var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];
var oddEvenCounts = numbers.reduce(function(counts, number) {
if (isOdd(number)) {
counts[odd]++;
} else {
counts[even]++;
}
}, {});
我正在寻找这段代码中的错误(仍在学习reduce方法;)) - 我哪里出错?
答案 0 :(得分:7)
带注释的工作代码:
var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];
function isOdd(n) {
return !!(n % 2);
}
var oddEvenCounts = numbers.reduce(function(counts, number) {
if (isOdd(number)) {
counts.odd++; // use dot notation or ['odd']
} else {
counts.even++; // use dot notation or ['even']
}
return counts; // return the accumulator
}, { odd: 0, even: 0 }); // set the initial values of odd and even
console.log(oddEvenCounts);
您可以使用括号表示法和三元运算符来缩短代码:
var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];
function isOdd(n) {
return !!(n % 2);
}
var oddEvenCounts = numbers.reduce(function(counts, number) {
counts[isOdd(number) ? 'odd' : 'even']++;
return counts;
}, { odd: 0, even: 0 });
console.log(oddEvenCounts);
答案 1 :(得分:1)
返回你的累加器:
var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];
var oddEvenCounts = numbers.reduce(function(counts, number) {
if (isOdd(number)) {
counts[odd]++;
} else {
counts[even]++;
}
return counts;
}, {});