函数，该函数将从SQL DB返回的行分组。

Question

下面包含了从SQL数据库返回的数据集，因为除了一个属性"name_alt“之外，您可以看到所有数据都保持不变。

[{
    "language_id": "tly",
    "language_name": "Talysh",
    "speakers_native": 800000,
    "total_speakers": 912000,
    "name_alt": "Taleshi",
    "country_main": "az"
}, {
    "language_id": "tly",
    "language_name": "Talysh",
    "speakers_native": 800000,
    "total_speakers": 912000,
    "name_alt": "Talish",
    "country_main": "az"
}, {
    "language_id": "tly",
    "language_name": "Talysh",
    "speakers_native": 800000,
    "total_speakers": 912000,
    "name_alt": "Talishi",
    "country_main": "az"
}, {
    "language_id": "tly",
    "language_name": "Talysh",
    "speakers_native": 800000,
    "total_speakers": 912000,
    "name_alt": "Talysh",
    "country_main": "az"
}, {
    "language_id": "tly",
    "language_name": "Talysh",
    "speakers_native": 800000,
    "total_speakers": 912000,
    "name_alt": "Talyshi",
    "country_main": "az"
}, {
    "language_id": "lez",
    "language_name": "Lezghian",
    "speakers_native": 171400,
    "total_speakers": 428400,
    "name_alt": "Kiurinsty",
    "country_main": "ru"
}, {
    "language_id": "lez",
    "language_name": "Lezghian",
    "speakers_native": 171400,
    "total_speakers": 428400,
    "name_alt": "Kiurinty",
    "country_main": "ru"
}, {
    "language_id": "lez",
    "language_name": "Lezghian",
    "speakers_native": 171400,
    "total_speakers": 428400,
    "name_alt": "Lezghi",
    "country_main": "ru"
}]

我希望将所有的name_alts组合成一个数组，所以我的数据如下所示：

[{
    "language_id": "tly",
    "language_name": "Talysh",
    "speakers_native": 800000,
    "total_speakers": 912000,
    "name_alt": ["Talesh", "Taleshi", "Talish", "Talishi", "Talyshi"],
    "country_main": "az"
}, {
    "language_id": "lez",
    "language_name": "Lezghian",
    "speakers_native": 171400,
    "total_speakers": 428400,
    "name_alt": ["Kiurinsty", "Kiurinty", "Lezghi", "Lezgi", "Lezgian", "Lezgin"],
    "country_main": "ru"
}]

我编写了这个函数，它接受数组，映射并过滤它，以返回我需要的数据集。它很有效，但我担心它不是很好的表演。它接受我想要压缩的数组(array)、我想按其分组的属性(mappedProperty)、对每一行都应该保持唯一的属性(uniqueProperty)，以及应该从分组属性集(mainProperty)中排除的属性。

你能复习一下吗？

function reduceRows(array, mappedProperty, uniqueProperty, mainProperty) {
    var currentUniqueProp;
    var objToReturn;
    var mappedObj = _.map(array, function(obj) {
            // we're grouping by the unique property (i.e. an ID), so if we have a new
            // unique property, we clone the object, and start adding the
            // mapped property (i.e. alternative names for a language) to the array
            //usage: u.reduceRows(data.data.languages, "name_alt", "language_id", "language_name")
            if (obj[uniqueProperty] != currentUniqueProp) {
                objToReturn = _.clone(obj)
                objToReturn[mappedProperty] = [obj[mappedProperty]]
                currentUniqueProp = obj[uniqueProperty]
            } else {
                // sometimes the grouped property has a value which is the same as the main property
                // i.e. the language_name is also included in the list of name_alts
                // We want to avoid this duplication
                if (obj[mappedProperty] != obj[mainProperty]) {
                    objToReturn[mappedProperty].push(obj[mappedProperty])
                }
            }
            return objToReturn;
        })
    var filteredObj = _.filter(mappedObj, function(obj) {
        //this function ensures that only one entry for each unique property
        //is returned
        if (obj[uniqueProperty] == currentUniqueProp) {
            return false
        } else {
            currentUniqueProp = obj[uniqueProperty]
            return true
        }
    })
    return filteredObj
}

