Desafio 1
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//dado o array de dictionary abaixo some o total dos valores | |
let dictionary = [{ '5': 50, '6': 60 }, { 'A': 10 }, { 'YEEZY': 30 }] | |
dictionary.reduce((acc, val) => acc.concat(Object.values(val)), []).reduce((acc, val) => acc + val, 0) |
const mapObject = (mapper, obj) =>
obj.reduce( (values, obj) => [...values, mapper(obj) ], [])
const sum = (a, b) => a + b
mapObject(Object.values, dictionary).reduce(sum, 0)
@halan Otima solução! Valeu
dictionary.map( Object.values ).reduce(sum, 0)
const mapObject = (mapper, obj) =>
obj.reduce( (values, obj) => [...values, mapper(obj) ], [])
const sum = (a, b) => a + b
const sumValues = (obj) =>
mapObject(Object.values, obj).reduce(sum, 0)
dictionary.map(sumValues).reduce(sum, 0)
const dictionary = [{ '5': 50, '6': 60 }, { 'A': 10 }, { 'YEEZY': 30 }],
sum = (valores) => valores.reduce((a, b) => a + b),
valores = dictionary.map(valor => sum(Object.values(valor))).reduce((a,b) => a + b)
Sem reduce
e sem callbacks (mas com eval
).
eval(JSON.stringify(dictionary).match(/\d+(?!")/g).join('+'));
const obj = [{ '5': 50, '6': 60 }, { 'A': 10 }, { 'YEEZY': 30 }]; const arrNumbers = () => obj.map(_a => Object.values(_a)); const onlyNumbers = () => [].concat(...arrNumbers()); const result = () => onlyNumbers().reduce((_a, _b) => _a + _b, 0);
function sum_items(list_items) {
var sum = 0;
list_items.map(
function (item) {
for (var key in item) {
sum += item[key];
}
}
);
return sum;
}
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halan commentedMar 9, 2017