Argthtjtr

So i'm relatively new to python but i have used loops in previous programs i have made but they were relatively simple.
I was wondering if there was a method that involved a loop or some other way to make this iteration more concise.

Basically is there a way for me to have a variable called 't' that holds values for t1,t2 and t3 in like a list which is then used in 'eq' and 'der' instead of copying and pasting the same piece of code?

I have already tried to look for an answer on here and other places but the solutions i have found so far don't seem to work with what i have got/i'm not python literate enough to understand them.

Thanks in advance.

import numpy as np

U235_Decay_Constant = 9.72e-10

U238_Decay_Constant = 1.54e-10





t0 = 4.1e9





eq = ((np.exp(U238_Decay_Constant*t0)-1)-(0.0167* 

((np.exp(U235_Decay_Constant*t0)-1)))-0.0094)

der = (U238_Decay_Constant*(np.exp(U238_Decay_Constant*t0)))-(0.0167* 

(U235_Decay_Constant*(np.exp(U235_Decay_Constant*t0))))



t1 = t0 - (eq/der)





eq = ((np.exp(U238_Decay_Constant*t1)-1)-(0.0167* 

((np.exp(U235_Decay_Constant*t1)-1)))-0.0094)

der = (U238_Decay_Constant*(np.exp(U238_Decay_Constant*t1)))-(0.0167* 

(U235_Decay_Constant*(np.exp(U235_Decay_Constant*t1))))



t2 = t1 - (eq/der)



eq = ((np.exp(U238_Decay_Constant*t2)-1)-(0.0167* 

((np.exp(U235_Decay_Constant*t2)-1)))-0.0094)

der = (U238_Decay_Constant*(np.exp(U238_Decay_Constant*t2)))-(0.0167* 

(U235_Decay_Constant*(np.exp(U235_Decay_Constant*t2))))



t3 = t2 - (eq/der)

print(t3)

Might be a little easier to read as:

import numpy as np

U235_Decay_Constant = 9.72e-10

U238_Decay_Constant = 1.54e-10





t = [4.1e9, None, None, None]



t[0] = 4.1e9

for i in range(3):

    eq = ((np.exp(U238_Decay_Constant*t[i])-1)-(0.0167*

    ((np.exp(U235_Decay_Constant*t[i])-1)))-0.0094)



    der = (U238_Decay_Constant*(np.exp(U238_Decay_Constant*t[0])))-(0.0167*

    (U235_Decay_Constant*(np.exp(U235_Decay_Constant*t[0]))))



    t[i+1] = t[i] - (eq/der)



print(t[3])

Yes, iteration can help here. Add your values to a list, then t? is the last value in the list so far; replacing your t? references with t[-1] gives:

t = [4.1e9]

for _ in range(3):

    eq = (

        (np.exp(U238_Decay_Constant * t[-1]) - 1)

        - (0.0167 * ((np.exp(U235_Decay_Constant * t[-1]) - 1)))

        - 0.0094

    )

    der = (U238_Decay_Constant * (np.exp(U238_Decay_Constant * t[-1]))) - (

        0.0167 * (U235_Decay_Constant * (np.exp(U235_Decay_Constant * t[-1])))

    )

    t.append(t[-1] - (eq / der))

The general principle is one of accumulation, where you produce the running output of a repeated application of a function. So the itertools.accumulate() function could help here too:

from itertools import accumulate, chain, repeat



def u238_decay(t, _):

    eq = (

        (np.exp(U238_Decay_Constant * t) - 1)

        - (0.0167 * ((np.exp(U235_Decay_Constant * t) - 1)))

        - 0.0094

    )

    der = (U238_Decay_Constant * (np.exp(U238_Decay_Constant * t))) - (

        0.0167 * (U235_Decay_Constant * (np.exp(U235_Decay_Constant * t)))

    )

    return t - (eq / der)



series = accumulate(chain([4.1e9], repeat(None)), u238_decay)

The above produces an unending series of decay values:

>>> series = accumulate(chain([4.1e9], repeat(None)), u238_decay)

>>> next(series)

4100000000.0

>>> next(series)

4081406102.7439713

>>> next(series)

4081163259.5641546

>>> next(series)

4081163218.6509323

>>> next(series)

4081163218.650931

You could look into creating a numpy universal function so you can do the same with the numpy.ufunc.accumulate() method.

However, I suspect that your formula can be re-cast to not depend on the previous input, but only as a formula of starting amount and t as time, at which point you can use full numpy vectorised calculations.