检查字典键是一个整数范围 [英] Checking for a dictionary key which is a range of integers

问题描述

我正在建立制造商字典。它看起来像这样：

I'm building a dictionary of manufacturers. It would look something like this:

mfgs =     {17491: 'DS', 6543: 'AC'}

在这种情况下，我需要表示一系列整数，这些整数都指向同一制造商。 （例如1-99都是由DC制造的）

In this scenario, I need to represent a range of integers that all refer to the same manufacturer. (e.g. 1 - 99 are all manufactured by DC)

我看到你可以创建一个表示范围的字典键。

I've seen that you can create a dictionary key in which a range is represented.

mfgs =     {17491: 'DS', 6543: 'AC', (1,99): 'DC'}

稍后，我将从外部文件中获取一个整数。根据遇到的值，我会将相应的制造商记录到另一个文件中。

Later on, I will have an integer grabbed from an external file. Depending upon the value encountered, I will log the appropriate manufacturer into another file.

我不清楚如何检查是否存在有效的键/值对（无论是数字还是数字范围）并记录未知（如果两者都不是）在没有构建扩展的情况下发现if / then会占用字典中定义的键。

I am unclear on how to check for the presence of a valid key/value pair (whether a number or a range of numbers) and log "Unknown" if neither is found without building an expanding if/then that accounts for keys defined in the dictionary.

似乎try / except是合适的，但如果遇到21，mfg [21]会失败，它不应该。

It seems like try/except is appropriate, but if 21 is encountered, mfg[21] fails and it shouldn't.

推荐答案

您需要的不仅仅是简单的散列图（即字典）查找。哈希映射查找特定键，而不是键是否在任何现有键的范围内。

You need more than just a simple hashmap (i.e. dictionary) lookup. Hashmaps lookup particular keys, not whether a key is in range of any of the existing keys.

您有两个简单的选项。

1. 如果您提前了解范围，请在查找之前将整数转换为范围：

1. If you know the ranges ahead of time, convert the integer into a range before looking it up:

def range_from_id(manufacturer_id):
    if 1 <= manufacturer_id <= 99:
        return (1, 99)
    return manufacturer_id

manufacturer_id = ... from file ...
manufacturer_range = range_from_id(manufacturer_id)
manufacturer = mfgs.get(manufacturer_range, "Unknown")

如果不这样做，则在单独的字典中跟踪范围，并遍历所有可能的值：

If you don't, then keep track of the ranges in a separate dictionary, and loop through all possible values:

mfgs = {17491: 'DS', ...}
mfg_ranges = {(1, 99): 'DC', ...}

def lookup_manufacturer(manufacturer_id):
    # direct look-up:
    if manufacturer_id in mfgs: 
        return mfgs[manufacturer_id]

    # range look-up:
    for (start, end), mfg in mfg_ranges.items():
        if start <= manufacturer_id <= end:
            return mfg

    return "Unknown"

如果速度是重要的是，请注意，此选项将采用 O（n），其中 n 是您拥有的范围数。更合适的方法是使用二进制树，如 amit在这里回答。这需要使用第三方库，如 bintrees 。

If speed is important, note that this option will take O(n), where n is the number of ranges you have. A more proper way to do it would be using binary trees, as amit answered here. This would require using a 3rd party library like bintrees.

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