《簡約的軟體開發思維：用 Functional Programming 重構程式》CH12 ~ CH13

Ch12. 利用函數走訪

Code Smell: 有相似的函式實作

隱性 -> 顯性函數
回呼函式

map()、filter()、reduce() 是三個常見的高階函數，讓 FP 的邏輯看起來更簡潔。

特性	for 迴圈	forEach()	map()
語法	for (let i = 0; …)	array.forEach(…)	array.map(…)
回傳值	無	無 (undefined)	新陣列
可否中斷	是 (break, continue)	否	否
修改原陣列	可手動控制	通常用於修改，但不回傳新陣列	不會修改（創建新陣列）
適用對象	任何可迭代物件	陣列	陣列
主要用途	高度控制，處理通用迭代	對每個元素執行副作用	將每個元素轉換為新形式，生成新陣列

1. `map()`

使用情境：將陣列中的每個元素轉換為新形式，生成新陣列。

蘋果 🍎 -> 包裝機 📦 -> 貼標的蘋果 🍎 💰

const arr = [1, 2, 3];
const newArr = arr.map((item) => {
  return item * 2;
});
console.log(newArr);

＊陣列的值可能為 null，使用 filter()、 ?. 來避免拋出錯誤。

練習 12-1：Mega Mart 寄送賀卡

customers 陣列包涵所有 customer 物件
使用 map() 產生賀卡，包涵 cursomer.firstName, cursomer.lastName, cursomer.address

const customers = [
  {
    firstName: "John",
    lastName: "Doe",
    address: "123 Main St",
    habbit: "reading",
  },
  {
    firstName: "Jane",
    lastName: "Smith",
    address: "456 Main St",
    habbit: "reading",
  },
  {
    firstName: "Jim",
    lastName: "Beam",
    address: "789 Main St",
    habbit: "reading",
  },
];

const cards = "YOUR ANSWER HERE";

2. `filter()`

使用情境：過濾陣列中的元素，生成新陣列。

一籃蘋果 🍎 🍏 -> 篩選 🔍 -> 好蘋果 🍎

業務情境： Mega Mart 想要與高消費力的顧客建立關係，因此需要過濾出高消費力的顧客。
總顧客 10 人 -> 高消費 4 人 / 低消費 6 人

function selectBestCustomers(customers) {
  return customers.filter((customer) => {
    return customer.total > 1000;
  });
}

如何整理過於複雜得篩選邏輯：回呼函式！

業務邏輯：「最近一年內有交易、總金額超過 $1000、且 email 有驗證、VIP 等級為 gold 或 platinum 的顧客」

function isHighSpender(customer) {
  return customer.total > 1000;
}

function isRecentBuyer(customer) {
  const oneYearAgo = new Date();
  oneYearAgo.setFullYear(oneYearAgo.getFullYear() - 1);
  return new Date(customer.lastPurchaseDate) > oneYearAgo;
}

function hasVerifiedEmail(customer) {
  return customer.emailVerified === true;
}

function isVipCustomer(customer) {
  return ["gold", "platinum"].includes(customer.vipLevel);
}

function selectBestCustomers(customers) {
  return customers.filter((customer) => {
    return (
      isHighSpender(customer) &&
      isRecentBuyer(customer) &&
      hasVerifiedEmail(customer) &&
      isVipCustomer(customer)
    );
  });
}

練習 12-2：Mega Mart 寄測試信件給顧客清單中的 1/3 使用者

測試名單 testGroup：顧客 id 可以整除 3 的顧客
非測試名單：其餘顧客

const customers = [
  { id: 1, name: "John" },
  { id: 2, name: "Jane" },
  { id: 3, name: "Jim" },
  { id: 4, name: "Jill" },
  { id: 5, name: "Jack" },
];
const testGroup = customers.filter("YOUR ANSWER HERE");

const nonTestGroup = customers.filter("YOUR ANSWER HERE");

3. `reduce()`

使用情境：將陣列中的元素累積為一個值。

一堆水果 🍎 🍏 -> 變成果汁 🍹

function sum(numbers) {
  return reduce(numbers, 0, function (total, number) {
    return total + number;
  });
}
/


// 將所有數字相乘
function product(numbers) {
  return reduce(numbers, 1, function (total, number) {
    return total * number;
  });
}

`reduce()` 的常見用途：

功能	概念說明	應用場景	核心特點	實際例子
復原 (Undo)	透過累積歷史狀態，可以回到之前的任何狀態	文字編輯器、繪圖軟體、遊戲存檔	狀態堆疊、歷史記錄	Word 的 Ctrl+Z 功能
重複 (Redo)	在復原後，可以重新執行被復原的操作	與復原配合使用的前進功能	雙向狀態管理	Word 的 Ctrl+Y 功能
時間移動除錯	記錄每一步的執行過程，可以跳到任意時間點	程式除錯、狀態分析、效能監控	時間軸記錄、快照	Redux DevTools 的時間旅行
審計軌跡	完整記錄所有操作的詳細資訊	系統安全、合規檢查、操作追蹤	不可變記錄、完整性	銀行交易記錄、系統日誌

按編：這邊算是自己找的例子，不在讀書會及書中提及。

// ==========================================
// 1. 復原 (Undo) - 純函數式文字編輯器
// ==========================================

// 純函數：執行單一編輯操作
const applyEdit = (content, edit) => {
  switch (edit.type) {
    case "INSERT":
      return (
        content.slice(0, edit.position) +
        edit.text +
        content.slice(edit.position)
      );
    case "DELETE":
      return (
        content.slice(0, edit.position) +
        content.slice(edit.position + edit.length)
      );
    case "REPLACE":
      return (
        content.slice(0, edit.position) +
        edit.newText +
        content.slice(edit.position + edit.oldText.length)
      );
    default:
      return content;
  }
};

// 純函數：建立編輯歷史狀態
const createEditHistory = (edits, initialContent = "") => {
  return edits.reduce(
    (history, edit) => {
      const previousContent = history[history.length - 1];
      const newContent = applyEdit(previousContent, edit);
      return [...history, newContent];
    },
    [initialContent]
  );
};

// 純函數：復原操作
const undo = (history, currentIndex) => ({
  content: history[Math.max(0, currentIndex - 1)],
  newIndex: Math.max(0, currentIndex - 1),
});

// 純函數：重複操作
const redo = (history, currentIndex) => ({
  content: history[Math.min(history.length - 1, currentIndex + 1)],
  newIndex: Math.min(history.length - 1, currentIndex + 1),
});

// 使用範例
const edits = [
  { type: "INSERT", position: 0, text: "Hello" },
  { type: "INSERT", position: 5, text: " World" },
  { type: "REPLACE", position: 6, oldText: "World", newText: "JavaScript" },
];

const history = createEditHistory(edits);
console.log("編輯歷史:", history);
// ['', 'Hello', 'Hello World', 'Hello JavaScript']

const undoResult = undo(history, 3);
console.log("復原後:", undoResult);
// { content: 'Hello World', newIndex: 2 }

// ==========================================
// 2. 重複 (Redo) - 純函數式狀態管理
// ==========================================

// 純函數：建立狀態管理器
const createStateManager = (actions, initialState = null) => {
  return actions.reduce(
    (stateHistory, action) => {
      const currentState = stateHistory.states[stateHistory.states.length - 1];
      const newState = action.reducer(currentState, action.payload);

      return {
        states: [...stateHistory.states, newState],
        actions: [...stateHistory.actions, action],
        currentIndex: stateHistory.states.length,
      };
    },
    {
      states: [initialState],
      actions: [],
      currentIndex: 0,
    }
  );
};

// 純函數：執行復原/重複操作
const timeTravel = (stateManager, targetIndex) => ({
  ...stateManager,
  currentIndex: Math.max(
    0,
    Math.min(stateManager.states.length - 1, targetIndex)
  ),
  currentState:
    stateManager.states[
      Math.max(0, Math.min(stateManager.states.length - 1, targetIndex))
    ],
});

// 使用範例 - 購物車狀態管理
const cartActions = [
  {
    type: "ADD_ITEM",
    payload: { id: 1, name: "iPhone", price: 999 },
    reducer: (state, item) => ({
      ...state,
      items: [...(state?.items || []), item],
      total: (state?.total || 0) + item.price,
    }),
  },
  {
    type: "ADD_ITEM",
    payload: { id: 2, name: "MacBook", price: 1999 },
    reducer: (state, item) => ({
      ...state,
      items: [...state.items, item],
      total: state.total + item.price,
    }),
  },
  {
    type: "REMOVE_ITEM",
    payload: { id: 1 },
    reducer: (state, payload) => {
      const newItems = state.items.filter((item) => item.id !== payload.id);
      const newTotal = newItems.reduce((sum, item) => sum + item.price, 0);
      return { ...state, items: newItems, total: newTotal };
    },
  },
];

const cartHistory = createStateManager(cartActions, { items: [], total: 0 });
console.log("購物車狀態歷史:", cartHistory.states);

// 復原到第1步
const undoToStep1 = timeTravel(cartHistory, 1);
console.log("復原到第1步:", undoToStep1.currentState);

// 重複到第2步
const redoToStep2 = timeTravel(cartHistory, 2);
console.log("重複到第2步:", redoToStep2.currentState);

// ==========================================
// 3. 時間移動除錯 - 純函數式執行追蹤
// ==========================================

// 純函數：建立除錯快照
const createDebugSnapshot = (
  step,
  operation,
  input,
  output,
  metadata = {}
) => ({
  step,
  timestamp: new Date().toISOString(),
  operation,
  input,
  output,
  metadata,
  performance: {
    executionTime: metadata.executionTime || 0,
    memoryUsage: metadata.memoryUsage || 0,
  },
});

// 純函數：執行可追蹤的操作序列
const executeWithTimeTravel = (operations, initialData) => {
  return operations.reduce(
    (debugState, operation, index) => {
      const startTime = performance.now();
      const result = operation.fn(debugState.currentData, operation.params);
      const endTime = performance.now();

      const snapshot = createDebugSnapshot(
        index + 1,
        operation.name,
        debugState.currentData,
        result,
        {
          params: operation.params,
          executionTime: endTime - startTime,
        }
      );

      return {
        currentData: result,
        snapshots: [...debugState.snapshots, snapshot],
        totalExecutionTime:
          debugState.totalExecutionTime + (endTime - startTime),
      };
    },
    {
      currentData: initialData,
      snapshots: [],
      totalExecutionTime: 0,
    }
  );
};

// 純函數：跳轉到特定時間點
const jumpToSnapshot = (debugState, stepNumber) => {
  const targetSnapshot = debugState.snapshots.find(
    (s) => s.step === stepNumber
  );
  return targetSnapshot ? targetSnapshot.output : debugState.currentData;
};

// 使用範例 - 數據處理管道
const dataOperations = [
  {
    name: "FILTER_ACTIVE_USERS",
    fn: (users) => users.filter((user) => user.active),
    params: { condition: "active === true" },
  },
  {
    name: "MAP_USER_SCORES",
    fn: (users) => users.map((user) => ({ ...user, score: user.points * 1.5 })),
    params: { multiplier: 1.5 },
  },
  {
    name: "SORT_BY_SCORE",
    fn: (users) => [...users].sort((a, b) => b.score - a.score),
    params: { order: "desc" },
  },
  {
    name: "TAKE_TOP_5",
    fn: (users) => users.slice(0, 5),
    params: { limit: 5 },
  },
];

const initialUsers = [
  { id: 1, name: "Alice", active: true, points: 100 },
  { id: 2, name: "Bob", active: false, points: 80 },
  { id: 3, name: "Charlie", active: true, points: 120 },
  { id: 4, name: "Diana", active: true, points: 90 },
  { id: 5, name: "Eve", active: true, points: 110 },
];

const debugResult = executeWithTimeTravel(dataOperations, initialUsers);
console.log("執行結果:", debugResult.currentData);
console.log("執行快照:", debugResult.snapshots);

