Question

考虑以下任务：

我们列出了不同欧洲城镇的日平均气温。

{ Hamburg: [14, 15, 16, 14, 18, 17, 20, 11, 21, 18, 19,11 ],
  Munich: [16, 17, 19, 20, 21, 23, 22, 21, 20, 19, 24, 23],
  Madrid: [24, 23, 20, 24, 24, 23, 21, 22, 24, 20, 24, 22],
  Stockholm: [16, 14, 12, 15, 13, 14, 14, 12, 11, 14, 15, 14],
  Warsaw: [17, 15, 16, 18, 20, 20, 21, 18, 19, 18, 17, 20] }

我们希望将这些城镇分为两组：“温暖”和“热”。 “温暖”应该是温度大于19的城镇，至少有3天。“热” 应该是每天温度超过19的城镇。

我最终做的是：

const _ = require('lodash');

let cities = {
    Hamburg: [14, 15, 16, 14, 18, 17, 20, 11, 21, 18, 19,11 ],
    Munich: [16, 17, 19, 20, 21, 23, 22, 21, 20, 19, 24, 23],
    Madrid: [24, 23, 20, 24, 24, 23, 21, 22, 24, 20, 24, 22],
    Stockholm: [16, 14, 12, 15, 13, 14, 14, 12, 11, 14, 15, 14],
    Warsaw: [17, 15, 16, 18, 20, 20, 21, 18, 19, 18, 17, 20]
};

let isHot = (degrees) => {
  return degrees > 19;
};

function getMinCategories(cities) {

    let res = {
        hot: [],
        warm: []
    };
    _.forEach(cities, function(val,key) {
      if(_.every(val, isHot)){
        res.hot.push(key);
      } else if(_.sumBy(val, degree => isHot(degree) ? 1 : 0) > 2){
        res.warm.push(key);
      }

    });
    return res;
}

console.log(getMinCategories(cities)); // prints { hot: [ 'Madrid' ], warm: [ 'Munich', 'Warsaw' ] } which is correct

有没有更优雅的方法来检查“至少3天温度＆gt; 19”而不是使用_.sumBy功能？也许使用_.some()？

Answer 1

我已经包含了一个香草js解决方案和一个lodash一个。

Vanilla JS

您可以使用过滤器计算符合条件的天数：

tempaturesArray.filter((t) => t > 19).length

你可以使用Array#reduce

来使用vanilla JS

const result = Object.keys(cities).reduce(( obj, city ) => {
  const days = cities[city].filter((t) => t > 19).length;
  const climate = days === cities[city].length ? 'hot' : (days >= 3 ? 'warm' : null);

  climate && obj[climate].push(city);

  return obj;
}, { hot: [], warm: []});

const cities = {
  Hamburg: [14, 15, 16, 14, 18, 17, 20, 11, 21, 18, 19,11 ],
  Munich: [16, 17, 19, 20, 21, 23, 22, 21, 20, 19, 24, 23],
  Madrid: [24, 23, 20, 24, 24, 23, 21, 22, 24, 20, 24, 22],
  Stockholm: [16, 14, 12, 15, 13, 14, 14, 12, 11, 14, 15, 14],
  Warsaw: [17, 15, 16, 18, 20, 20, 21, 18, 19, 18, 17, 20]
};

const result = Object.keys(cities).reduce(( obj, city ) => {
  const days = cities[city].filter((t) => t > 19).length;
  const climate = days === cities[city].length ? 'hot' : (days >= 3 ? 'warm' : null);
  
  climate && obj[climate].push(city);
  
  return obj;
}, { hot: [], warm: []});
        
console.log(result);

<强> lodash

使用_.mapValues()将数组转换为冷/热/热字符串，然后使用_.invertBy()将值切换为键，并收集数组中国家/地区的名称。我已使用_.sumBy()来计算天数，但删除了_.every()，因此一次通过将计算两种气候：

const result = _(cities)
  .mapValues((temperature, country) => {
    const days = _.sumBy(temperature, (t) => t > 19);
    return days === temperature.length ? 'hot' : (days >= 3 ? 'warm' : 'cold');
  })
  .invertBy()
  .pick(['warm', 'hot'])
  .value();

const cities = {
  Hamburg: [14, 15, 16, 14, 18, 17, 20, 11, 21, 18, 19,11 ],
  Munich: [16, 17, 19, 20, 21, 23, 22, 21, 20, 19, 24, 23],
  Madrid: [24, 23, 20, 24, 24, 23, 21, 22, 24, 20, 24, 22],
  Stockholm: [16, 14, 12, 15, 13, 14, 14, 12, 11, 14, 15, 14],
  Warsaw: [17, 15, 16, 18, 20, 20, 21, 18, 19, 18, 17, 20]
};

const result = _(cities)
  .mapValues((temperature, country) => {
    const days = _.sumBy(temperature, (t) => t > 19);
    return days === temperature.length ? 'hot' : (days >= 3 ? 'warm' : 'cold');
  })
  .invertBy()
  .pick(['warm', 'hot'])
  .value();
        
console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>

Answer 2

您可以像这样使用pickBy：

＆＃13;

var cities = {
  "Hamburg":   [14,15,16,14,18,17,20,11,21,18,19,11],
  "Munich":    [16,17,19,20,21,23,22,21,20,19,24,23],
  "Madrid":    [24,23,20,24,24,23,21,22,24,20,24,22],
  "Stockholm": [16,14,12,15,13,14,14,12,11,14,15,14],
  "Warsaw":    [17,15,16,18,20,20,21,18,19,18,17,20]
}

var hotCities = _(cities)
  .pickBy(temps => _(temps).map(x => x > 19).every())
  .value();
var warmCities = _(cities)
  .pickBy((temps, name) => hotCities[name] === undefined) // already hot
  .pickBy(temps => _(temps).filter(x => x > 19).size() >= 3)
  .value();

console.log(hotCities)
console.log(warmCities)

＆＃13;

<script src="https://cdn.jsdelivr.net/lodash/4.17.2/lodash.min.js"></script>

＆＃13;

Answer 3

如何创建一个mixin来封装'至少x号在集合中的逻辑'必须为真'：

_.mixin( { 'atLeast': function(list, times, predicate){
    return _.filter(list, function(item){
        return predicate(item);
    }).length >= times;
}});

然后可用于过滤热门城市：

// helper function to work out if a city is hot
let hotCity = city => _.atLeast(city, 3, isHot)

let hotCities = _.chain(cities)
    .pickBy(hotCity)
    .keys()
    .value();

Answer 4

使用过滤器：

Work

Answer 5

您可以使用_.reduce和_.filter

执行此操作

_.reduce(cities, (result, temps, town) => {
    return _.chain(temps)
        .filter((temp) => {
            return temp > 19;
        })
        .thru((filtTemps) => {
            let kind;
            kind = filtTemps.length >= 3 ? 'warm' : kind;
            kind = filtTemps.length === temps.length ? 'hot' : kind;
            if (!_.isUndefined(kind)) {
                 result[kind].push(town);
            }
            return result;
        })
        .value();

}, {
    hot: [],
    warm: []
});

lodash - 带有条件/计数器的_.some（）

5 个答案: