zoukankan      html  css  js  c++  java
  • [Javascript] Multiply Two Arrays over a Function in JavaScript

    Just like the State ADT an Array is also an Applicative Functor. That means we can do the same tricks with liftA2 with Array that we have been doing with State.

    While the Applicative aspect of State allows use to combine multiple stateful transitions over a function, Array allows us to create a new Array that contains the results of calling every permutation of each element in two arrays over a function. We will use this ability to pull from two separate locations in our AppStateand generate an Array of Cards.

    liftA2 from crocks.js "Ever see yourself wanting to map a binary or trinary function, but map only allows unary functions? Both of these functions allow you to pass in your function as well as the number of Applicatives (containers that provide bothof and apfunctions) you need to get the mapping you are looking for."

    lift from Ramda.js "lifts" a function of arity > 1 so that it may "map over" a list, Function or other object that satisfies the FantasyLand Apply spec.

    // Crocks.js
    
    const buildCard = curry((color, shape) => console.log(color, shape) || ({
        id: `${color}-${shape}`,
        color,
        shape
    }));
    // buildCards :: [String] -> [String] -> [Card]
    const buildCards = liftA2(buildCard);
    
    
    // Ramda.js
    
    const build = lift(buildCard);

     --- 

    const {prop,assoc, pick, State, identity, omit, curry, filter, converge,map, composeK, liftA2, equals, constant,option, chain, mapProps, find, propEq, isNumber, compose, safe} = require('crocks');
    const  {get, modify, of} = State; 
    const {lift} = require('ramda');
    
    const state = {
        colors: [ 'orange', 'green', 'blue', 'yellow' ],
        shapes: [ 'square', 'triangle', 'circle' ]
      };
    
    const getState = (key) => get(prop(key))
    const getColors = () => getState('colors').map(option([]));
    const getShapes = () => getState('shapes').map(option([]));
    
    const buildCard = curry((color, shape) => console.log(color, shape) || ({
        id: `${color}-${shape}`,
        color,
        shape
    }));
    // buildCards :: [String] -> [String] -> [Card]
    const buildCards = liftA2(buildCard);
    const generateCards = converge(
        liftA2(buildCards),
        getColors,
        getShapes
    )
    console.log(
        generateCards()
            .evalWith(state)
    )
    
    const build = lift(buildCard);
    console.log('build', build([1,2,3], ['c', 'd']))
  • 相关阅读:
    LeetCode子集问题
    面试题-求最大字典区间
    链表快速排序
    树的非递归遍历
    快速排序非递归实现
    leetcode217 python3 72ms 存在重复元素
    leetcode121 C++ 12ms 买股票的最佳时机 只能买卖一次
    leetcode1 python3 76ms twoSum 360面试题
    leetcode485 python3 88ms 最大连续1的个数
    leetcode119 C++ 0ms 杨辉三角2
  • 原文地址:https://www.cnblogs.com/Answer1215/p/10277748.html
Copyright © 2011-2022 走看看