我有以下for循环，当我使用splice()删除一个项目时，我得到'seconds'是未定义的。我可以检查它是否未定义，但我觉得可能有一种更优雅的方式来做到这一点。他们的愿望是简单地删除一个项目，然后继续前进。
for (i = 0, len = Auction.auctions.length; i < len; i++) {
auction = Auction.auctions[i];
Auction.auctions[i]['seconds'] --;
if (auction.seconds < 0) {
Auction.auctions.splice(i, 1);
}
}
当前回答
你可以浏览一下，然后使用shift()
其他回答
尝试在循环时将数组中继到newArray:
var auctions = Auction.auctions;
var auctionIndex;
var auction;
var newAuctions = [];
for (
auctionIndex = 0;
auctionIndex < Auction.auctions.length;
auctionIndex++) {
auction = auctions[auctionIndex];
if (auction.seconds >= 0) {
newAuctions.push(
auction);
}
}
Auction.auctions = newAuctions;
如果你正在使用ES6+ -为什么不直接使用数组。过滤方法?
Auction.auctions = Auction.auctions.filter((auction) => {
auction['seconds'] --;
return (auction.seconds > 0)
})
注意，在过滤器迭代过程中修改数组元素只对对象有效，而对基元值数组无效。
试试吧
RemoveItems.forEach((i, j) => {
OriginalItems.splice((i - j), 1);
});
你可以浏览一下，然后使用shift()
这是这个简单线性时间问题的一个简单线性时间解。
当我运行这个代码片段时，n = 100万，每次调用filterInPlace()需要0.013到0.016秒。一个二次解(例如，公认的答案)将需要它的一百万倍左右。
// Remove from array every item such that !condition(item). function filterInPlace(array, condition) { var iOut = 0; for (var i = 0; i < array.length; i++) if (condition(array[i])) array[iOut++] = array[i]; array.length = iOut; } // Try it out. A quadratic solution would take a very long time. var n = 1*1000*1000; console.log("constructing array..."); var Auction = {auctions: []}; for (var i = 0; i < n; ++i) { Auction.auctions.push({seconds:1}); Auction.auctions.push({seconds:2}); Auction.auctions.push({seconds:0}); } console.log("array length should be "+(3*n)+": ", Auction.auctions.length) filterInPlace(Auction.auctions, function(auction) {return --auction.seconds >= 0; }) console.log("array length should be "+(2*n)+": ", Auction.auctions.length) filterInPlace(Auction.auctions, function(auction) {return --auction.seconds >= 0; }) console.log("array length should be "+n+": ", Auction.auctions.length) filterInPlace(Auction.auctions, function(auction) {return --auction.seconds >= 0; }) console.log("array length should be 0: ", Auction.auctions.length)
注意，这只是修改原始数组，而不是创建一个新数组;这样做是有好处的，例如，在数组是程序的单一内存瓶颈的情况下;在这种情况下，您不希望创建另一个相同大小的数组，即使是临时的。
