从 2 个列表中查找相关系数

Question

我正在开发一个项目，该项目在给定几个数据列表时具有许多功能。我已经分离了列表，并且定义了一些我知道可以正常工作的函数，即平均函数和标准差函数。我的问题是在测试我的列表时，我得到了正确的均值、正确的标准差，但相关系数不正确。我的数学可以在这里关闭吗？我只需要用 Python 的标准库找到相关系数。

我的密码：

def correlCo(someList1, someList2):

    # First establish the means and standard deviations for both lists.
    xMean = mean(someList1)
    yMean = mean(someList2)
    xStandDev = standDev(someList1)
    yStandDev = standDev(someList2)
    zList1 = []
    zList2 = []

    # Create 2 new lists taking (a[i]-a's Mean)/standard deviation of a
    for x in someList1:
        z1 = ((float(x)-xMean)/xStandDev)
        zList1.append(z1)

    for y in someList2:
        z2 = ((float(y)-yMean)/yStandDev)
        zList2.append(z2)

    # Mapping out the lists to be float values instead of string     
    zList1 = list(map(float,zList1))
    zList2 = list(map(float,zList2))
    # Multiplying each value from the lists
    zFinal = [a*b for a,b in zip(zList1,zList2)]
    totalZ = 0
    # Taking the sum of all the products
    for a in zFinal:
        totalZ += a
    # Finally calculating correlation coefficient
    r = (1/(len(someList1) - 1)) * totalZ

    return r

样品运行：

我有一个 [1,2,3,4,4,8] 和 [3,3,4,5,8,9] 的列表

我期望 r = 0.8848 的正确答案，但得到 r = .203727

编辑：包括我所做的均值和标准差函数。

def mean(someList):
    total = 0
    for a in someList:
        total += float(a)
    mean = total/len(someList)
    return mean

def standDev(someList):
    newList = []
    sdTotal = 0
    listMean = mean(someList)
    for a in someList:
        newNum = (float(a) - listMean)**2
        newList.append(newNum)

    for z in newList:
        sdTotal += float(z)

    standardDeviation = sdTotal/(len(newList))

    return standardDeviation

Answer 1

皮尔逊相关系数可以用numpy的corrcoef来计算。

import numpy
numpy.corrcoef(list1, list2)[0, 1]

Answer 2

皮尔逊相关系数

代码（修改）

def mean(someList):
    total = 0
    for a in someList:
        total += float(a)
    mean = total/len(someList)
    return mean
def standDev(someList):
    listMean = mean(someList)
    dev = 0.0
    for i in range(len(someList)):
        dev += (someList[i]-listMean)**2
    dev = dev**(1/2.0)
    return dev
def correlCo(someList1, someList2):

    # First establish the means and standard deviations for both lists.
    xMean = mean(someList1)
    yMean = mean(someList2)
    xStandDev = standDev(someList1)
    yStandDev = standDev(someList2)
    # r numerator
    rNum = 0.0
    for i in range(len(someList1)):
        rNum += (someList1[i]-xMean)*(someList2[i]-yMean)

    # r denominator
    rDen = xStandDev * yStandDev

    r =  rNum/rDen
    return r

print(correlCo([1,2,3,4,4,8], [3,3,4,5,8,9]))

输出

0.884782972876

Answer 3

你的标准差是错误的。你忘了取平方根。 您实际上返回的是方差而不是该函数的标准偏差。 @死亡痘

Answer 4

通常根据标准差公式，你应该在sqrrt之前将dev除以样本数（列表的长度）。对吧？ 我是说： dev += ((someList[i]-listMean)**2)/len(someList)

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