查找值在范围列表中的位置的 Pythonic 方法 [英] Pythonic way of finding where a value is in a list of ranges

问题描述

我有一个开始和结束值字典的列表:

I have a list of start and end value dicts:

a = [
      {'start': 1, 'end': 3},
      {'start': 5, 'end': 6},
      {'start': 8, 'end': 10}
    ]

列表保证有序，没有任何重叠.给定一个值，我需要找到它在列表中的位置.如果它在定义的范围之一内，我需要取回该值.否则，我需要按顺序获得下一个开始"值.例如，给定 b = 2，我会取回 2 值，因为它在第一个范围内.但是，给定值 b = 7 我会返回 8 因为它是从 7 开始的下一个范围的起始值.

The list is guaranteed to be in order, and not have any overlap. Given a value, I need to find where in the list it falls. If it is in one of the defined ranges, I need to get the value back. Otherwise, I need the next 'start' value in sequence. e.g., given b = 2, I would get back the value 2 because it is in the first range. However, given the value b = 7 I would get back 8 as it is the start value for the next range in sequence from 7.

我已经写了一些代码来做到这一点:

I have written some code to do this:

def where_in_range(v, val_ranges):
  # Check if v comes before all ranges
  if v < val_ranges[0]['start']:
    return val_ranges[0]['start']

  # Check if v is within one of the ranges
  for r in val_ranges:
    if v >= r['start'] and v <= r['end']:
      return v

  # Check if v is between ranges
  for i in range(0, len(val_ranges)-1):
    if v > val_ranges[i]['end'] and v < val_ranges[i+1]['start']:
      return val_ranges[i+1]['start']

  return None

但它似乎非常非pythonic.我愿意使用一个轻量级的包，但是像 Pandas 一样重的东西已经出来了，因为它被嵌入到了 django 应用程序中.关于如何使这更加 Pythonic 的任何想法?

But it seems quite non-pythonic. I am willing to use a lightweight package, but something as heavy as pandas is out, as this is being embedded in a django application. Any thoughts on how to make this more pythonic?

推荐答案

只需单独检查每个范围.范围内的值会自动处理，任何范围前的值也是如此.

Just check each range individually. Inter-range values are handled automatically, as are any pre-range values.

def where_in_range(v, val_ranges):
    for d in val_ranges:
        if v < d['start']:
            return d['start']
        elif v < d['end']:
            return v

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