Answer 1

有几件事：

• 你不需要下划线。节点支持ECMAScript 5特性，如Array.prototype.map和Array.prototype.filter。将_.map(array, function)替换为array.map(function)等。
• 命名您的内联函数。这对可读性更好，如果代码出错，如果不知道哪个匿名函数，就不会得到(anonymous function)。
• map用于将一个数组映射到另一个数组，如果您只想迭代一个数组并运行一个函数，则需要Array.prototype.forEach。

我会采取另一种方法来做这件事。

1. 计算出唯一的条目(除了name_alt)
2. 迭代每个唯一条目的“组”，并将每个reduce分别迭代到单个对象。
3. 将每个新创建的对象推入一个新数组中。

以下是我的看法：

// ## Get an array of language_ids to be used as a unique key. 
// ## I'm assuming here that if language_id is equal, the rest (aside from name_alt) is also equal.
var uniqueKeys = data
    .map(function pluckLanguageId(obj) { return obj.language_id; }) 
    //Now your array looks like ["a", "a", "a", "b", "b" ,"b"]
    .filter(function makeUnique(language, index, array) { return array.indexOf(language) === index; }); 
    //Now your array looks like ["a", "b"]

// ## Map each of the keys to a formatted object:
var result = uniqueKeys.map(function mapKeysToFormattedObjects(key) {
    //Take the entire data array
    return data
        //Filter it by language_id
        .filter(function getOnlyDataWithKey(item) { return item.language_id === key; }) 
        //Then combine it into one object
        .reduce(function combineToOneObject(previous, current) { 
            if (typeof(previous.name_alt) === "string") { //In this first object, the name_alt would be a string.
                previous.name_alt = [previous.name_alt]; //So make it into an array!
            }
            previous.name_alt.push(current.name_alt); //Push the current name_alt into the previous one.
            return previous;
        }); //Result is an object with all of the name_alts in an array
})

这是假设原始数据在data变量中。如果有什么你不明白的，请告诉我，我会尽力解释的。

注意我是如何链接我的调用映射，过滤和减少。map和filter都返回一个数组，因此对这个新处理的数组执行链式调用。

一种更理想的编写方法是将所有的东西分解成函数，然后有一个oneliner：

function pluckLanguageId(obj) { return obj.language_id; }
function makeUnique(language, index, array) { return array.indexOf(language) === index; }

function mapKeysToFormattedObjects(key) {
    return data
        .filter(function getOnlyDataWithKey(item) { return item.language_id === key; }) 
        .reduce(function combineToOneObject(previous, current) { 
            if (typeof(previous.name_alt) === "string") { //In this first object, the name_alt would be a string.
                previous.name_alt = [previous.name_alt]; //So make it into an array!
            }
            previous.name_alt.push(current.name_alt); //Push the current name_alt into the previous one.
            return previous;
        }); //Result is an object with all of the name_alts in an array
}

var result = data.map(pluckLanguageId).filter(makeUnique).map(mapKeysToFormattedObjects);
//One liner!

这段代码可以通过返回函数和诸如此类的函数来进一步改进，但是以第一个例子作为参考。

Answer 2

因为javascript有一个else if语句，所以您应该使用它来删除一个级别的缩进。

if (obj[uniqueProperty] != currentUniqueProp) {
    objToReturn = _.clone(obj)
    objToReturn[mappedProperty] = [obj[mappedProperty]]
    currentUniqueProp = obj[uniqueProperty]
} else if (obj[mappedProperty] != obj[mainProperty]) {
    // sometimes the grouped property has a value which is the same as the main property
    // i.e. the language_name is also included in the list of name_alts
    // We want to avoid this duplication

    objToReturn[mappedProperty].push(obj[mappedProperty])
}  

underscore滤波方法可以简化为

var filteredObj = _.filter(mappedObj, function(obj) {
    //this function ensures that only one entry for each unique property
    //is returned
    if (obj[uniqueProperty] == currentUniqueProp) {
        return false
    }
    currentUniqueProp = obj[uniqueProperty]
    return true
})

如果条件的计算结果为true，则方法将返回，因此不需要else。

您应该始终在需要变量的地方声明它们。所以var objToReturn应该在这里声明

var mappedObj = _.map(array, function(obj) {
        var objToReturn;
        // we're grouping by the unique property