列表的标准偏差

问题:列表的标准偏差

我想找到几个(Z)列表的第一,第二,…个数字的均值和标准差。例如,我有

A_rank=[0.8,0.4,1.2,3.7,2.6,5.8]
B_rank=[0.1,2.8,3.7,2.6,5,3.4]
C_Rank=[1.2,3.4,0.5,0.1,2.5,6.1]
# etc (up to Z_rank )...

现在,我要获取的均值和std *_Rank[0],的均值和std *_Rank[1]
(即:所有(A..Z)_rank列表中第一个数字
的均值和std;来自的第二个数字的均值和std所有(A..Z)_rank列表;
第三个数字的均值和std …;等等)。

I want to find mean and standard deviation of 1st, 2nd,… digits of several (Z) lists. For example, I have

A_rank=[0.8,0.4,1.2,3.7,2.6,5.8]
B_rank=[0.1,2.8,3.7,2.6,5,3.4]
C_Rank=[1.2,3.4,0.5,0.1,2.5,6.1]
# etc (up to Z_rank )...

Now I want to take the mean and std of *_Rank[0], the mean and std of *_Rank[1], etc.
(ie: mean and std of the 1st digit from all the (A..Z)_rank lists;
the mean and std of the 2nd digit from all the (A..Z)_rank lists;
the mean and std of the 3rd digit…; etc).


回答 0

从Python 3.4 / PEP450开始statistics module,标准库中提供了一个,该库提供了一种stdev用于计算像您这样的可迭代对象的标准偏差的方法

>>> A_rank = [0.8, 0.4, 1.2, 3.7, 2.6, 5.8]
>>> import statistics
>>> statistics.stdev(A_rank)
2.0634114147853952

Since Python 3.4 / PEP450 there is a statistics module in the standard library, which has a method stdev for calculating the standard deviation of iterables like yours:

>>> A_rank = [0.8, 0.4, 1.2, 3.7, 2.6, 5.8]
>>> import statistics
>>> statistics.stdev(A_rank)
2.0634114147853952

回答 1

我将A_Rank等人放入二维NumPy数组中,然后使用numpy.mean()numpy.std()计算均值和标准差:

In [17]: import numpy

In [18]: arr = numpy.array([A_rank, B_rank, C_rank])

In [20]: numpy.mean(arr, axis=0)
Out[20]: 
array([ 0.7       ,  2.2       ,  1.8       ,  2.13333333,  3.36666667,
        5.1       ])

In [21]: numpy.std(arr, axis=0)
Out[21]: 
array([ 0.45460606,  1.29614814,  1.37355985,  1.50628314,  1.15566239,
        1.2083046 ])

I would put A_Rank et al into a 2D NumPy array, and then use numpy.mean() and numpy.std() to compute the means and the standard deviations:

In [17]: import numpy

In [18]: arr = numpy.array([A_rank, B_rank, C_rank])

In [20]: numpy.mean(arr, axis=0)
Out[20]: 
array([ 0.7       ,  2.2       ,  1.8       ,  2.13333333,  3.36666667,
        5.1       ])

In [21]: numpy.std(arr, axis=0)
Out[21]: 
array([ 0.45460606,  1.29614814,  1.37355985,  1.50628314,  1.15566239,
        1.2083046 ])

回答 2

这是一些纯Python代码,可用于计算均值和标准差。

以下所有代码均基于statisticsPython 3.4+中的模块。

def mean(data):
    """Return the sample arithmetic mean of data."""
    n = len(data)
    if n < 1:
        raise ValueError('mean requires at least one data point')
    return sum(data)/n # in Python 2 use sum(data)/float(n)

def _ss(data):
    """Return sum of square deviations of sequence data."""
    c = mean(data)
    ss = sum((x-c)**2 for x in data)
    return ss

def stddev(data, ddof=0):
    """Calculates the population standard deviation
    by default; specify ddof=1 to compute the sample
    standard deviation."""
    n = len(data)
    if n < 2:
        raise ValueError('variance requires at least two data points')
    ss = _ss(data)
    pvar = ss/(n-ddof)
    return pvar**0.5

注意:为提高浮点求和时的准确性,该statistics模块使用了自定义函数,_sum而不是sum我使用的内置函数。

现在我们有例如:

>>> mean([1, 2, 3])
2.0
>>> stddev([1, 2, 3]) # population standard deviation
0.816496580927726
>>> stddev([1, 2, 3], ddof=1) # sample standard deviation
0.1

Here’s some pure-Python code you can use to calculate the mean and standard deviation.

All code below is based on the statistics module in Python 3.4+.

def mean(data):
    """Return the sample arithmetic mean of data."""
    n = len(data)
    if n < 1:
        raise ValueError('mean requires at least one data point')
    return sum(data)/n # in Python 2 use sum(data)/float(n)

def _ss(data):
    """Return sum of square deviations of sequence data."""
    c = mean(data)
    ss = sum((x-c)**2 for x in data)
    return ss

def stddev(data, ddof=0):
    """Calculates the population standard deviation
    by default; specify ddof=1 to compute the sample
    standard deviation."""
    n = len(data)
    if n < 2:
        raise ValueError('variance requires at least two data points')
    ss = _ss(data)
    pvar = ss/(n-ddof)
    return pvar**0.5

Note: for improved accuracy when summing floats, the statistics module uses a custom function _sum rather than the built-in sum which I’ve used in its place.