// 跳轉到第2步查看狀態
const step2State = jumpToSnapshot(debugResult, 2);
console.log("第2步狀態:", step2State);

// ==========================================
// 4. 審計軌跡 - 純函數式操作日誌
// ==========================================

// 純函數：建立審計記錄
const createAuditEntry = (
  action,
  user,
  resource,
  oldValue,
  newValue,
  metadata = {}
) => ({
  id: crypto.randomUUID ? crypto.randomUUID() : Math.random().toString(36),
  timestamp: new Date().toISOString(),
  action,
  user: {
    id: user.id,
    name: user.name,
    role: user.role,
    ip: user.ip || "unknown",
  },
  resource: {
    type: resource.type,
    id: resource.id,
    name: resource.name,
  },
  changes: {
    before: oldValue,
    after: newValue,
  },
  metadata: {
    userAgent: metadata.userAgent || "unknown",
    sessionId: metadata.sessionId || "unknown",
    ...metadata,
  },
  hash: `${action}-${user.id}-${resource.id}-${Date.now()}`,
});

// 純函數：處理業務操作並生成審計軌跡
const executeBusinessOperations = (operations, initialState) => {
  return operations.reduce(
    (auditState, operation) => {
      const oldValue = auditState.currentState[operation.resourceId];
      const newValue = operation.transformer(oldValue, operation.payload);

      const auditEntry = createAuditEntry(
        operation.action,
        operation.user,
        operation.resource,
        oldValue,
        newValue,
        operation.metadata
      );

      return {
        currentState: {
          ...auditState.currentState,
          [operation.resourceId]: newValue,
        },
        auditTrail: [...auditState.auditTrail, auditEntry],
      };
    },
    {
      currentState: initialState,
      auditTrail: [],
    }
  );
};

// 純函數：查詢審計軌跡
const queryAuditTrail = (auditTrail, filters = {}) => {
  return auditTrail.filter((entry) => {
    return Object.entries(filters).every(([key, value]) => {
      if (key === "user") return entry.user.id === value;
      if (key === "action") return entry.action === value;
      if (key === "resource") return entry.resource.type === value;
      if (key === "dateFrom")
        return new Date(entry.timestamp) >= new Date(value);
      if (key === "dateTo") return new Date(entry.timestamp) <= new Date(value);
      return true;
    });
  });
};

// 使用範例 - 銀行帳戶操作審計
const bankOperations = [
  {
    action: "DEPOSIT",
    user: {
      id: "user123",
      name: "John Doe",
      role: "customer",
      ip: "192.168.1.1",
    },
    resource: { type: "account", id: "acc001", name: "Checking Account" },
    resourceId: "acc001",
    payload: { amount: 1000 },
    transformer: (currentBalance, payload) =>
      (currentBalance || 0) + payload.amount,
    metadata: { branch: "downtown", teller: "jane_smith" },
  },
  {
    action: "WITHDRAW",
    user: {
      id: "user123",
      name: "John Doe",
      role: "customer",
      ip: "192.168.1.1",
    },
    resource: { type: "account", id: "acc001", name: "Checking Account" },
    resourceId: "acc001",
    payload: { amount: 200 },
    transformer: (currentBalance, payload) => currentBalance - payload.amount,
    metadata: { atm: "atm_007", card: "**** 1234" },
  },
  {
    action: "TRANSFER",
    user: {
      id: "user123",
      name: "John Doe",
      role: "customer",
      ip: "192.168.1.1",
    },
    resource: { type: "account", id: "acc001", name: "Checking Account" },
    resourceId: "acc001",
    payload: { amount: 300, toAccount: "acc002" },
    transformer: (currentBalance, payload) => currentBalance - payload.amount,
    metadata: { transferType: "internal", recipient: "acc002" },
  },
];

const bankAudit = executeBusinessOperations(bankOperations, { acc001: 0 });
console.log("帳戶最終狀態:", bankAudit.currentState);
console.log("完整審計軌跡:", bankAudit.auditTrail);

