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What is the best way to efficiently compute the teager energy kurtosis using list comprehension?

I would like to calculate the Teager Energy Kurtosis in a function in Python 3.8. I think this should also work with list comprehension.
I tried it with the following code, but I get an error message that the numpy object is not iterable. The variable data contains a list with measured values from an accelerometer.
def EO(data):
numerator = pow(len(data),2)*sum((pow(((pow(data[i+1],2) - pow(data[i],2))-(sum(pow(data[i+1],2) - pow(data[i],2))/len(data))),4)) for i in range(len(data)-1))
denominator = pow(sum(pow(((pow(data[i+1],2) - pow(data[i],2))-(sum(pow(data[i+1],2) - pow(data[i],2))/len(data))),2) for i in range(len(data)-1)),2)
energy_operator = numerator/denominator
return energy_operator
What is the general approach for implementing such formulas where you have to iterate multiple times, also of course with regard to efficiency. The dataset from which the values are to be calculated contains 133329 entries.
I guess the main problem is that the sum of the denominator contains another sum which has to be formed first. How to do that ?. Without list comprehension I would iterate through the whole dataset twice with a for loop to first get the average value and with that calculate the rest in the second iteration. The readability of this is then of course gone.
Any suggestions are welcome !
Cheers,
Gerrit
EDIT:
This is the working code without using list comprehension:
def EO_5(data):
summe = 0
num_sum = 0
den_sum = 0
for i in range(1,len(data)-1):
summe += pow(data[i],2)-((data[i-1])*(data[i+1]))
ave = summe/len(data)
for i in range(1,len(data)-1):
num_sum += pow((pow(data[i],2)-((data[i-1])*(data[i+1])))-ave,4)
den_sum += pow((pow(data[i],2)-((data[i-1])*(data[i+1])))-ave,2)
numerator = (len(data)-1)*num_sum
denominator = pow(den_sum,2)
return numerator/denominator

sum(pow(data[i+1],2) - pow(data[i],2))
I think that's the (a) problem. The argument to sum is basically just a number, when it should be something list-like (iterable).
The other problem is that this is badly over-golfed. Lines that run on that long are frowned on, etc.
The other other problem is that the math expressed in the two code-blocks you've shared don't seem to match. The first, which isn't working, seems to more closely follow what's in the image you linked, but IDK if that means it's "correct". Do you have a better reference for "Teager Energy Kurtosis".
I haven't tested this in any way, but it's pretty much how I'd simplify the code you said is working.
def EO_5(data):
n = len(data) - 1)
deltas = tuple(
pow(x, 2) - (before * after)
for (before, x, after)
in zip(data[:-2], data[1:-1], data[2:])
)
ave = sum(deltas) / len(data)
num_sum = sum(pow(d - ave, 4) for d in deltas)
den_sum = sum(pow(d - ave, 2) for d in deltas)
numerator = n * num_sum
denominator = pow(den_sum, 2)
return numerator / denominator
If you're having problems with performance, you may be able to get numpy to leverage vector operations to make this even more streamlined, but I have limited experience with that.

Here it is.
def EO_5(data):
ave = (sum([i**2 for i in data[1:-1]])-sum([i*j for i,j in zip(data[:-2],data[2:])]))/len(data)
num = (sum([(j**2-i*k-ave)**4 for i,j,k in zip(data[:-2],data[1:-1],data[2:])]))*(len(data)-1)
den = (sum([(j**2-i*k-ave)**2 for i,j,k in zip(data[:-2],data[1:-1],data[2:])]))**2
return num/den

Project Euler #25 - Can the performance be improved further?

The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
a = 1
b = 1
i = 2
while(1):
c = a + b
i += 1
length = len(str(c))
if length == 1000:
print(i)
break
a = b
b = c
I got the answer(works fast enough). Just looking if there's a better way for this question

If you've answered the question, you'll find plenty of explanations on answers in the problem thread. The solution you posted is pretty much okay. You may get a slight speedup by simply checking that your c>=10^999 at every step instead of first converting it to a string.
The better method is to use the fact that when the Fibonacci numbers become large enough, the Fibonacci numbers converge to round(phi**n/(5**.5)) where phi=1.6180... is the golden ratio and round(x) rounds x to the nearest integer. Let's consider the general case of finding the first Fibonacci number with more than m digits. We are then looking for n such that round(phi**n/(5**.5)) >= 10**(m-1)
We can easily solve that by just taking the log of both sides and observe that
log(phi)*n - log(5)/2 >= m-1 and then solve for n.
If you're wondering "well how do I know that it has converged by the nth number?" Well, you can check for yourself, or you can look online.
Also, I think questions like these either belong on the Code Review SE or the Computer Science SE. Even Math Overflow might be a good place for Project Euler questions, since many are rooted in number theory.

Your solution is completely fine for #25 on project euler. However, if you really want to optimize for speed here you can try to calculate fibonacci using the identities I have written about in this blog post: https://sloperium.github.io/calculating-the-last-digits-of-large-fibonacci-numbers.html
from functools import lru_cache
#lru_cache(maxsize=None)
def fib4(n):
if n <= 1:
return n
if n % 2:
m = (n + 1) // 2
return fib4(m) ** 2 + fib4(m - 1) ** 2
else:
m = n // 2
return (2 * fib4(m - 1) + fib4(m)) * fib4(m)
def binarySearch( length):
first = 0
last = 10**5
found = False
while first <= last and not found:
midpoint = (first + last) // 2
length_string = len(str(fib4(midpoint)))
if length_string == length:
return midpoint -1
else:
if length < length_string:
last = midpoint - 1
else:
first = midpoint + 1
print(binarySearch(1000))
This code tests about 12 times faster than your solution. (it does require an initial guess about max size though)

How do I implement summation and array iteration correctly based on Pseudo code. PYTHON Relaxation Method

I am trying to implement relaxation iterative solver for a project. The function we create should intake two inputs: Matrix A, and Vector B, and should return iterative vectors X that Approximate solution Ax = b.
Pseudo Code from the book is here:
enter image description here
I am new to Python so I am struggling quite a bit with implementing this method. Here is my code:
def SOR_1(A,b):
k=1
n = len(A)
xo = np.zeros_like(b)
x = np.zeros_like(b)
omega = 1.25
while (k <= N):
for i in range(n-1):
x[i] = (1.0-omega)*xo[i] + (1.0/A[i][i])[omega(-np.sum(A[i][j]*x[j]))
-np.sum(A[i][j]*xo[j] + b[i])]
if ( np.linalg.norm(x - xo) < 1e-9):
print (x)
k = k + 1.0
for i in range(n-1):
xo[i] = x[i]
return x
My question is how do I implement the for loop and generating the arrays correctly based off of the Pseudo Code.

Welcome to Python!
Variables in Python are case sensitive so n is defined but N is not defined. If they are supposed to be different variables, I don't see what your value is for N.
You are off to a good start but the following line is still psuedocode for the most part:
x[i] = (1.0-omega)*xo[i] + (1.0/A[i][i])[omega(-np.sum(A[i][j]*x[j]))
-np.sum(A[i][j]*xo[j] + b[i])]
In the textbook's pseudocode square brackets are being used as a grouping symbol but in Python, they are reserved for creating and accessing lists (which is what python calls arrays). Also, there is no implicit multiplication in Python so you have to write things like (1 + 2)*(3*(4+5)) rather than (1 + 2)[3(4+5)]
The other major issue is that you don't define j. You probably need a for loop that would either look like:
for j in range(1, i):
or if you want to do it inline
sum(A[i][j]*x[j] for j in range(1, i))
Note that range has two arguments, where to start and what value to stop before so range(1, i) is equivalent to the summation from 1 to i - 1
I think you are struggling with that line because there's far too much going on in that line. See if you can figure out parts of it using separate variables or offload some of the work to separate functions.
something like: x[i] =a + b * c * d() - e() but give a,b c, d and e meaningful names. You'd then have to correctly set each variable and define each function but at least you are trying to solve separate problems rather than one huge complex one.
Also, make sure you have your tabs correct. k = k + 1.0 should not be inside that for loop, just inside the while loop.
Coding is an iterative process. First get the while loop working. Don't try to do anything in it (except print out the variable so you can see that it is working). Next get the for loop working inside the while loop (again, just printing the variables). Next get (1.0-omega)*xo[i] working. Along the way, you'll discover and solve issues such as (1.0-omega)*xo[i] will evaluate to 0 because xo is a NumPy list initiated with all zeros.
You'd start with something like:
k = 1
N = 3
n = 3
xo = [1, 2, 3]
while (k <= N):
for i in range(n):
print(k, i)
omega = 1.25
print((1.0-omega)*xo[i])
k += 1
And slowly work more and more of the relaxation solver in until you have everything working.

Python Pi approximation

So I have to approximate Pi with following way: 4*(1-1/3+1/5-1/7+1/9-...). Also it should be based on number of iterations. So the function should look like this:
>>> piApprox(1)
4.0
>>> piApprox(10)
3.04183961893
>>> piApprox(300)
3.13825932952
But it works like this:
>>> piApprox(1)
4.0
>>> piApprox(10)
2.8571428571428577
>>> piApprox(300)
2.673322240709928
What am I doing wrong? Here is the code:
def piApprox(num):
pi=4.0
k=1.0
est=1.0
while 1<num:
k+=2
est=est-(1/k)+1/(k+2)
num=num-1
return pi*est

This is what you're computing:
4*(1-1/3+1/5-1/5+1/7-1/7+1/9...)
You can fix it just by adding a k += 2 at the end of your loop:
def piApprox(num):
pi=4.0
k=1.0
est=1.0
while 1<num:
k+=2
est=est-(1/k)+1/(k+2)
num=num-1
k+=2
return pi*est
Also the way you're counting your iterations is wrong since you're adding two elements at the time.
This is a cleaner version that returns the output that you expect for 10 and 300 iterations:
def approximate_pi(rank):
value = 0
for k in xrange(1, 2*rank+1, 2):
sign = -(k % 4 - 2)
value += float(sign) / k
return 4 * value
Here is the same code but more compact:
def approximate_pi(rank):
return 4 * sum(-float(k%4 - 2) / k for k in xrange(1, 2*rank+1, 2))

