如何使用 Lodash 从嵌套对象集合中获取图像数组?
How can i get an array of images from a collection of nested objects with Lodash?
{
"shop": {
"homebackground": "http://padmenu.s3.amazonaws.com/15/11/2014/05/08/2ec2ff61-d6a0-11e3-8857-10ddb1e6e201.jpg",
"name": {
"tr": "My Shop"
},
"menus": [{
"name": {
"en": "Menu"
},
"children": [{
"name": {
"en_US": "Category"
},
"images": [
"http://www.progressivedental-ellenlimdds.com/wp-content/uploads/2014/06/red-wine.jpg"
],
"children": [{
"name": {
"en_US": "Item"
},
"images": [
"http://res.cloudinary.com/finedine/image/upload/c_fill,g_center,h_600/v1435916818/WIne-Bottle_uz03a0.jpg",
"http://media.riepenau.com/wines/17973_b.jpg",
"http://lorempixel.com/400/400/food/3",
"http://lorempixel.com/400/400/food/4",
"http://lorempixel.com/400/400/food/5",
"http://lorempixel.com/400/400/food/6",
"http://lorempixel.com/400/400/food/7"
]
}]
}]
}]
}
}
我想 select 商店 "children" 对象中的所有 "images" 数组。
我如何使用 Lodash 库来做到这一点?
输出应该是一个由图像 url 组成的数组:
["url1","url2","url3"]
伪代码
你可以用一点递归来解决这个问题:
- 获取子列表。
- 从
pluck
的子列表中提取所有图像。
- 对所有后代重复步骤 1。
- 合并所有结果并展平。
核心代码
function deepExtract(collection, childKey, property) {
var exists = _.negate(_.isEmpty);
var children = _.chain(collection).pluck(childKey).filter(exists).flatten();
if (_.isEmpty(children.value())) {
return [];
}
var images = children.pluck(property).value();
var descendantImages = deepExtract(children.value(), childKey, property);
return _.flatten(images.concat(descendantImages));
};
var tree = _.chain(data).get('shop.menus').value();
var images = deepExtract(tree, 'children', 'images');
演示
var data = {
"shop": {
"homebackground": "http://padmenu.s3.amazonaws.com/15/11/2014/05/08/2ec2ff61-d6a0-11e3-8857-10ddb1e6e201.jpg",
"name": {
"tr": "My Shop"
},
"menus": [{
"name": {
"en": "Menu"
},
"children": [{
"name": {
"en_US": "Category"
},
"images": [
"http://www.progressivedental-ellenlimdds.com/wp-content/uploads/2014/06/red-wine.jpg"
],
"children": [{
"name": {
"en_US": "Item"
},
"images": [
"http://res.cloudinary.com/finedine/image/upload/c_fill,g_center,h_600/v1435916818/WIne-Bottle_uz03a0.jpg",
"http://media.riepenau.com/wines/17973_b.jpg",
"http://lorempixel.com/400/400/food/3",
"http://lorempixel.com/400/400/food/4",
"http://lorempixel.com/400/400/food/5",
"http://lorempixel.com/400/400/food/6",
"http://lorempixel.com/400/400/food/7"
]
}]
}]
}]
}
};
function deepExtract(collection, childKey, property) {
var exists = _.negate(_.isEmpty);
var children = _.chain(collection).pluck(childKey).filter(exists).flatten();
if (_.isEmpty(children.value())) {
return [];
}
var images = children.pluck(property).value();
var descendantImages = deepExtract(children.value(), childKey, property);
return _.flatten(images.concat(descendantImages));
};
var tree = _.chain(data).get('shop.menus').value();
log(deepExtract(tree, 'children', 'images'));
// Helper method to output to screen
function log(value) {
document.getElementById("output").innerHTML += JSON.stringify(value, null, 2) + "\n"
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.0/lodash.min.js"></script>
<pre id="output"></pre>
解决此问题的最简单方法是递归地提取 children 及其后代。重点在getImages()
函数中;其中它将所有 children 数组展平在一个级别,提取每个图像数组并压缩所有项目以删除未定义的值(由 children 引起的没有图像),然后展平图像数组并准备连接。递归的停止点是当前 children
没有图像时,返回一个空数组。如果找到图像,那么我们递归地连接所有可能的后代图像。至于我们如何获得后代,我们使用与获得 images
数组相同的链接序列,但使用 children
作为采摘键。
function getImages(children) {
var images = _(children).flatten().pluck('images').compact().flatten().value();
if(_.isEmpty(images)) {
return [];
}
var descendants = _(children).flatten().pluck('children').compact().flatten().value();
return images.concat(getImages(descendants));
}
function getShopImages(data) {
var children = _.pluck(data.shop.menus, 'children');
return getImages(children);
}
console.log(getShopImages(data));
我在这里找到了我的问题的替代解决方案:
var children = _(shop.menus[0].children)
.thru(function(coll) {
return _.union(coll, _.pluck(coll, 'children'));
})
.flatten();
var images = _.chain(children).pluck('images').flattenDeep().compact().uniq().value();