// 查詢特定用戶的操作記錄
const userOperations = queryAuditTrail(bankAudit.auditTrail, {
  user: "user123",
});
console.log("用戶 user123 的操作記錄:", userOperations);

// 查詢所有轉帳操作
const transferOperations = queryAuditTrail(bankAudit.auditTrail, {
  action: "TRANSFER",
});
console.log("所有轉帳操作:", transferOperations);

// ==========================================
// 額外：組合式使用範例
// ==========================================

// 結合四種功能的完整狀態管理器
const createCompleteStateManager = (actions, initialState) => {
  return actions.reduce(
    (manager, action, index) => {
      const startTime = performance.now();
      const oldState = manager.states[manager.states.length - 1];
      const newState = action.reducer(oldState, action.payload);
      const endTime = performance.now();

      // 建立快照 (時間移動除錯)
      const snapshot = createDebugSnapshot(
        index + 1,
        action.type,
        oldState,
        newState,
        { executionTime: endTime - startTime }
      );

      // 建立審計記錄
      const auditEntry = createAuditEntry(
        action.type,
        action.user || { id: "system", name: "System" },
        action.resource || {
          type: "state",
          id: "app",
          name: "Application State",
        },
        oldState,
        newState
      );

      return {
        // 狀態歷史 (支援復原/重複)
        states: [...manager.states, newState],
        currentIndex: manager.states.length,

        // 除錯快照 (時間移動除錯)
        snapshots: [...manager.snapshots, snapshot],

        // 審計軌跡
        auditTrail: [...manager.auditTrail, auditEntry],

        // 效能統計
        totalExecutionTime: manager.totalExecutionTime + (endTime - startTime),
      };
    },
    {
      states: [initialState],
      currentIndex: 0,
      snapshots: [],
      auditTrail: [],
      totalExecutionTime: 0,
    }
  );
};

// 使用範例
const appActions = [
  {
    type: "LOGIN",
    payload: { userId: "user123", sessionId: "sess_abc" },
    reducer: (state, payload) => ({
      ...state,
      user: payload,
      isLoggedIn: true,
    }),
    user: { id: "user123", name: "John Doe", role: "user" },
  },
  {
    type: "UPDATE_PROFILE",
    payload: { name: "John Smith", email: "john.smith@example.com" },
    reducer: (state, payload) => ({
      ...state,
      user: { ...state.user, ...payload },
    }),
    user: { id: "user123", name: "John Doe", role: "user" },
  },
];

const completeManager = createCompleteStateManager(appActions, {
  user: null,
  isLoggedIn: false,
});

console.log("完整狀態管理器:", {
  currentState: completeManager.states[completeManager.currentIndex],
  stateHistory: completeManager.states,
  debugSnapshots: completeManager.snapshots,
  auditTrail: completeManager.auditTrail,
  performance: { totalTime: completeManager.totalExecutionTime },
});

章節提問

解決 null 的方法?

使用不具有 null 的值
使用 filter 過濾掉 null

map() 和 filter() 的差異?

map() 會將每個元素轉換為新形式，生成新陣列。
filter() 會過濾掉不符合條件的元素，生成新陣列。

Ch13. 串連函數式工具

章節重點

將函數式工具串起來成為多步驟的鏈式操作 (chain) 來處理複雜的任務
每個步驟是一項簡單操作，易讀、易撰寫。
為了寫函式鏈的下一步，有時需先產生新資料或擴增既有的資料。
- 將隱性訊息，表示成顯性資料。

