Pearson相关函数回归南
Pearson相关函数回归南
提问于 2016-02-19 15:08:01
我正在学习“编程集体智能”中的练习，但我使用的是JavaScript。我的Pearson相关算法有点麻烦。以下是功能：
function rec(object1, object2) {
var sum1 = 0;
var sum2 = 0;
var squareSum1 = 0;
var squareSum2 = 0;
var productsSum = 0;
var i;
var commonKeys = commonProperties(object1, object2);
for (i = 0; i >= commonKeys.length; i += 1) {
sum1 += object1[commonKeys[i]];
sum2 += object2[commonKeys[i]];
squareSum1 += Math.pow(object1[commonKeys[i]], 2);
squareSum2 += Math.pow(object2[commonKeys[i]], 2);
productsSum += object1[commonKeys[i]] * object2[commonKeys[i]];
}
var num1 = productsSum - (sum1 * sum2 / commonKeys.length);
var num2 = Math.sqrt((squareSum1 - (Math.pow(sum1, 2) / commonKeys.length)) * (squareSum2 - (Math.pow(sum2, 2) / commonKeys.length)));
return num1 / num2;
}Full JSFiddle是这里。我已经遍历了JSLint，这就是为什么它可能有点混乱的原因。有人知道怎么回事吗？
回答 2
Stack Overflow用户
回答已采纳
发布于 2016-02-19 15:41:09
在条件"for“中，您犯了一个小错误，变量"i”永远不会超过commonKeys.length。
function rec(object1, object2) {
var sum1 = 0;
var sum2 = 0;
var squareSum1 = 0;
var squareSum2 = 0;
var productsSum = 0;
var i;
var commonKeys = commonProperties(object1, object2);
for (i = 0; i < commonKeys.length; i += 1) {
sum1 += object1[commonKeys[i]];
sum2 += object2[commonKeys[i]];
squareSum1 += Math.pow(object1[commonKeys[i]], 2);
squareSum2 += Math.pow(object2[commonKeys[i]], 2);
productsSum += object1[commonKeys[i]] * object2[commonKeys[i]];
}
var num1 = productsSum - (sum1 * sum2 / commonKeys.length);
var num2 = Math.sqrt((squareSum1 - (Math.pow(sum1, 2) / commonKeys.length)) * (squareSum2 - (Math.pow(sum2, 2) / commonKeys.length)));
return num1 / num2;
}https://jsfiddle.net/98uoy87u/2/
它工作得很好，给出"-1“作为答案。
再见。
Stack Overflow用户
发布于 2016-02-19 15:45:27
根据我的注释，您永远不会以i < commonKeys.length的形式输入for循环。更正此错误后，您的算法是正确的。我已经用rec(janeSmith, johnSmith)验证了这一点。Excel从脚本返回0.650791373，而从脚本返回0.6507913734559685。
function intersection_destructive(a, b) {
var result = [];
while (a.length > 0 && b.length > 0) {
if (a[0] < b[0]) {
a.shift();
} else if (a[0] > b[0]) {
b.shift();
} else /* they're equal */ {
result.push(a.shift());
b.shift();
}
}
return result;
}
function commonProperties(object1, object2) {
var keys1 = Object.keys(object1);
var keys2 = Object.keys(object2);
return intersection_destructive(keys1, keys2);
}
var johnSmith = {
"Zoolander": 2.5,
"Batman Begins": 3.5,
"Deadpool": 4.5,
"Thor": 1.5
};
var janeSmith = {
"Zoolander": 4.5,
"Batman Begins": 3,
"Deadpool": 5,
"Thor": 2.5,
"The Avengers": 4,
"The Internship": 2.5
};
var johnDoe = {
"Zoolander": 4,
"The Internship": 3,
"Batman Begins": 4.5,
"Thor": 5
};
function rec(object1, object2) {
var sum1 = 0;
var sum2 = 0;
var squareSum1 = 0;
var squareSum2 = 0;
var productsSum = 0;
var i;
var commonKeys = commonProperties(object1, object2);
for (i = 0; i < commonKeys.length; i += 1) {
sum1 += object1[commonKeys[i]];
sum2 += object2[commonKeys[i]];
squareSum1 += Math.pow(object1[commonKeys[i]], 2);
squareSum2 += Math.pow(object2[commonKeys[i]], 2);
productsSum += object1[commonKeys[i]] * object2[commonKeys[i]];
}
var num1 = productsSum - ((sum1 * sum2) / (commonKeys.length));
var num2 = Math.sqrt(((squareSum1 - (Math.pow(sum1, 2)) / commonKeys.length)) * ((squareSum2 - (Math.pow(sum2, 2)) / commonKeys.length)));
return num1 / num2;
}
var value = rec(janeSmith, johnSmith);
document.getElementById('value').innerHTML = value;<h1>
Value:
</h1>
<p id="value">
</p>
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/35508443
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