Is there are more pythonic way to write a while loop that only updates a variable?

Question

I have this while loop, and I was wondering if their is a more pythonic way to write it:

k = 1
while np.sum(s[0:k]) / s_sum < retained_variance:
    k += 1

s is a numpy vector. Thanks!

Answer 1

I'd say it's pretty pythonic: explicit is better than implicit.

Answer 2

Probably not the most efficient solution, but fast if most of the array needs to be searched:

import numpy as np

ss = np.cumsum(s)  # array with cumulative sum
k = ss.searchsorted(retained_variance*s_sum) # exploit that ss is monotonically increasing

EDIT: Simon pointed out that

k = np.cumsum(s).searchsorted(retained_variance*s_sum) + 1

is the value which corresponds to the question.

Answer 3

The last line of the following code will find the value of k in one line:

import numpy as np
import math

s = np.array([1,2,3,4,5,6,7,8,9])
s_sum = 1
retained_variance = 4.3

k = 1
while np.sum(s[0:k]) / s_sum < retained_variance:
    k += 1

print (k)

print (np.ceil(np.interp(retained_variance,s.cumsum()/s_sum,range(1,len(s)+1))))

Answer 4

this is more like Haskell than python:

>>> from itertools import count, dropwhile
>>> pred = lambda k: np.sum(s[:k]) / s_sum < retained_variance
>>> head = next  # just to look more like haskell
>>> head(dropwhile(pred, count()))

---

edit: this will be more efficient:

>>> from itertools import dropwhile, accumulate
>>> pred = lambda x: x[1] / s_sum < retained_variance
>>> head = next
>>> head(dropwhile(pred, enumerate(accumulate(s), start=1)))[0]