss(sigma=varience(), mu=mean(), x = random.uniform((float("inf"))*-1, float("inf"))).
I'm pretty sure you don't actually want to call varience and mean on this line, just reference them. So also change that line to make_gauss(sigma=varience, mu=mean, x = random.uniform((float("inf"))*-1, float("inf")))
I've never used incremental PCA which exists in sklearn and I'm a bit confused about it's parameters and not able to find a good explanation of them.
I see that there is batch_size in the constructor, but also, when using partial_fit method you can again pass only a part of your data, I've found the following way:
n = df.shape[0]
chunk_size = 100000
iterations = n//chunk_size
ipca = IncrementalPCA(n_components=40, batch_size=1000)
for i in range(0, iterations):
ipca.partial_fit(df[i*chunk_size : (i+1)*chunk_size].values)
ipca.partial_fit(df[iterations*chunk_size : n].values)
Now, what I don't understand is the following - when using partial fit, does the batch_size play any role at all, or not? And how are they related?
Moreover, if both are considered, how should I change their values properly, when wanting to increase the precision while increasing memory footprint (and the other way around, decrease the memory consumption for the price of decreased accuracy)?
The docs say:
batch_size : int or None, (default=None)
The number of samples to use for each batch. Only used when calling fit...
This param is not used within partial_fit, where the batch-size is controlled by the user.
Bigger batches will increase memory-consumption, smaller ones will decrease it.
This is also written in the docs:
This algorithm has constant memory complexity, on the order of batch_size, enabling use of np.memmap files without loading the entire file into memory.
Despite some checks and parameter-heuristics, the whole fit-function looks like this:
for batch in gen_batches(n_samples, self.batch_size_):
self.partial_fit(X[batch], check_input=False)
Here is some an incremental PCA code based on https://github.com/kevinhughes27/pyIPCA which is an implementation of CCIPCA method.
import scipy.sparse as sp
import numpy as np
from scipy import linalg as la
import scipy.sparse as sps
from sklearn import datasets
class CCIPCA:
def __init__(self, n_components, n_features, amnesic=2.0, copy=True):
self.n_components = n_components
self.n_features = n_features
self.copy = copy
self.amnesic = amnesic
self.iteration = 0
self.mean_ = None
self.components_ = None
self.mean_ = np.zeros([self.n_features], np.float)
self.components_ = np.ones((self.n_components,self.n_features)) / \
(self.n_features*self.n_components)
def partial_fit(self, u):
n = float(self.iteration)
V = self.components_
# amnesic learning params
if n <= int(self.amnesic):
w1 = float(n+2-1)/float(n+2)
w2 = float(1)/float(n+2)
else:
w1 = float(n+2-self.amnesic)/float(n+2)
w2 = float(1+self.amnesic)/float(n+2)
# update mean
self.mean_ = w1*self.mean_ + w2*u
# mean center u
u = u - self.mean_
# update components
for j in range(0,self.n_components):
if j > n: pass
elif j == n: V[j,:] = u
else:
# update the components
V[j,:] = w1*V[j,:] + w2*np.dot(u,V[j,:])*u / la.norm(V[j,:])
normedV = V[j,:] / la.norm(V[j,:])
normedV = normedV.reshape((self.n_features, 1))
u = u - np.dot(np.dot(u,normedV),normedV.T)
self.iteration += 1
self.components_ = V / la.norm(V)
return
def post_process(self):
self.explained_variance_ratio_ = np.sqrt(np.sum(self.components_**2,axis=1))
idx = np.argsort(-self.explained_variance_ratio_)
self.explained_variance_ratio_ = self.explained_variance_ratio_[idx]
self.components_ = self.components_[idx,:]
self.explained_variance_ratio_ = (self.explained_variance_ratio_ / \
self.explained_variance_ratio_.sum())
for r in range(0,self.components_.shape[0]):
d = np.sqrt(np.dot(self.components_[r,:],self.components_[r,:]))
self.components_[r,:] /= d
You can test it with
import pandas as pd, ccipca
df = pd.read_csv('iris.csv')
df = np.array(df)[:,:4].astype(float)
pca = ccipca.CCIPCA(n_components=2,n_features=4)
S = 10
print df[0, :]
for i in range(150): pca.partial_fit(df[i, :])
pca.post_process()
The resulting eigenvectors / values will not exaactly be the same as the batch PCA. Results are approximate, but they are useful.