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This is a script I have written to calculate the population standard deviation. I feel that this can be simplified and also be made more pythonic.

from math import sqrt

def mean(lst):
    """calculates mean"""
    sum = 0
    for i in range(len(lst)):
        sum += lst[i]
    return (sum / len(lst))

def stddev(lst):
    """calculates standard deviation"""
    sum = 0
    mn = mean(lst)
    for i in range(len(lst)):
        sum += pow((lst[i]-mn),2)
    return sqrt(sum/len(lst)-1)

numbers = [120,112,131,211,312,90]

print stddev(numbers)
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The easiest way to make mean() more pythonic is to use the sum() built-in function.

def mean(lst):
    return sum(lst) / len(lst)

Concerning your loops on lists, you don't need to use range(). This is enough:

for e in lst:
   sum += e

Other comments:

  • You don't need parentheses around the return value (check out PEP 8 when you have a doubt about this).
  • Your docstrings are useless: it's obvious from the name that it calculates the mean. At least make them more informative ("returns the mean of lst").
  • Why do you use "-1" in the return for stddev? Is that a bug?
  • You are computing the standard deviation using the variance: call that "variance", not sum!
  • You should type pow(e-mn,2), not pow((e-mn),2). Using parentheses inside a function call could make the reader think he's reading a tuple (eg. pow((e,mn),2) is valid syntax)
  • You shouldn't use pow() anyway, ** is enough.

This would give:

def stddev(lst):
    """returns the standard deviation of lst"""
    variance = 0
    mn = mean(lst)
    for e in lst:
        variance += (e-mn)**2
    variance /= len(lst)

    return sqrt(variance)

It's still way too verbose! Since we're handling lists, why not using list comprehensions?

def stddev(lst):
    """returns the standard deviation of lst"""
    mn = mean(lst)
    variance = sum([(e-mn)**2 for e in lst]) / len(lst)
    return sqrt(variance)

This is not perfect. You could add tests using doctest. Obviously, you should not code those functions yourself, except in a small project. Consider using Numpy for a bigger project.

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  • \$\begingroup\$ Thank you Cygal for your answer. I realize things like tests and validation need to be added, but I think you put me in the right direction. \$\endgroup\$ – Animesh Feb 22 '12 at 8:23
  • \$\begingroup\$ @mad, I realize you're not able to comment due to your reputation, but if you see a problem in a post and want to fix it, you'll either have to be patient and wait until you have 50 reputation or go out, answer a question, and get five upvotes (or ask a good question and get 10). Please don't try to circumvent the system. Third-party edits should only edit the content of the post (as opposed to formatting, grammar, spelling, pasting in content from links etc.) with explicit approval from the poster. \$\endgroup\$ – Nic Hartley Dec 30 '15 at 20:09
  • \$\begingroup\$ Looks like you forgot to divide the variance by N before taking the sqrt in the last/least verbose example. \$\endgroup\$ – Cody A. Ray Sep 29 '16 at 16:41
  • \$\begingroup\$ @CodyA.Ray Your Rev 2 corrected the result, but it was not the right fix. \$\endgroup\$ – 200_success Sep 29 '16 at 19:19
  • \$\begingroup\$ @200_success can you elaborate? Yeah, variance is the wrong variable name there. I could've just divided in the "return" line. But the equation seems correct for non-sampled std dev: libweb.surrey.ac.uk/library/skills/Number%20Skills%20Leicester/… \$\endgroup\$ – Cody A. Ray Sep 29 '16 at 22:09
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You have some serious calculation errors…


Assuming that this is Python 2, you also have bugs in the use of division: if both operands of / are integers, then Python 2 performs integer division. Possible remedies are:

(Assuming that this is Python 3, you can just use statistics.stdev().


The formula for the sample standard deviation is

$$ s = \sqrt{\frac{\sum_{i=1}^{n}\ (x_i - \bar{x})^2}{n - 1}}$$

In return sqrt(sum/len(lst)-1), you have an error with the precedence of operations. It should be

return sqrt(float(sum) / (len(lst) - 1))
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  • \$\begingroup\$ Source for formula? \$\endgroup\$ – Agostino Mar 26 '15 at 23:40
  • \$\begingroup\$ @Agostino It's basically common knowledge in statistics. \$\endgroup\$ – 200_success Mar 26 '15 at 23:42

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