以下是個人在工作中收藏總結(jié)的一些關(guān)于javascript數(shù)組方法reduce的相關(guān)代碼片段，后續(xù)遇到其他使用這個函數(shù)的場景，將會陸續(xù)添加，這里作為備忘。
javascript數(shù)組那么多方法，為什么我要單挑reduce方法，一個原因是我對這個方法掌握不夠，不能夠用到隨心所欲。另一個方面，我也感覺到了這個方法的龐大魅力，在許多的場景中發(fā)揮著神奇的作用。
理解reduce函數(shù)
reduce() 方法接收一個函數(shù)作為累加器（accumulator），數(shù)組中的每個值（從左到右）開始縮減，最終為一個值。
arr.reduce([callback, initialValue])
看如下例子：
let arr = [1, 2, 3, 4, 5]; // 10代表初始值，p代表每一次的累加值，在第一次為10 // 如果不存在初始值，那么p第一次值為1 // 此時累加的
片段一：字母游戲
const anagrams = str => {
if (str.length <= 2) {
return str.length === 2 ? [str, str[1] + str[0]] : str;
}
return str.split("").reduce((acc, letter, i) => {
return acc.concat(anagrams(str.slice(0, i) + str.slice(i + 1)).map(val => letter + val));
}, []);
}
anagrams("abc"); // 結(jié)果會是什么呢？reduce負(fù)責(zé)篩選出每一次執(zhí)行的首字母，遞歸負(fù)責(zé)對剩下字母的排列組合。
片段二：累加器
const sum = arr => arr.reduce((acc, val) => acc + val, 0); sum([1, 2, 3]);
片段三：計數(shù)器
const countOccurrences = (arr, value) => arr.reduce((a, v) => v === value ? a + 1 : a + 0, 0); countOccurrences([1, 2, 3, 2, 2, 5, 1], 1);
循環(huán)數(shù)組，每遇到一個值與給定值相等，即加1，同時將加上之后的結(jié)果作為下次的初始值。
片段四：函數(shù)柯里化
函數(shù)柯里化的目的就是為了儲存數(shù)據(jù)，然后在最后一步執(zhí)行。
const curry = (fn, arity = fn.length, ...args) => arity <= args.length ? fn(...args) : curry.bind(null, fn, arity, ...args); curry(Math.pow)(2)(10); curry(Math.min, 3)(10)(50)(2);
通過判斷函數(shù)的參數(shù)取得當(dāng)前函數(shù)的length(當(dāng)然也可以自己指定)，如果所傳的參數(shù)比當(dāng)前參數(shù)少，則繼續(xù)遞歸下面，同時儲存上一次傳遞的參數(shù)。
片段五：數(shù)組扁平化
const deepFlatten = arr => arr.reduce((a, v) => a.concat(Array.isArray(v) ? deepFlatten(v) : v), []); deepFlatten([1, [2, [3, 4, [5, 6]]]]);
片段六：生成菲波列契數(shù)組
const fibonacci = n => Array(n).fill(0).reduce((acc, val, i) => acc.concat(i > 1 ? acc[i - 1] + acc[i - 2] : i), []); fibonacci(5);
片段七：管道加工器
const pipe = (...funcs) => arg => funcs.reduce((acc, func) => func(acc), arg);
pipe(btoa, x => x.toUpperCase())("Test");
通過對傳遞的參數(shù)進(jìn)行函數(shù)加工，之后將加工之后的數(shù)據(jù)作為下一個函數(shù)的參數(shù)，這樣層層傳遞下去。
片段八：中間件
const dispatch = action => {
console.log('action', action);
return action;
}
const middleware1 = dispatch => {
return action => {
console.log("middleware1");
const result = dispatch(action);
console.log("after middleware1");
return result;
}
}
const middleware2 = dispatch => {
return action => {
console.log("middleware2");
const result = dispatch(action);
console.log("after middleware2");
return result;
}
}
const middleware3 = dispatch => {
return action => {
console.log("middleware3");
const result = dispatch(action);
console.log("after middleware3");
return result;
}
}
const compose = middlewares => middlewares.reduce((a, b) => args => a(b(args)))
const middlewares = [middleware1, middleware2, middleware3];
const afterDispatch = compose(middlewares)(dispatch);
const testAction = arg => {
return { type: "TEST_ACTION", params: arg };
};
afterDispatch(testAction("1111"));
redux中經(jīng)典的compose函數(shù)中運(yùn)用了這種方式，通過對中間件的重重層疊，在真正發(fā)起action的時候觸發(fā)函數(shù)執(zhí)行。
片段九：redux-actions對state的加工片段
// redux-actions/src/handleAction.js
const handleAction = (type, reducer, defaultState) => {
const types = type.toString();
const [nextReducer, throwReducer] = [reducer, reducer];
return (state = defaultState, action) => {
const { type: actionType } = action;
if (!actionType || types.indexOf(actionType.toString()) === -1) {
return state;
}
return (action.error === true ? throwReducer : nextReducer)(state, action);
}
}
// reduce-reducers/src/index.js
const reduceReducer = (...reducers) => {
return (previous, current) => {
reducers.reduce((p, r) => r(p, current), previous);
}
}
// redux-actions/src/handleActions.js
const handleActions = (handlers, defaultState, { namespace } = {}) => {
