k*****u 发帖数: 1688 | 1 【 以下文字转载自 JobHunting 讨论区 】
发信人: killniu (I shall be your eyes!), 信区: JobHunting
标 题: 请教个javascript的问题
发信站: BBS 未名空间站 (Sun Oct 2 22:41:56 2016, 美东)
如果有一个list的data,[[v1, w1, t1], [v2, w2, t2], [v3, w3, t3], [v4, w4, t4
]...]
怎么能把它变成两个list, s1 = [[v1, t1], [v2, t2], [v3, t3], [v4, t4]...], s2
= [[w1, t1], [w2, t2], [w3, t3], [w4, t4]...]
如果用$.each(),应该怎么做
谢谢 | s***o 发帖数: 2191 | 2 yourArray.map(a => {...}) | m*********a 发帖数: 3299 | 3 var newArr = arr.map(function(a){ return [a[0],a[2]];}); | g***a 发帖数: 271 | 4 data =[[v1, w1, t1], [v2, w2, t2], [v3, w3, t3], [v4, w4, t4
]...];
$.each(function(tri) {
s1.push([tri[0], tri[2]]);
s2.push([tri[1], tri[2]]);
}
output:
s1 = [[v1, t1], [v2, t2], [v3, t3], [v4, t4]...],
s2 = [[w1, t1], [w2, t2], [w3, t3], [w4, t4]...]
哈哈哈,so easy | T*******x 发帖数: 8565 | 5 Python:
z=zip(*data)
s1=zip(z[0],z[2])
s2=zip(z[1],z[2])
【在 g***a 的大作中提到】 : data =[[v1, w1, t1], [v2, w2, t2], [v3, w3, t3], [v4, w4, t4 : ]...]; : $.each(function(tri) { : s1.push([tri[0], tri[2]]); : s2.push([tri[1], tri[2]]); : } : output: : s1 = [[v1, t1], [v2, t2], [v3, t3], [v4, t4]...], : s2 = [[w1, t1], [w2, t2], [w3, t3], [w4, t4]...] : 哈哈哈,so easy
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