从数组中获取最近的数字

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我有一个从负1000到+1000的数字,还有一个数组,里面有数字。就像这样:

[2, 42, 82, 122, 162, 202, 242, 282, 322, 362]

我想要得到的数字更改为数组中最近的数。

例如,我得到80我希望它能得到82...

提问于
用户回答回答于

ES5版本:

var counts = [4, 9, 15, 6, 2],
  goal = 5;

var closest = counts.reduce(function(prev, curr) {
  return (Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});

console.log(closest);

用户回答回答于

下面是应该转换为任何过程语言的伪代码:

array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
number = 112
print closest (number, array)

def closest (num, arr):
    curr = arr[0]
    foreach val in arr:
        if abs (num - val) < abs (num - curr):
            curr = val
    return curr

它简单地计算出给定数字和每个数组元素之间的绝对差异,并返回一个差异最小的数组元素。

对于示例值:

number = 112  112  112  112  112  112  112  112  112  112
array  =   2   42   82  122  162  202  242  282  322  362
diff   = 110   70   30   10   50   90  130  170  210  250
                         |
                         +-- one with minimal absolute difference.

作为概念的证明,下面是我用来演示这一点的Python代码:

def closest (num, arr):
    curr = arr[0]
    for index in range (len (arr)):
        if abs (num - arr[index]) < abs (num - curr):
            curr = arr[index]
    return curr

array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362]
number = 112
print closest (number, array)

如果你在Javascript中需要它,请参阅下面的完整HTML文件,它演示了实际的功能:

<html>
    <head></head>
    <body>
        <script language="javascript">
            function closest (num, arr) {
                var curr = arr[0];
                var diff = Math.abs (num - curr);
                for (var val = 0; val < arr.length; val++) {
                    var newdiff = Math.abs (num - arr[val]);
                    if (newdiff < diff) {
                        diff = newdiff;
                        curr = arr[val];
                    }
                }
                return curr;
            }
            array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
            number = 112;
            alert (closest (number, array));
        </script>
    </body>
</html>

现在请记住,如果你的数据项被排序(可以从示例数据中推断,但您没有显式地声明它),那么可能会有提高效率的余地。例如,可以使用二进制搜索来查找最接近的项。

如果你想要这样做(并且可以保证数组按升序排序),这是一个很好的起点:

<html>
    <head></head>
    <body>
        <script language="javascript">
            function closest (num, arr) {
                var mid;
                var lo = 0;
                var hi = arr.length - 1;
                while (hi - lo > 1) {
                    mid = Math.floor ((lo + hi) / 2);
                    if (arr[mid] < num) {
                        lo = mid;
                    } else {
                        hi = mid;
                    }
                }
                if (num - arr[lo] <= arr[hi] - num) {
                    return arr[lo];
                }
                return arr[hi];
            }
            array = [2, 42, 82, 122, 162, 202, 242, 282, 322, 362];
            number = 112;
            alert (closest (number, array));
        </script>
    </body>
</html>

它基本上是用包套和检查中间值,使每一次迭代的解空间减少一半,这是一个经典的方法。O(log N)算法,而上面的顺序搜索是O(N):

0  1  2   3   4   5   6   7   8   9  <- indexes
2 42 82 122 162 202 242 282 322 362  <- values
L             M                   H  L=0, H=9, M=4, 162 higher, H<-M
L     M       H                      L=0, H=4, M=2, 82 lower/equal, L<-M
      L   M   H                      L=2, H=4, M=3, 122 higher, H<-M
      L   H                          L=2, H=3, difference of 1 so exit
          ^
          |
          H (122-112=10) is closer than L (112-82=30) so choose H

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