crossfilter中的reduceAdd,reduceSum,reduceRemove函数是什么？它们应该如何使用？

Question

简单来说有人能解释如何减少功能与它的参数reduceAdd,reduceSum,reduceRemove工程crossfilter？

Answer 1

请记住,map reduce会按特定维度的键减少数据集.例如,让我们使用带有记录的crossfilter实例:

[
    { name: "Gates",      age: 57,   worth: 72000000000, gender: "m" },
    { name: "Buffet",     age: 59,   worth: 58000000000, gender: "m" },
    { name: "Winfrey",    age: 83,   worth:  2900000000, gender: "f"   },
    { name: "Bloomberg",  age: 71,   worth: 31000000000, gender: "m"  },
    { name: "Walton",     age: 64,   worth: 33000000000, gender: "f"  },
]

和尺寸名称,年龄,价值和性别.我们将使用reduce方法减少性别维度.

首先,我们定义reduceAdd,reduceRemove和reduceInitial回调方法.

reduceInitial返回具有缩小对象和初始值形式的对象.它不需要参数.

function reduceInitial() {
    return {
        worth: 0,
        count: 0
    };
}

reduceAdd定义当记录被"过滤"为特定键的缩减对象时发生的情况.第一个参数是简化对象的瞬态实例.第二个对象是当前记录.该方法将返回增强的瞬态缩减对象.

function reduceAdd(p, v) {
    p.worth = p.worth + v.worth;
    p.count = p.count + 1;
    return p;
}

reduceRemove与reduceAdd(至少在这个例子中)相反.它采用相同的参数reduceAdd.这是必需的,因为在过滤记录时更新组减少,有时需要从先前计算的组减少中删除记录.

function reduceRemove(p, v) {
    p.worth = p.worth - v.worth;
    p.count = p.count - 1;
    return p;
}

调用reduce方法如下所示:

mycf.dimensions.gender.reduce(reduceAdd, reduceRemove, reduceInitial)

要查看减少的值,请使用该all方法.要查看前n个值,请使用该top(n)方法.

mycf.dimensions.gender.reduce(reduceAdd, reduceRemove, reduceInitial).all()

返回的数组将(应该)如下所示:

[
    { key: "m", value: { worth: 161000000000, count: 3 } },
    { key: "f", value: { worth:  35000000000, count: 2 } },
]

减少数据集的目标是通过首先按公共密钥对记录进行分组,然后将这些分组的维度减少为每个密钥的单个值来派生新数据集.在这种情况下,我们按性别分组,并通过添加共享相同密钥的记录值来减少该分组的价值维度.

其他reduceX方法是reduce方法的便捷方法.

对于这个例子来说,这reduceSum是最合适的替代品.

mycf.dimensions.gender.reduceSum(function(d) {
    return d.worth;
});

调用all返回的分组将(应该)看起来像:

[
    { key: "m", value: 161000000000 },
    { key: "f", value: 35000000000 },
]

reduceCount 将计算记录

mycf.dimensions.gender.reduceCount();

调用all返回的分组将(应该)看起来像:

[
    { key: "m", value: 3 },
    { key: "f", value: 2 },
]

希望这可以帮助 :)

资料来源:https://github.com/square/crossfilter/wiki/API-Reference

Answer 2

http://blog.rusty.io/2012/09/17/crossfilter-tutorial/

var livingThings = crossfilter([
  // Fact data.
  { name: “Rusty”,  type: “human”, legs: 2 },
  { name: “Alex”,   type: “human”, legs: 2 },
  { name: “Lassie”, type: “dog”,   legs: 4 },
  { name: “Spot”,   type: “dog”,   legs: 4 },
  { name: “Polly”,  type: “bird”,  legs: 2 },
  { name: “Fiona”,  type: “plant”, legs: 0 }
]);

例如,我家里有多少生物？

为此,我们将调用groupAll便捷函数,它将所有记录选择到一个组中,然后调用该函数,该reduceCount函数创建记录计数.

// How many living things are in my house?
var n = livingThings.groupAll().reduceCount().value();
console.log("There are " + n + " living things in my house.") // 6

现在让我们算一下我家里所有的腿.同样,我们将使用该groupAll函数来获取单个组中的所有记录,然后我们调用该 reduceSum函数.这将把价值加在一起.有什么价值？好吧,我们想要腿,所以让我们传递一个函数来提取并返回事实中的腿数.

// How many total legs are in my house?
var legs = livingThings.groupAll().reduceSum(function(fact) {
  return fact.legs;
}).value()
console.log("There are " + legs + " legs in my house.")

reduceCountfunction创建记录计数.
reduceSumfunction是这些记录的总和值.