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

试试吧

RemoveItems.forEach((i, j) => {
    OriginalItems.splice((i - j), 1);
});

当您执行.splice()时,数组正在重新索引,这意味着当一个索引被删除时,您将跳过一个索引,并且缓存的.length已过时。

要修复它,你要么需要在.splice()后面递减i,要么简单地反向迭代…

var i = Auction.auctions.length
while (i--) {
    ...
    if (...) { 
        Auction.auctions.splice(i, 1);
    } 
}

这样,重新索引就不会影响迭代中的下一项,因为索引只影响从当前点到数组末尾的项,并且迭代中的下一项低于当前点。

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。

这是一个很常见的问题。解决方案是反向循环:

for (var i = Auction.auctions.length - 1; i >= 0; i--) {
    Auction.auctions[i].seconds--;
    if (Auction.auctions[i].seconds < 0) { 
        Auction.auctions.splice(i, 1);
    }
}

如果你把它们从末端取出来也没关系因为下标会在逆向过程中保留下来。

举两个例子:

一个例子

// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    for (var l = temp_products_images.length; l--;) {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}

两个例子

// Remove from Listing the Items Checked in Checkbox for Delete
let temp_products_images = store.state.c_products.products_images
if (temp_products_images != null) {
    let l = temp_products_images.length
    while (l--)
    {
        // 'mark' is the checkbox field
        if (temp_products_images[l].mark == true) {
            store.state.c_products.products_images.splice(l,1);         // THIS WORKS
            // this.$delete(store.state.c_products.products_images,l);  // THIS ALSO WORKS
        }
    }
}