但是,您选择这样做,反向启动和倒计时最简单。这也取决于你的数组是否稀疏,如果你希望它保持稀疏。最简单的就是创建一个可重用的函数和你自己的库。你可以做到这一点。如果您将compress
设置为true,那么您的数组将变为连续而非稀疏数组。该函数将删除所有匹配的值,并返回一个已删除元素的数组。
的Javascript
function is(x, y) {
if (x === y) {
if (x === 0) {
return 1/x === 1/y;
}
return true;
}
var x1 = x,
y1 = y;
return x !== x1 && y !== y1;
}
function removeMatching(array, value /*, compress (default = false)*/) {
var removed = [],
compress = arguments[2],
index,
temp,
length;
if (typeof compress !== "boolean") {
compress = false;
}
if (compress) {
temp = [];
length = array.length;
index = 0;
while (index < length) {
if (array.hasOwnProperty(index)) {
temp.push(array[index]);
}
index += 1;
}
} else {
temp = array;
}
index = 0;
length = temp.length;
while (index < length) {
if (temp.hasOwnProperty(index) && is(temp[index], value)) {
if (compress) {
removed.push(temp.splice(index, 1)[0]);
} else {
removed.push(temp[index]);
delete temp[index];
}
}
index += 1;
}
if (compress) {
array.length = 0;
index = 0;
length = temp.length;
while (index < length) {
if (temp.hasOwnProperty(index)) {
array.push(temp[index]);
}
index += 1;
}
}
return removed;
}
var test = [];
test[1] = 1;
test[50] = 2;
test[100] = NaN;
test[101] = NaN;
test[102] = NaN;
test[200] = null;
test[300] = undefined;
test[400] = Infinity;
test[450] = NaN;
test[500] = -Infinity;
test[1000] = 3;
console.log(test);
console.log(removeMatching(test, NaN));
console.log(test);
console.log(removeMatching(test, Infinity, true));
console.log(test);
输出
[1: 1, 50: 2, 100: NaN, 101: NaN, 102: NaN, 200: null, 300: undefined, 400: Infinity, 450: NaN, 500: -Infinity, 1000: 3]
[NaN, NaN, NaN, NaN]
[1: 1, 50: 2, 200: null, 300: undefined, 400: Infinity, 500: -Infinity, 1000: 3]
[Infinity]
[1, 2, null, undefined, -Infinity, 3]
在jsfiddle
你检查这[http://stackoverflow.com/questions/9882284/looping-through-array-and-删除项目,没有破环for]回答? –
你可以让你的向前迭代的第一个例子,并把您的文章递减'i'直接在'.splice()'调用:'myarray.splice(我 - ,1);' –