使用比较时如何简化log if/elif语句

Question

我正在尝试简化这个：

        if num < 9:
            y = 1
        elif num < 17:
            y = 2
        elif num < 25:
            y = 3
        elif num < 33:
            y = 4
        elif num < 41:
            y = 5
        elif num < 49:
            y = 6
        elif num < 57:
            y = 7
        else:
            y = 8

我还没有找到方法 - 有人可以帮我吗？

Answer 1

我想到的唯一想法是创建一个以范围作为键的字典。

ranges = {range(0,9):1,range(9,17):2,range(17,25):3,range(25,33):4,range(33,41):5,range(41,49):6,range(49,57):7}
number = 10
for k,v in ranges.items():
    if number in k:
        print(v)

在这种情况下，输出是 2

Answer 2

尝试枚举？

num=32
x=[9,17,25,33,41,48,57]
for nums,data in enumerate(x):
    if data>num:
        y=nums+1
        break
    else:
        pass

print(y)

Answer 3

跨越一组已排序的任意边界可以通过以下方式完成：

all_bounds = [9,17,25,33,41,49,57]
y = len(all_bounds) + 1     # case when all tests fail
for ix, bound in enumerate(all_bounds):
    if num < bound:
        y = ix + 1
        break

如评论中所述，如果有关于如何得出边界的规则，那么最好使用它 - 但前提是它是一个明确的规则，最好对它的产生有一定的了解。不要在任意数据上强行拟合模式。

对于大量的边界值，我会用二分法搜索正确的值；对于这个例子，这是不值得的。

Answer 4

您可以像这样遍历条件：

num = 32

limits = (57, 49, 41, 33, 25, 17, 9,) 
# ^ traversing in reverse because then
# the chance that we don't have to go
# through the entire loop is greater.

for possible_y,  limit in zip(range(7, 0, -1), limits):
    # zip takes 2 or more iterables and returns
    # tuples of the current iteration for each iterable
    # so e.g (7, 57,) is returned for the first iteration. 

    if num < limit:
        y = possible_y
        break
else:
    # else after a for loop in Python is only executed if 
    # no break statement was encountered in the outer-
    # most loop.
    y = 8

使用比较时如何简化log if/elif语句

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