鏈式操作：將多個函數串接在一起，形成一個新的函數。

當業務情境變得複雜，需要將 map()、filter()、reduce() 串接在一起，形成一個新的函數。

13.1 計算高消費力的最高消費金額

總顧客: 10 人 -> 高消費 4 人 / 低消費 6 人
需要找到高消費力，且消費次數 3 次以上的顧客，並計算其最高消費金額

function biggestPurchaseBestCustomers(customers) {
  var bestCustomers = filter(customers, function (customer) {
    return customer.purchases.length > 3;
  });

  var biggestPurchases = map(bestCustomers, function (customer) {
    return reduce(
      customer.purchases,
      { total: 0 },
      function (biggestSoFar, purchase) {
        if (biggestSoFar.total > purchase.total) {
          return biggestSoFar;
        } else {
          return purchase;
        }
      }
    );
  });

  return biggestPurchases;
}

問題：巢狀結構不好理解

命名高階函數

function biggestPurchaseBestCustomers(customers) {
  var bestCustomers = selectBestCustomers(customers); // 拆成高階函數
  var biggestPurchases = getBiggestPurchase(bestCustomers); // 拆成高階函數
  return biggestPurchases;
}

function isBestCustomer(customer) {
  return customer.purchases.length > 3;
}

function findBiggestPurchase(biggestSoFar, purchase) {
  if (biggestSoFar.total > purchase.total) {
    return biggestSoFar;
  } else {
    return purchase;
  }
}

function getBiggestPurchase(customer) {
  return reduce(customer.purchases, { total: 0 }, findBiggestPurchase);
}

缺點：如果直接命名成高階函數，無法重複使用。

依照步驟，命名回呼函數

function biggestPurchaseBestCustomers(customers) {
  var bestCustomers = filter(customers, isBestCustomer);
  var biggestPurchases = map(bestCustomers, getBiggestPurchase);
  return biggestPurchases;
}

// 將複雜的業務邏輯拆成回呼函數，供 filter()、map()、reduce() 使用
function isBestCustomer(customer) {
  return customer.purchases.length > 3;
}

function findBiggestPurchase(biggestSoFar, purchase) {
  if (biggestSoFar.total > purchase.total) {
    return biggestSoFar;
  } else {
    return purchase;
  }
}

function getBiggestPurchase(customer) {
  return reduce(customer.purchases, { total: 0 }, findBiggestPurchase);
}

13.5 找出只消費過一次的顧客，以陣列回傳顧客的 email

可以使用哪些 javascript 高階函數？

點我看答案

答案是：你可以用 map()、filter()

完整程式碼

點我看答案

function getFirstTimersEmails(customers) {
  var firstTimers = filter(customers, function (customer) {
    return customer.purchases.length === 1;
  });

  var firstTimersEmails = map(firstTimers, function (customer) {
    return customer.email;
  });

  return firstTimersEmails;
}

流融合(stream fusion)

var names = map(customers, getFullName);
var nameLengths = map(names, getLength);

流融合：

var nameLengths = map(map(customers, getFullName), getLength);

13.6 當 for loop 難以重構時

理解並重寫

先讀懂 for loop 的目的
再用函數式工具重構

依照線索進行重構、細化步驟

無法讀懂 for loop 的邏輯
根據 for loop 展示的低階訊息
轉換成函式鏈

情境問題: 已經有 for loop，要如何重構？

var answer = [];

var window = 5;

for (var i = 0; i < numbers.length; i++) {
  var sum = 0;
  var count = 0;

  for (var w = 0; w < window; w++) {
    sum += numbers[i + w];
    count++;
  }

  answer.push(sum / count);
}

如何重構？

點我看提示

外層迴圈：外層迴圈將 array 陣列的每個元素取出做處理
- 使用 map()
內層迴圈：內層迴圈將 array 陣列的每個元素累進成一個值
- 使用 reduce()

重構後的程式碼

// 可以重複使用的函數
function range(start, end) {
  var ret = [];

  for (var i = start; i < end; i++) {
    ret.push(i);
  }

  return ret;
}

// 基於 functional programming 的程式碼
var window = 5;

var indices = range(0, numbers.length);
var subarrays = map(indices, function (i) {
  return range(i, i + window);
});

var answer = map(subarrays, average);

走訪丟失的購物車

var itemAdded = ["shirt", "shoes", "pants","shirts"];

var shippingCart = reduce(itemAdded, {}, addOne);

function addOne(shippingCart, item) {
  if(!cart[item]){
   return additem...
  }
}

各種練習題們

13.13 情境：顧客購物車資料遺失，網站以陣列形式記錄所有曾出現過的商品

var itemAdded = ["shirt", "shoes", "pants", "shirts"];

情境問題：使用者的資料丟失，是否可以用上述的陣列，重建購物車？

點我看答案

var itemAdded = ["shirt", "shoes", "pants", "shirts"];

var shippingCart = reduce(itemAdded, {}, addOne);

function addOne(shippingCart, item) {
  if(!cart[item]){
   return additem...
  }
}

練習 13-16~19：棒球練習賽系列

情境問題：Mega Mart 每年都會派員參加職業棒球比賽，教練會根據員工名單生成員工建議位置清單，並將評估資料存入 evaluations 陣列。

┌─────────────────┐
│ Megamart 員工名單 │
└─────────┬───────┘
          │
          ↓
┌─────────────────┐
│ recommendations │
│   (推薦名單)     │
└─────────┬───────┘
          │
          ↓
┌─────────────────┐
│ evaluations 陣列 │
│   (評估資料)     │
└─────────┬───────┘
          │
          ↓
┌─────────────────┐     ┌─────────────────┐
│   參賽人員名單   │────→│   先發選手名單    │
└─────────────────┘     └─────────┬───────┘
                                  ↑
                                  │
                        ┌─────────────────┐
                        │  教練給每個人的   │
                        │      建議       │
                        └─────────────────┘

var evaluations = [
  {
    name: "John",
    position: "catcher",
    score: 2,
  },
  {
    name: "Jane",
    position: "pitcher",
    score: 3,
  },
  {
    name: "Jim",
    position: "pitcher",
    score: 1,
  },
];

var roaster = {
  pitcher: "Jane",
  catcher: "John",
};

如何根據 evaluations 陣列，生成 roster 物件？
其他問題略，請見書本。

章節提問

麼是流融合？

將 map、filter 等涵式串接再一起，避免過多中繼程式

Grokking Simplicity FP CH12 ~ CH13

張三

目錄

《簡約的軟體開發思維：用 Functional Programming 重構程式》CH12 ~ CH13

目錄

Ch12. 利用函數走訪

1. `map()`

2. `filter()`

如何整理過於複雜得篩選邏輯：回呼函式！

3. `reduce()`

`reduce()` 的常見用途：

章節提問

Ch13. 串連函數式工具

章節重點

鏈式操作：將多個函數串接在一起，形成一個新的函數。

13.1 計算高消費力的最高消費金額

問題：巢狀結構不好理解

13.5 找出只消費過一次的顧客，以陣列回傳顧客的 email

流融合(stream fusion)

13.6 當 for loop 難以重構時

重構後的程式碼

各種練習題們

13.13 情境：顧客購物車資料遺失，網站以陣列形式記錄所有曾出現過的商品

練習 13-16~19：棒球練習賽系列

章節提問

探索更多精彩內容

Grokking Simplicity FP CH12 ~ CH13

張三

目錄

目錄

《簡約的軟體開發思維：用 Functional Programming 重構程式》CH12 ~ CH13

目錄

Ch12. 利用函數走訪

1. map()

2. filter()

如何整理過於複雜得篩選邏輯：回呼函式！

3. reduce()

reduce() 的常見用途：

章節提問

Ch13. 串連函數式工具

章節重點

鏈式操作：將多個函數串接在一起，形成一個新的函數。

13.1 計算高消費力的最高消費金額

問題：巢狀結構不好理解

13.5 找出只消費過一次的顧客，以陣列回傳顧客的 email

流融合(stream fusion)

13.6 當 for loop 難以重構時

重構後的程式碼

各種練習題們

13.13 情境 ：顧客購物車資料遺失，網站以陣列形式記錄所有曾出現過的商品

練習 13-16~19：棒球練習賽系列

章節提問

探索更多精彩內容

1. `map()`

2. `filter()`

3. `reduce()`

`reduce()` 的常見用途：

13.13 情境：顧客購物車資料遺失，網站以陣列形式記錄所有曾出現過的商品