Now we have for example:

>>> mean([1, 2, 3])
2.0
>>> stddev([1, 2, 3]) # population standard deviation
0.816496580927726
>>> stddev([1, 2, 3], ddof=1) # sample standard deviation
0.1

回答 3

在Python 2.7.1中,您可以使用numpy.std()以下方法计算标准差:

  • 人口标准:仅使用numpy.std()数据列表之外的其他参数即可。
  • 示例std:您需要将ddof(即Delta自由度)设置为1,如以下示例所示:

numpy.std(<您的列表>,ddof = 1

计算中使用的除数为N-ddof,其中N表示元素数。默认情况下,ddof为零。

它计算样本std而不是总体std。

In Python 2.7.1, you may calculate standard deviation using numpy.std() for:

  • Population std: Just use numpy.std() with no additional arguments besides to your data list.
  • Sample std: You need to pass ddof (i.e. Delta Degrees of Freedom) set to 1, as in the following example:

numpy.std(< your-list >, ddof=1)

The divisor used in calculations is N – ddof, where N represents the number of elements. By default ddof is zero.

It calculates sample std rather than population std.


回答 4

在python 2.7中,您可以使用NumPy numpy.std()给出总体标准差

在Python 3.4中statistics.stdev()返回样本标准偏差。该pstdv()功能是一样的numpy.std()

In python 2.7 you can use NumPy’s numpy.std() gives the population standard deviation.

In Python 3.4 statistics.stdev() returns the sample standard deviation. The pstdv() function is the same as numpy.std().


回答 5

使用python,以下是几种方法:

import statistics as st

n = int(input())
data = list(map(int, input().split()))

方法1-使用功能

stdev = st.pstdev(data)

方法2:计算方差并求平方根

variance = st.pvariance(data)
devia = math.sqrt(variance)

方法3:使用基本数学

mean = sum(data)/n
variance = sum([((x - mean) ** 2) for x in X]) / n
stddev = variance ** 0.5

print("{0:0.1f}".format(stddev))

注意:

  • variance 计算样本总体的方差
  • pvariance 计算整个人口的方差
  • 相似的差异stdevpstdev

Using python, here are few methods:

import statistics as st

n = int(input())
data = list(map(int, input().split()))

Approach1 – using a function

stdev = st.pstdev(data)

Approach2: calculate variance and take square root of it

variance = st.pvariance(data)
devia = math.sqrt(variance)

Approach3: using basic math

mean = sum(data)/n
variance = sum([((x - mean) ** 2) for x in X]) / n
stddev = variance ** 0.5

print("{0:0.1f}".format(stddev))

Note:

  • variance calculates variance of sample population
  • pvariance calculates variance of entire population
  • similar differences between stdev and pstdev

回答 6

纯python代码:

from math import sqrt

def stddev(lst):
    mean = float(sum(lst)) / len(lst)
    return sqrt(float(reduce(lambda x, y: x + y, map(lambda x: (x - mean) ** 2, lst))) / len(lst))

pure python code:

from math import sqrt

def stddev(lst):
    mean = float(sum(lst)) / len(lst)
    return sqrt(float(reduce(lambda x, y: x + y, map(lambda x: (x - mean) ** 2, lst))) / len(lst))

回答 7

其他答案涵盖了如何在python中充分执行std dev,但没有人解释如何进行您所描述的怪异遍历。

我将假设AZ是整个人口。如果没有,请参阅Ome关于如何从样本推断的答案。

因此,要获得每个列表的第一位数字的标准差/均值,您将需要如下所示:

#standard deviation
numpy.std([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])

#mean
numpy.mean([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])

为了缩短代码并将其通用化为第n个数字,请使用我为您生成的以下函数:

def getAllNthRanks(n):
    return [A_rank[n], B_rank[n], C_rank[n], D_rank[n], E_rank[n], F_rank[n], G_rank[n], H_rank[n], I_rank[n], J_rank[n], K_rank[n], L_rank[n], M_rank[n], N_rank[n], O_rank[n], P_rank[n], Q_rank[n], R_rank[n], S_rank[n], T_rank[n], U_rank[n], V_rank[n], W_rank[n], X_rank[n], Y_rank[n], Z_rank[n]] 

现在,您可以像这样简单地从AZ获取所有n个位置的stdd和均值:

#standard deviation
numpy.std(getAllNthRanks(n))

#mean
numpy.mean(getAllNthRanks(n))

The other answers cover how to do std dev in python sufficiently, but no one explains how to do the bizarre traversal you’ve described.

I’m going to assume A-Z is the entire population. If not see Ome‘s answer on how to inference from a sample.

So to get the standard deviation/mean of the first digit of every list you would need something like this:

#standard deviation
numpy.std([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])

#mean
numpy.mean([A_rank[0], B_rank[0], C_rank[0], ..., Z_rank[0]])

To shorten the code and generalize this to any nth digit use the following function I generated for you:

def getAllNthRanks(n):
    return [A_rank[n], B_rank[n], C_rank[n], D_rank[n], E_rank[n], F_rank[n], G_rank[n], H_rank[n], I_rank[n], J_rank[n], K_rank[n], L_rank[n], M_rank[n], N_rank[n], O_rank[n], P_rank[n], Q_rank[n], R_rank[n], S_rank[n], T_rank[n], U_rank[n], V_rank[n], W_rank[n], X_rank[n], Y_rank[n], Z_rank[n]] 

Now you can simply get the stdd and mean of all the nth places from A-Z like this:

#standard deviation
numpy.std(getAllNthRanks(n))

#mean
numpy.mean(getAllNthRanks(n))