Question

Hello I am trying to calculate a bunch of standard deviations of a list about 20,000 values long. Here is an example of my code:

from statistics import stdev

def main():
    a = [x for x in range(0,20000)]
    b = []

    for x in range(2, len(a) + 2):
        b.append(stdev(a[:x]))

    print(b)

main()

This method is extremely slow and I am trying to figure out a way to make it more efficient. Any help is appreciated. Thank you.

[Done] exited with code=null in 820.376 seconds

Answer 1

看起来你想要一个不断扩展的标准偏差，为此我会使用pandas库和pandas.Series.expanding方法：

In [156]: main()[:5]
Out[156]: 
[0.7071067811865476,
 1.0,
 1.2909944487358056,
 1.5811388300841898,
 1.8708286933869707]

In [157]: pd.Series(range(20000)).expanding().std()[:5]
Out[157]: 
0         NaN
1    0.707107
2    1.000000
3    1.290994
4    1.581139
dtype: float64

您可以轻松切掉第一个元素并转换为列表，如果您愿意：

In [158]: pd.Series(range(20000)).expanding().std()[1:6].tolist()
Out[158]: 
[0.7071067811865476,
 1.0,
 1.2909944487358056,
 1.5811388300841898,
 1.8708286933869707]

虽然Series是一个更有用的数据类型，用于处理时间序列而不是列表，并且肯定更高效：

In [159]: %timeit pd.Series(range(20000)).expanding().std()
1.07 ms ± 30.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answer 2

您可以跟踪平方值和值的总和：

from math import sqrt

a = range(0,20000)

def sdevs(a):
    sds = [0]
    n = 1
    sum_x = a[0]
    sum_x_squared = a[0]**2

    for x in a[1:]:
        sum_x += x
        sum_x_squared += x**2
        n += 1
        # as noted by @Andrey Tyukin, statistics.stdev returns
        # the unbiased estimator, hence the n/(n-1)
        sd = sqrt(n/(n-1)*(sum_x_squared/n - (sum_x/n)**2))
        sds.append(sd)
    return sds

sds = sdevs(a)
print(sds[10000])
# 2887.184355042123

在10年前的PC上大约需要24毫秒。

More efficient way to calculate standard deviation of a large list in Python

2 个答案: