从2个列表中查找相关系数

Question

我正在开发一个项目，当给出几个数据列表时，它有许多功能。我已经把这些列表分开了，我已经定义了一些我知道正确工作的函数，它们是一个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，正确的标准偏差，但错误的相关系数。我的数学可以在这里吗？我需要只用Python的标准库找到相关系数。

我的代码：

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

SAMPLE RUN：

我有一份[1,2,3,4,4,8]和[3,3,4,5,8,9]

的清单

我期望r = 0.8848的正确答案，但得到r = .203727

编辑：包含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

Pearson Correlation Coefficient

代码（已修改）

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 2

Pearson相关性可以用numpy＆＃39; corrcoef来计算。

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

Answer 3

您的标准偏差是错误的。你忘了带广场。 您实际上是返回方差而不是该函数的标准偏差。 @DeathPox

Answer 4

通常根据标准偏差公式，你应该在sqrrt it.Right之前将dev除以样本数（列表的长度）。 我的意思是： dev + =（（someList [i] -listMean）** 2）/ len（someList）

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