在python当中的字典能让使用者快速找到对应的数据值，但是如果想要实现这样的一个字典：

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_dict = {
    [1,5):"a",
    [5,10):"b"
}

能让操作者在仅知道一个数值就可以索引到范围对应的那个value，如_dict[3]输出"a"该怎么做呢？

解决办法

可以借助python库bisect对字典进行重写解决该问题，bisect相关信息就不过多阐述了，简单来说就是通过排序算法，定位指定值在范围内的位置，再通过索引找到位置。代码如下：

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from bisect import bisect_left, bisect_right
from collections.abc import MutableMapping

class RangeBisection(MutableMapping):
    def __init__(self, map=None):
        self._upper = []
        self._lower = []
        self._values = []
        if map is not None:
            self.update(map)

    def __len__(self):
        return len(self._values)

    def __getitem__(self, point_or_range):
        if isinstance(point_or_range, tuple):
            low, high = point_or_range
            i = bisect_right(self._upper, high)
            point = low
        else:
            point = point_or_range
            i = bisect_right(self._upper, point)
        if i >= len(self._values) or self._lower[i] > point:
            raise IndexError(point_or_range)
        return self._values[i]

    def __setitem__(self, r, value):
        lower, upper = r
        i = bisect_right(self._upper, upper)
        if i < len(self._values) and self._lower[i] < upper:
            raise IndexError('No overlaps permitted')
        self._upper.insert(i, upper)
        self._lower.insert(i, lower)
        self._values.insert(i, value)

    def __delitem__(self, r):
        lower, upper = r
        i = bisect_right(self._upper, upper)
        if self._upper[i] != upper or self._lower[i] != lower:
            raise IndexError('Range not in map')
        del self._upper[i]
        del self._lower[i]
        del self._values[i]

    def __iter__(self):
        yield from zip(self._lower, self._upper)

d = RangeBisection({
    (0.0, 0.1): "a",
    (0.1, 0.3): "b",
    (0.3, 0.55): "c",
    (0.55, 0.7): "d",
    (0.7, 1.0): "e",})
d[0.0]