输出"images"是一个图像数组。
{
"shop": {
"homebackground": "http://padmenu.s3.amazonaws.com/15/11/2014/05/08/2ec2ff61-d6a0-11e3-8857-10ddb1e6e201.jpg",
"name": {
"tr": "My Shop"
},
"menus": [{
"name": {
"en": "Menu"
},
"children": [{
"name": {
"en_US": "Category"
},
"images": [
"http://www.progressivedental-ellenlimdds.com/wp-content/uploads/2014/06/red-wine.jpg"
],
"children": [{
"name": {
"en_US": "Item"
},
"images": [
"http://res.cloudinary.com/finedine/image/upload/c_fill,g_center,h_600/v1435916818/WIne-Bottle_uz03a0.jpg",
"http://media.riepenau.com/wines/17973_b.jpg",
"http://lorempixel.com/400/400/food/3",
"http://lorempixel.com/400/400/food/4",
"http://lorempixel.com/400/400/food/5",
"http://lorempixel.com/400/400/food/6",
"http://lorempixel.com/400/400/food/7"
]
}]
}]
}]
}
}
我想 select 商店 "children" 对象中的所有 "images" 数组。 我如何使用 Lodash 库来做到这一点? 输出应该是一个由图像 url 组成的数组: ["url1","url2","url3"]
伪代码
你可以用一点递归来解决这个问题:
- 获取子列表。
- 从
pluck
的子列表中提取所有图像。 - 对所有后代重复步骤 1。
- 合并所有结果并展平。
核心代码
function deepExtract(collection, childKey, property) {
var exists = _.negate(_.isEmpty);
var children = _.chain(collection).pluck(childKey).filter(exists).flatten();
if (_.isEmpty(children.value())) {
return [];
}
var images = children.pluck(property).value();
var descendantImages = deepExtract(children.value(), childKey, property);
return _.flatten(images.concat(descendantImages));
};
var tree = _.chain(data).get('shop.menus').value();
var images = deepExtract(tree, 'children', 'images');
演示
var data = {
"shop": {
"homebackground": "http://padmenu.s3.amazonaws.com/15/11/2014/05/08/2ec2ff61-d6a0-11e3-8857-10ddb1e6e201.jpg",
"name": {
"tr": "My Shop"
},
"menus": [{
"name": {
"en": "Menu"
},
"children": [{
"name": {
"en_US": "Category"
},
"images": [
"http://www.progressivedental-ellenlimdds.com/wp-content/uploads/2014/06/red-wine.jpg"
],
"children": [{
"name": {
"en_US": "Item"
},
"images": [
"http://res.cloudinary.com/finedine/image/upload/c_fill,g_center,h_600/v1435916818/WIne-Bottle_uz03a0.jpg",
"http://media.riepenau.com/wines/17973_b.jpg",
"http://lorempixel.com/400/400/food/3",
"http://lorempixel.com/400/400/food/4",
"http://lorempixel.com/400/400/food/5",
"http://lorempixel.com/400/400/food/6",
"http://lorempixel.com/400/400/food/7"
]
}]
}]
}]
}
};
function deepExtract(collection, childKey, property) {
var exists = _.negate(_.isEmpty);
var children = _.chain(collection).pluck(childKey).filter(exists).flatten();
if (_.isEmpty(children.value())) {
return [];
}
var images = children.pluck(property).value();
var descendantImages = deepExtract(children.value(), childKey, property);
return _.flatten(images.concat(descendantImages));
};
var tree = _.chain(data).get('shop.menus').value();
log(deepExtract(tree, 'children', 'images'));
// Helper method to output to screen
function log(value) {
document.getElementById("output").innerHTML += JSON.stringify(value, null, 2) + "\n"
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.0/lodash.min.js"></script>
<pre id="output"></pre>
解决此问题的最简单方法是递归地提取 children 及其后代。重点在getImages()
函数中;其中它将所有 children 数组展平在一个级别,提取每个图像数组并压缩所有项目以删除未定义的值(由 children 引起的没有图像),然后展平图像数组并准备连接。递归的停止点是当前 children
没有图像时,返回一个空数组。如果找到图像,那么我们递归地连接所有可能的后代图像。至于我们如何获得后代,我们使用与获得 images
数组相同的链接序列,但使用 children
作为采摘键。
function getImages(children) {
var images = _(children).flatten().pluck('images').compact().flatten().value();
if(_.isEmpty(images)) {
return [];
}
var descendants = _(children).flatten().pluck('children').compact().flatten().value();
return images.concat(getImages(descendants));
}
function getShopImages(data) {
var children = _.pluck(data.shop.menus, 'children');
return getImages(children);
}
console.log(getShopImages(data));
我在这里找到了我的问题的替代解决方案:
var children = _(shop.menus[0].children)
.thru(function(coll) {
return _.union(coll, _.pluck(coll, 'children'));
})
.flatten();
var images = _.chain(children).pluck('images').flattenDeep().compact().uniq().value();
输出"images"是一个图像数组。