《集体智慧编程》中的这个 python 函数有什么问题?

发布于 2024-08-04 07:16:04 字数 924 浏览 4 评论 0 原文

这是有问题的函数。它计算 p1 和 p2 的皮尔逊相关系数,该系数应该是 -1 到 1 之间的数字。

当我将其与真实用户数据一起使用时,它有时会返回大于 1 的数字,如下例所示:

def sim_pearson(prefs,p1,p2):
    si={}
    for item in prefs[p1]: 
        if item in prefs[p2]: si[item]=1

    if len(si)==0: return 0

    n=len(si)

    sum1=sum([prefs[p1][it] for it in si])
    sum2=sum([prefs[p2][it] for it in si])

    sum1Sq=sum([pow(prefs[p1][it],2) for it in si])
    sum2Sq=sum([pow(prefs[p2][it],2) for it in si]) 

    pSum=sum([prefs[p1][it]*prefs[p2][it] for it in si])

    num=pSum-(sum1*sum2/n)
    den=sqrt((sum1Sq-pow(sum1,2)/n)*(sum2Sq-pow(sum2,2)/n))

    if den==0: return 0

    r=num/den

    return r

critics = {
    'user1':{
        'item1': 3,
        'item2': 5,
        'item3': 5,
        },

    'user2':{
        'item1': 4,
        'item2': 5,
        'item3': 5,
        }
}

print sim_pearson(critics, 'user1', 'user2', )

1.15470053838

This is the function in question. It calculates the Pearson correlation coefficient for p1 and p2, which is supposed to be a number between -1 and 1.

When I use this with real user data, it sometimes returns a number greater than 1, like in this example:

def sim_pearson(prefs,p1,p2):
    si={}
    for item in prefs[p1]: 
        if item in prefs[p2]: si[item]=1

    if len(si)==0: return 0

    n=len(si)

    sum1=sum([prefs[p1][it] for it in si])
    sum2=sum([prefs[p2][it] for it in si])

    sum1Sq=sum([pow(prefs[p1][it],2) for it in si])
    sum2Sq=sum([pow(prefs[p2][it],2) for it in si]) 

    pSum=sum([prefs[p1][it]*prefs[p2][it] for it in si])

    num=pSum-(sum1*sum2/n)
    den=sqrt((sum1Sq-pow(sum1,2)/n)*(sum2Sq-pow(sum2,2)/n))

    if den==0: return 0

    r=num/den

    return r

critics = {
    'user1':{
        'item1': 3,
        'item2': 5,
        'item3': 5,
        },

    'user2':{
        'item1': 4,
        'item2': 5,
        'item3': 5,
        }
}

print sim_pearson(critics, 'user1', 'user2', )

1.15470053838

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评论(4

七分※倦醒 2024-08-11 07:16:04

看起来您可能意外地使用了整数除法。我进行了以下更改,您的函数返回了 1.0

num=pSum-(1.0*sum1*sum2/n)
den=sqrt((sum1Sq-1.0*pow(sum1,2)/n)*(sum2Sq-1.0*pow(sum2,2)/n))

请参阅 PEP 238 了解有关 Python 中除法运算符的更多信息。修复上述代码的另一种方法是:

from __future__ import division

It looks like you may be unexpectedly using integer division. I made the following change and your function returned 1.0:

num=pSum-(1.0*sum1*sum2/n)
den=sqrt((sum1Sq-1.0*pow(sum1,2)/n)*(sum2Sq-1.0*pow(sum2,2)/n))

See PEP 238 for more information on the division operator in Python. An alternate way of fixing your above code is:

from __future__ import division
萌无敌 2024-08-11 07:16:04

好吧,我花了一分钟时间阅读代码,但似乎如果您将输入数据更改为浮点数,它就会起作用

Well it took me a minute to read over the code but it seems if you change your input data to floats it will work

晚风撩人 2024-08-11 07:16:04

整数除法令人困惑。如果你将 n 设置为浮点数,它就会起作用:

n=float(len(si))

Integer division is confusing it. It works if you make n a float:

n=float(len(si))
画中仙 2024-08-11 07:16:04

好吧,我无法完全找到您的函数中的逻辑有什么问题,所以我只是使用皮尔逊系数的定义重新实现它:

from math import sqrt

def sim_pearson(p1,p2):
    keys = set(p1) | set(p2)
    n = len(keys)

    a1 = sum(p1[it] for it in keys) / n
    a2 = sum(p2[it] for it in keys) / n

#    print(a1, a2)

    sum1Sq = sum((p1[it] - a1) ** 2 for it in keys)
    sum2Sq = sum((p2[it] - a2) ** 2 for it in keys) 

    num = sum((p1[it] - a1) * (p2[it] - a2) for it in keys)
    den = sqrt(sum1Sq * sum2Sq)

#    print(sum1Sq, sum2Sq, num, den)
    return num / den

critics = {
    'user1':{
        'item1': 3,
        'item2': 5,
        'item3': 5,
        },

    'user2':{
        'item1': 4,
        'item2': 5,
        'item3': 5,
        }
}

assert 0.999 < sim_pearson(critics['user1'], critics['user1']) < 1.0001

print('Your example:', sim_pearson(critics['user1'], critics['user2']))
print('Another example:', sim_pearson({1: 1, 2: 2, 3: 3}, {1: 4, 2: 0, 3: 1}))

请注意,在您的示例中,皮尔逊系数只是 1.0 因为向量 (-4/3, 2/3, 2/3) 和 (-2/3, 1/3, 1/3) 是平行的。

Well, I wasn't exactly able to find what's wrong with the logic in your function, so I just reimplemented it using the definition of Pearson coefficient:

from math import sqrt

def sim_pearson(p1,p2):
    keys = set(p1) | set(p2)
    n = len(keys)

    a1 = sum(p1[it] for it in keys) / n
    a2 = sum(p2[it] for it in keys) / n

#    print(a1, a2)

    sum1Sq = sum((p1[it] - a1) ** 2 for it in keys)
    sum2Sq = sum((p2[it] - a2) ** 2 for it in keys) 

    num = sum((p1[it] - a1) * (p2[it] - a2) for it in keys)
    den = sqrt(sum1Sq * sum2Sq)

#    print(sum1Sq, sum2Sq, num, den)
    return num / den

critics = {
    'user1':{
        'item1': 3,
        'item2': 5,
        'item3': 5,
        },

    'user2':{
        'item1': 4,
        'item2': 5,
        'item3': 5,
        }
}

assert 0.999 < sim_pearson(critics['user1'], critics['user1']) < 1.0001

print('Your example:', sim_pearson(critics['user1'], critics['user2']))
print('Another example:', sim_pearson({1: 1, 2: 2, 3: 3}, {1: 4, 2: 0, 3: 1}))

Note that in your example the Pearson coefficient is just 1.0 since vectors (-4/3, 2/3, 2/3) and (-2/3, 1/3, 1/3) are parallel.

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