js中的filter就是过滤的意思，比如，我们以什么样的方式进行过滤，得到我们想要的结果。

对，我们要的就是这个结果。

给定一个数组，我们要的是Burger(汉堡)

const restaurants = [    {        name: "Dan's Hamburgers",        price: 'Cheap',        cuisine: 'Burger',    },    {        name: "Austin's Pizza",        price: 'Cheap',        cuisine: 'Pizza',    },    {        name: "Via 313",        price: 'Moderate',        cuisine: 'Pizza',    },    {        name: "Bufalina",        price: 'Expensive',        cuisine: 'Pizza',    },    {        name: "P. Terry's",        price: 'Cheap',        cuisine: 'Burger',    },    {        name: "Hopdoddy",        price: 'Expensive',        cuisine: 'Burger',    },    {        name: "Whataburger",        price: 'Moderate',        cuisine: 'Burger',    },    {        name: "Chuy's",        cuisine: 'Tex-Mex',        price: 'Moderate',    },    {        name: "Taquerias Arandina",        cuisine: 'Tex-Mex',        price: 'Cheap',    },    {        name: "El Alma",        cuisine: 'Tex-Mex',        price: 'Expensive',    },    {        name: "Maudie's",        cuisine: 'Tex-Mex',        price: 'Moderate',    },];

我们分析这个对象数组，发现cuisine: 'Burger',接下来我们进行求值

const isBurger = ({cuisine}) => cuisine === 'Burger';const burgerJoints = restaurants.filter(isBurger);

开始的时候我们找了Burger，接下来我们要找除了Burger之的东西

只需要找到的值不是Burger就可以了

const isNotBurger = ({cuisine}) => cuisine !== 'Burger';

从restaurant数组中取就是

const isNotBurger = restaurant => !isBurger(restaurant);

如果我们此时要找Burger和Pizza

那就是

const isBurger = ({cuisine}) => cuisine === 'Burger';const isPizza  = ({cuisine}) => cuisine === 'Pizza';

这两者的共通点是我们可以用一个函数存着他们，然后我们想找哪个的时候，就去调用哪个

const isCuisine = comparison => ({cuisine}) => cuisine === comparison;const isBurger  = isCuisine('Burger');const isPizza   = isCuisine('Pizza');

如果我们不找吃的，准备找价格怎么做呢?

const isPrice = comparison => ({price}) => price === comparison;const isCheap = isPrice('Cheap');const isExpensive = isPrice('Expensive');

我们可以看到当我们找吃的时候和找价格的时候，代码存在一些公共点，此时我们又可以提取成为函数

const isKeyEqualToValue = key => value => object => object[key] === value;const isCuisine = isKeyEqualToValue('cuisine');const isPrice = isKeyEqualToValue('price');const isBurger = isCuisine('Burger');const isPizza = isCuisine('Pizza');const isCheap = isPrice('Cheap');const isExpensive = isPrice('Expensive');

我们我们要找便宜的汉堡那就是

const cheapBurgers = restaurants.filter(isCheap).filter(isBurger);

或者是

const isCheapBurger = restaurant => isCheap(restaurant) && isBurger(restaurant);const isCheapPizza = restaurant => isCheap(restaurant) && isPizza(restaurant);

我们提取重复的代码就是

const both = (predicate1, predicate2) => value =>  predicate1(value) && predicate2(value);const isCheapBurger = both(isCheap, isBurger);const isCheapPizza = both(isCheap, isPizza);const cheapBurgers = restaurants.filter(isCheapBurger);const cheapPizza = restaurants.filter(isCheapPizza);

如果没有就是

const either = (predicate1, predicate2) => value =>  predicate1(value) || predicate2(value);const isDelicious = either(isBurger, isPizza);const deliciousFood = restaurants.filter(isDelicious);

如果想要两种以上的食物就是

const isDelicious = restaurant =>  [isPizza, isBurger, isBbq].some(predicate => predicate(restaurant));const isCheapAndDelicious = restaurant =>  [isDelicious, isCheap].every(predicate => predicate(restaurant));

我们进行抽象就是

const isEvery = predicates => value =>  predicates.every(predicate => predicate(value));const isAny = predicates => value =>  predicates.some(predicate => predicate(value));const isDelicious = isAny([isBurger, isPizza, isBbq]);const isCheapAndDelicious = isEvery([isCheap, isDelicious]);

本文提炼自https://zcfy.cc/article/level-up-your-filter-game

by我还差的很远很远