[英]Random numbers and floor vs round function
为什么我使用随机数生成器并且范围为0 - 9我没有得到与将其与floor功能相结合的相同均匀分布?
Math.floor(Math.random() * 10)
gives quite uniform distribution, while Math.round(Math.random() * 10)
doesn't. Math.floor(Math.random() * 10)
给出了非常均匀的分布,而Math.round(Math.random() * 10)
则没有。
Math.floor() returns 0 for any value in the range [0, 1) (1 exclusive), 1 for any value in the range [1, 2), etc. Math.floor()为范围[0,1)中的任何值(1不包括)返回0,对于范围[1,2]中的任何值返回1,等等。
So if we have equal chances of getting a number in one of these ranges, we will get an equal distribution of 0's and 1's. 因此,如果我们在这些范围之一中获得数字的机会相等,我们将得到0和1的平均分布。
Math.round(), however, returns 0 for values under 0.5, 1 for values under 1.5, etc. 但是,Math.round()对于0.5以下的值返回0,对于1.5以下的值返回1等。
So we actually have half the chances of getting a 0, as only values from 0 to 0.5 will round to 0. 所以我们实际上有一半获得0的机会,因为只有从0到0.5的值将舍入为0。
╔═════════╦═════════╦═════════╗
║ Range ║ floor() ║ round() ║
╠═════════╬═════════╬═════════╣
║ [0, 1) ║ 0 ║ 0 or 1 ║
║ [1, 2) ║ 1 ║ 1 or 2 ║
║ [2, 3) ║ 2 ║ 2 or 3 ║
║ ... ║ ... ║ ... ║
║ [9, 10) ║ 9 ║ 9 or 10 ║
╚═════════╩═════════╩═════════╝
I really like to trigger 31-bit int instead of Math.floor()
, by doing a binary "or 0" operation. 我真的想通过执行二进制“或0”操作来触发31位int而不是
Math.floor()
。 It makes it slightly faster (stack vs heap,) seems a tad neater, and does the same thing, in a practical sense. 它使得它稍微更快(堆栈与堆,)似乎有点整洁,并在实际意义上做同样的事情。 Here is an example that gets a random element of an array:
这是一个获取数组的随机元素的示例:
var ar=[1,2,3,4,5,6,7,8,9,0,'a','b','c','d']
console.log(ar[(Math.random() * ar.length) |0])
This might seem like premature optimization, but as I said, I like how it looks, and it takes less typing than Math.floor()
. 这似乎是过早优化,但正如我所说,我喜欢它的外观,并且它比
Math.floor()
需要更少的输入。 Using Math.round()
would require an extra step (because the spread is 1
to ar.length
not 0
to ar.length-1
) 使用
Math.round()
需要一个额外的步骤(因为扩展为1
到ar.length
而不是0
到ar.length-1
)
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