Question

在这里谈论Python 3。

我希望将一个数字四舍五入到一组可以变化的值

假设value_set = [x, y, z]并且为了示例x, y, z = 1, 3.12, 4我正在寻找一个将给定浮点数舍入到最接近数字的函数

custom_round(0) --> 1

custom_round(2.7) --> 3.12

请注意，它应该是通用的，value_set长度也会变化

Answer 1

当密钥是x-n的绝对值（x是列表中的每个项目）时，您可以使用min函数查找列表中的最小值。

value_set = [1, 3.12, 4]

def return_closest(n):
  return min(value_set, key=lambda x:abs(x-n))

number_to_check = 3
print (return_closest(number_to_check))

>>> 3.12

Answer 2

您可以先对列表进行排序，然后使用二进制搜索：

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您可以构建from bisect import bisect_left class CustomRound: def __init__(self,iterable): self.data = sorted(iterable) def __call__(self,x): data = self.data ndata = len(data) idx = bisect_left(data,x) if idx <= 0: return data[0] elif idx >= ndata: return data[ndata-1] x0 = data[idx-1] x1 = data[idx] if abs(x-x0) < abs(x-x1): return x0 return x1之类的：

CustomRound

并简单地称之为：

values = [1,3.12,4]
custom_round = CustomRound(values)

此方法适用于 O（log n）用于舍入和 O（n log n）用于构建。因此，您将投入一些额外的时间来构建>>> custom_round(0) 1 >>> custom_round(0.5) 1 >>> custom_round(1.5) 1 >>> custom_round(2.5) 3.12 >>> custom_round(3.12) 3.12 >>> custom_round(3.9) 4 >>> custom_round(4.1) 4 >>> custom_round(4.99) 4，但如果您经常调用它，它最终会在舍入个别数字时获得回报。

将数字舍入到给定的一组值

2 个答案: