从数组中删除某些值的最快方法

Question

我有一个看起来像这样的数组：

const data =  [
    ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '04b073133c7843248a7a3dbc968f75a0', 'network1', 'affiliate',
        1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575],
    ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '4420cc109ec54214b68edc906b18e44a', 'network1', 'affiliate',
        1141338.0, 18164.0, 0.75, 0.67, 5.58, 0.5625, 0.129375, 0.691875, 4.185, 0.96255, 5.14755, 0.5025, 0.115575, 0.618075],
    ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '90a7cbf1cf4e4043889626c4119d4b4d', 'network1', 'affiliate',
        1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575],
    ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '0f04f1ff385541d3a8d9ea2f0d85482b', 'network1', 'affiliate',
        1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575],
    ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '0f04f1ff385541d3a8d9ea2f0d85482b', 'network1', 'affiliate',
        1232113, 1232133, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
];

我想要实现（但不知道如何）是删除所有看起来像这个数组中最后一个条目的条目（以最快的方式），例如[Wed Sep 20 09:00:00 GMT+02:00 2017, SKYSCANNER, 0f04f1ff385541d3a8d9ea2f0d85482b, network1, affiliate, 1232113, 1232133, , 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

基本上，如果第 8 个位置之后的所有值都为零，则不需要该条目并且必须删除该条目。

通常在我的情况下，这样的数组可以有 5-15k 个条目，所以我想知道实现这一目标的最快方法是什么？ 有人可以提供一个工作片段吗？

谢谢！

Answer 1

以下在我的旧机器上需要 11 毫秒。 正如 Cerbrus 所写，只需使用过滤器，15k 并不多：

 const data = [ ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '04b073133c7843248a7a3dbc968f75a0', 'network1', 'affiliate', 1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '4420cc109ec54214b68edc906b18e44a', 'network1', 'affiliate', 1141338.0, 18164.0, 0.75, 0.67, 5.58, 0.5625, 0.129375, 0.691875, 4.185, 0.96255, 5.14755, 0.5025, 0.115575, 0.618075], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '90a7cbf1cf4e4043889626c4119d4b4d', 'network1', 'affiliate', 1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '0f04f1ff385541d3a8d9ea2f0d85482b', 'network1', 'affiliate', 1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '0f04f1ff385541d3a8d9ea2f0d85482b', 'network1', 'affiliate', 1232113, 1232133, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]; let test = []; for (let i = 0; i < 3000; i += 1) { test = test.concat(data); } const start = performance.now(); const result = test.filter( // reject entry, if not // every value from 8th to last // is 0 (entry) => !entry.slice(8).every((val) => 0 === val) ); console.log(performance.now() - start);

Answer 2

如果您想要快速代码，只需检查单独的值。

or ( || ) 将短路，因此e[7] - e[18]中具有值的第一个条目将使该行通过测试，这意味着filter将移至下一行。

 const data = [ ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '04b073133c7843248a7a3dbc968f75a0', 'network1', 'affiliate', 1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '4420cc109ec54214b68edc906b18e44a', 'network1', 'affiliate', 1141338.0, 18164.0, 0.75, 0.67, 5.58, 0.5625, 0.129375, 0.691875, 4.185, 0.96255, 5.14755, 0.5025, 0.115575, 0.618075], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '90a7cbf1cf4e4043889626c4119d4b4d', 'network1', 'affiliate', 1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '0f04f1ff385541d3a8d9ea2f0d85482b', 'network1', 'affiliate', 1141338.0, 18164.0, 0.08, 0.07, 0.62, 0.06, 0.0138, 0.0738, 0.465, 0.10695, 0.57195, 0.0525, 0.012075, 0.064575], ['Wed Sep 20 09:00:00 GMT+02:00 2017', 'SKYSCANNER', '0f04f1ff385541d3a8d9ea2f0d85482b', 'network1', 'affiliate', 1232113, 1232133, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]; let test = []; for (let i = 0; i < 3000; i += 1) { test = test.concat(data); } const start = performance.now(); const result = test.filter(e => e[7]||e[8]||e[9]||e[10]||e[11]||e[12]||e[13]||e[14]||e[15]||e[16]||e[17]||e[18]); console.log(performance.now() - start);

（从 Yoshi 的回答中借用的时间代码）

（对我来说，）它的运行速度是 Yoshi 使用的slice / every组合的两倍，但可读性/可维护性只有一半。

您是否更喜欢可读的代码，或者您是否想将所有指南都抛到窗外以追求原始速度，这取决于您。