Important edit:
whoever expects this approximation to yield PI -- quote from Wikipedia:
It converges quite slowly, though – after 500,000 terms, it produces
only five correct decimal digits of π
Original answer:
This is an educational example. You try to use a shortcut and attempt to implement the "oscillating" sign of the summands by handling two steps for k in the same iteration. However, you adjust k only by one step per iteration.
Usually, in math at least, an oscillating sign is achieved with (-1)**i. So, I have chosen this for a more readable implementation:
def pi_approx(num_iterations):
k = 3.0
s = 1.0
for i in range(num_iterations):
s = s-((1/k) * (-1)**i)
k += 2
return 4 * s
As you can see, I have changed your approach a bit, to improve readability. There is no need for you to check for num in a while loop, and there is no particular need for your pi variable. Your est actually is a sum that grows step by step, so why not call it s ("sum" is a built-in keyword in Python). Just multiply the sum with 4 in the end, according to your formula.
Test:
>>> pi_approx(100)
3.1514934010709914
The convergence, however, is not especially good:
>>> pi_approx(100) - math.pi
0.009900747481198291
Your expected output is flaky somehow, because your piApprox(300) (should be 3.13825932952, according to your) is too far away from PI. How did you come up with that? Is that possibly affected by an accumulated numerical error?
Edit
I would not trust the book too much in regard of what the function should return after 10 and 300 iterations. The intermediate result, after 10 steps, should be rather free of numerical errors, indeed. There, it actually makes a difference whether you take two steps of k at the same time or not. So this most likely is the difference between my pi_approx(10) and the books'. For 300 iterations, numerical error might have severely affected the result in the book. If this is an old book, and they have implemented their example in C, possibly using single precision, then a significant portion of the result may be due to accumulation of numerical error (note: this is a prime example for how bad you can be affected by numerical errors: a repeated sum of small and large values, it does not get worse!).
What counts is that you have looked at the math (the formula for PI), and you have implemented a working Python version of approximating that formula. That was the learning goal of the book, so go ahead and tackle the next problem :-).

def piApprox(num):
pi=4.0
k=3.0
est=1.0
while 1<num:
est=est-(1/k)+1/(k+2)
num=num-1
k+=4
return pi*est
Also for real task use math.pi

Here is a slightly simpler version:
def pi_approx(num_terms):
sign = 1. # +1. or -1.
pi_by_4 = 1. # first term
for div in range(3, 2 * num_terms, 2): # 3, 5, 7, ...
sign = -sign # flip sign
pi_by_4 += sign / div # add next term
return 4. * pi_by_4
which gives
>>> for n in [1, 10, 300, 1000, 3000]:
... print(pi_approx(n))
4.0
3.0418396189294032
3.1382593295155914
3.140592653839794
3.1412593202657186

While all of these answers are perfectly good approximations, if you are using the Madhava-Leibniz Series than you should arrive at ,"an approximation of π correct to 11 decimal places as 3.14159265359" within in first 21 terms according to this website: https://en.wikipedia.org/wiki/Approximations_of_%CF%80
Therefore, a more accurate solution could be any variation of this:
import math
def estimate_pi(terms):
ans = 0.0
for k in range(terms):
ans += (-1.0/3.0)**k/(2.0*k+1.0)
return math.sqrt(12)*ans
print(estimate_pi(21))
Output: 3.141592653595635

Solving functional equations programmatically

Given:
F(F(n)) = n
F(F(n + 2) + 2) = n
F(0) = 1
where n is a non-negative integer. F(129) = ?
How can we solve such kind of functional equations programmatically? My programming language of choice is Python.

Functional equations, in their most general terms, are really really hard. It is no coincidence that pretty much every international mathematics competition features one of them, usually looking about as innocent as the one you've written. The methods of solving them vary from plain induction to infinite-dimensional Banach space analysis and a generic programming approach to solving them is very unlikely.
In this particular case, here's a straight-forward approach:
Suppose for any two integers m, n we have F(m) = F(n) = k. But then m = F(F(m)) = F(k) = F(F(n)) = n : therefore m = n and F never takes the same value on two different inputs. But we know that F(F(n)) = n = F(F(n+2)+2) - therefore F(n) and F(n+2)+2 must be the same number - which is to say, F(n+2) == F(n) - 2 == F(n-2) - 4 = ... . Now we know F(0) = 1, so F(1) = F(F(0)) = 0. But then F(129) = F(127) - 2 = F(125) - 4 = ... = F(1) - 128 = -128
So there is your solution - but a mechanical algorithm for solving any variation just doesn't exist.

Based on what #ivancho and #aaronasterling have said, I have been able to write this program that should do the trick:
def f(n):
if not n:
return 1
else:
return f(n-2)-2
>>> f(4)
-3
Comment if this is not what you're after