[TOC] ## 习题1 ``` // 示例数据 var CARS = [ {name: "Ferrari FF", horsepower: 660, dollar_value: 700000, in_stock: true}, {name: "Spyker C12 Zagato", horsepower: 650, dollar_value: 648000, in_stock: false}, {name: "Jaguar XKR-S", horsepower: 550, dollar_value: 132000, in_stock: false}, {name: "Audi R8", horsepower: 525, dollar_value: 114200, in_stock: false}, {name: "Aston Martin One-77", horsepower: 750, dollar_value: 1850000, in_stock: true}, {name: "Pagani Huayra", horsepower: 700, dollar_value: 1300000, in_stock: false} ]; // 练习 1: // ============ // 使用 _.compose() 重写下面这个函数。提示：_.prop() 是 curry 函数 var isLastInStock = function(cars) { var last_car = _.last(cars); return _.prop('in_stock', last_car); }; // 使用 compose var isLastInStock = _.compose(_.prop('in_stock'), _.last); 练习 2: // 使用 _.compose()、_.prop() 和 _.head() 获取第一个 car 的 name var nameOfFirstCar = _.compose(_.prop('name'), _.head); / 彩蛋 1: // ============ // 使用 compose 重构 availablePrices var availablePrices = function(cars) { var available_cars = _.filter(_.prop('in_stock'), cars); return available_cars.map(function(x){ return accounting.formatMoney(x.dollar_value); }).join(', '); }; // 解答 var formatPrice = _.compose(accounting.formatMoney,_.prop("dollar_value")) var availablePrices = compose(_.join, _.map(formatPrice),_.filter,_.prop('in_stock')) ```