javascript 循环遍历多个数组，组合成单个数组，维护子数组内的索引
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Loop through multiple arrays, combining to single array,maintaining indexes within the subarray
提问by TommyBs
Bit of a long winded title but I'll try to explain what I want to do. Basically I've got a number of arrays that I want to combine into a single array. The trouble is, I need to loop through the items in the subarrays and add them 1 at a time and maintain the order. The end goal is I want to display the data back paged. I've got a simple example below that I will use to try and convey what I mean. This isn't an alphabetical sort i.e hshould not be before iin the bottom example.
有点冗长的标题，但我会尽力解释我想做什么。基本上我有许多数组，我想将它们组合成一个数组。问题是，我需要遍历子数组中的项目并一次添加 1 并保持顺序。最终目标是我想显示分页的数据。我在下面有一个简单的例子，我将用它来尝试传达我的意思。这不是按字母顺序排序，即在底部示例中h不应在i之前。
So in my example I know I want 3 pages of results. The first page should have 4 items, the second page 4 items and the third only 1 item.
所以在我的例子中，我知道我想要 3 页的结果。第一个页面应该有 4 个项目，第二个页面应该有 4 个项目，第三个只有 1 个项目。
I can do the final paging myself as I will have an array of all inner items at the end of it "mix", but I can't work out how to loop through my arrays and add them in how I need.
我可以自己做最后的分页，因为我将在它“混合”的末尾有一个包含所有内部项目的数组，但我无法弄清楚如何遍历我的数组并按照我需要的方式添加它们。
I've got the page variable upfront but I'm not sure how to structure the loop. I think I basically need to loop through each subarray and pop() the first item off, then loop through the next one, pop() the first item and so forth. But somewhere I need to check how many items are left in each subarray.
我预先获得了页面变量，但我不确定如何构建循环。我想我基本上需要遍历每个子数组并 pop() 关闭第一项，然后遍历下一个， pop() 第一项等等。但是在某处我需要检查每个子数组中还剩下多少项。
For instance if I only had array "one" I would in theory have 2 pages the first containing a,c,e,i and the second only k, this one is obviously simple enough as I just check the length of the only array.
例如，如果我只有数组“一个”，理论上我会有 2 页，第一页包含 a、c、e、i，第二页只有 k，这显然很简单，因为我只检查唯一数组的长度。
But if I added in another array "third" [1,2,3,4,5] then I would expect the mix array to be ['a','b',1,'c','d',2...etc]; Each of these arrays could in theory have different lengths so then I would obviously skip an empty value.
但是如果我添加另一个数组“第三个”[1,2,3,4,5] 那么我希望混合数组是 ['a','b',1,'c','d',2 ...等等]; 这些数组中的每一个理论上都可以有不同的长度，所以我显然会跳过一个空值。
var one = ['a','c','e','i','k'];
var two = ['b','d','f','h'];
var all = [one,two];
var pagecount = 3;
var mix = [];
for(var i = 0; i< all.length; i++){
var area = all[i];
}
// End result should be mix = ['a','b','c','d','e','f','i','h','k'];
I've tried to word this as best as I can, but I'm struggling to get my head around how to explain this myself! Unfortunately in the real world I have no control over the data/size of the data arrays.
我已经尽我所能地表达了这一点，但我正在努力弄清楚如何自己解释这一点！不幸的是，在现实世界中，我无法控制数据数组的数据/大小。
Any questions or if something is not clear then please leave a comment.
如有任何问题或不清楚的地方，请发表评论。
采纳答案by Andrew Clark
The following should work:
以下应该工作：
for (var i = 0; all.length !== 0; i++) {
var j = 0;
while (j < all.length) {
if (i >= all[j].length) {
all.splice(j, 1);
} else {
mix.push(all[j][i]);
j += 1;
}
}
}
On each iteration of the outer loop we increase iby one, this will be the index in each array to grab an item from. For the inner loop we will do one of the following:
在外循环的每次迭代中，我们增加i一，这将是每个数组中要从中获取项目的索引。对于内部循环，我们将执行以下操作之一：
- If the index
iis beyond the maximum index for the arrayall[j]we are done with that array so it is removed usingall.splice(j, 1). We do not advancejbecauseall[j]will refer to the next array after the previous element at that location was removed. - Otherwise we add the item
all[j][i]tomixand increasejby one to move to the next array on the next iteration.
- 如果索引
i超出数组的最大索引，all[j]我们就完成了该数组的处理，因此使用all.splice(j, 1). 我们不前进，j因为all[j]将在删除该位置的前一个元素后引用下一个数组。 - 否则，我们将项目添加
all[j][i]到mix并增加j一个以在下一次迭代中移动到下一个数组。
The outer loop doesn't stop until there are no arrays left in all, which will happen when ihas exceeded the length of the longest array.
外循环直到 中没有数组时才会停止all，当i超过最长数组的长度时会发生这种情况。
For example with three arrays all of different lengths:
例如，对于三个长度不同的数组：
var one = [1, 2, 3, 4];
var two = ['a', 'b'];
var three = ['U', 'V', 'W', 'X', 'Y', 'Z'];
var all = [one, two, three];
var mix = [];
// after running the above loop mix will have the following contents:
// [1, "a", "U", 2, "b", "V", 3, "W", 4, "X", "Y", "Z"]
回答by Guffa
Loop through the inner index in the outer loop, and the arrays in the inner loop:
循环遍历外循环中的内索引和内循环中的数组：
for (var i = 0, cont = true; cont; i++) {
cont = false;
for (j = 0; j < all.length; j++) {
if (i < all[j].length) {
mix.push(all[j][i]);
cont = true;
}
}
}
回答by jonhopkins
It looks like you want a simple merge function. You could do it like this
看起来您想要一个简单的合并功能。你可以这样做
var one = ['a','c','e','i','k'];
var two = ['b','d','f','h'];
var mix = new Array();
var merging = true;
var index = 0;
while (merging) {
merging = false;
if (index < one.length) {
mix[mix.length] = one[index];
merging = true;
}
if (index < two.length) {
mix[mix.length] = two[index];
merging = true;
}
// add if blocks for arrays three, four, etc...
index++;
}
This can be expanded for any number of arrays by simply adding another if block inside the loop
这可以通过简单地在循环内添加另一个 if 块来扩展任意数量的数组