// reducers的扁平化
const flattenedReducerMap = flattenReducerMap(handles, namespace);
// 每一種ACTION下對應(yīng)的reducer處理方式
const reducers = Reflect.ownkeys(flattenedReducerMap).map(type => handleAction(
type,
flattenedReducerMap[type],
defaultState
));
// 狀態(tài)的加工器，用于對reducer的執(zhí)行
const reducer = reduceReducers(...reducers);
// reducer觸發(fā)
return (state = defaultState, action) => reducer(state, action);
}
const reducers = {
totalInEuros: (state, item) => {
return state.euros += item.price * 0.897424392;
},
totalInYen: (state, item) => {
return state.yens += item.price * 113.852;
}
};
const manageReducers = reducers => {
return (state, item) => {
return Object.keys(reducers).reduce((nextState, key) => {
reducers[key](state, item);
return state;
}, {})
}
}
const bigTotalPriceReducer = manageReducers(reducers);
const initialState = { euros: 0, yens: 0 };
const items = [{ price: 10 }, { price: 120 }, { price: 1000 }];
const totals = items.reduce(bigTotalPriceReducer, initialState);
let school = {
name: 'Hope middle school',
created: '2001',
classes: [
{
name: '三年二班',
teachers: [
{ name: '張二蛋', age: 26, sex: '男', actor: '班主任' },
{ name: '王小妞', age: 23, sex: '女', actor: '英語老師' }
]
},
{
name: '明星班',
teachers: [
{ name: '歐陽娜娜', age: 29, sex: '女', actor: '班主任' },
{ name: '李易峰', age: 28, sex: '男', actor: '體育老師' },
{ name: '楊冪', age: 111, sex: '女', actor: '藝術(shù)老師' }
]
}
]
};
// 常規(guī)做法
school.classes &&
school.classes[0] &&
school.classes[0].teachers &&
school.classes[0].teachers[0] &&
school.classes[0].teachers[0].name
// reduce方法
const get = (p, o) => p.reduce((xs, x) => (xs && xs[x] ? xs[x] : null), o);
get(['classes', 0, 'teachers', 0, 'name'], school); // 張二蛋
const groupBy = (arr, func) =>
arr.map(typeof func === 'function' ? func : val => val[func]).reduce((acc, val, i) => {
acc[val] = (acc[val] || []).concat(arr[i]);
return acc;
}, {});
groupBy([6.1, 4.2, 6.3], Math.floor);
groupBy(['one', 'two', 'three'], 'length'); map計算出所有的鍵值，然后再根據(jù)建值進(jìn)行歸類
片段十三：對象過濾
const pick = (obj, arr) =>
arr.reduce((acc, curr) => (curr in obj && (acc[curr] = obj[curr]), acc), {});pick({ a: 1, b: '2', c: 3 }, ['a', 'c']);
根據(jù)給出的鍵值來遍歷，比較對象中是否存在相同鍵值的的值，然后通過逗號表達(dá)式把賦值后的對象賦給下一個的初始值
片段十四：數(shù)組中刪除指定位置的值
const remove = (arr, func) =>
Array.isArray(arr)
? arr.filter(func).reduce((acc, val) => {
arr.splice(arr.indexOf(val), 1);
return acc.concat(val);
}, []) : [];
const arr = [1, 2, 3, 4];
remove(arr, n => n % 2 == 0);
首先根據(jù)filter函數(shù)過濾出數(shù)組中符合條件的值，然后使用reduce在原數(shù)組中刪除符合條件的值，可以得出最后arr的值變成了[1, 3]
片段十五：promise按照順序執(zhí)行
const runPromisesInSeries = ps => ps.reduce((p, next) => p.then(next), Promise.resolve());
const delay = d => new Promise(r => setTimeout(r, d));
const print = args => new Promise(r => r(args));
runPromisesInSeries([() => delay(1000), () => delay(2000), () => print('hello')]);片段十六：排序
const orderBy = (arr, props, orders) =>
[...arr].sort((a, b) =>
props.reduce((acc, prop, i) => {
if (acc === 0) {
const [p1, p2] = orders && orders[i] === 'desc' ? [b[prop], a[prop]] : [a[prop], b[prop]];
acc = p1 > p2 ? 1 : p1 < p2 ? -1 : 0;
}
return acc;
}, 0)
);
const users = [{ name: 'fred', age: 48 }, { name: 'barney', age: 36 }, { name: 'fly', age: 26 }];
orderBy(users, ['name', 'age'], ['asc', 'desc']);
orderBy(users, ['name', 'age']);片段十七：選擇
const select = (from, selector) =>
selector.split('.').reduce((prev, cur) => prev && prev[cur], from);
const obj = { selector: { to: { val: 'val to select' } } };
select(obj, 'selector.to.val');聲明：本網(wǎng)頁內(nèi)容旨在傳播知識，若有侵權(quán)等問題請及時與本網(wǎng)聯(lián)系，我們將在第一時間刪除處理。TEL:177 7030 7066 E-MAIL:11247931@qq.com
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