Question

我的if语句太多，很难理解字典系统。

这似乎很容易解决，但我尝试过的一切都会使它变得更糟。

    """
    Find the place where the person was born.

    Possible locations are following: Kuressaare, Tartu, Tallinn, Kohtla-Järve, Narva, Pärnu,
    Paide, Rakvere, Valga, Viljandi, Võru and undefined. Lastly if the number is incorrect the function must return
    the following 'Wrong input!'
    :param birth_number: int
    :return: str
    """
    return 
    if not is_valid_birth_number(birth_number):
        return "Wrong input!"
    if 10 >= birth_number >= 1:
        return "Kuressaare"
    elif 20 >= birth_number >= 11:
        return "Tartu"
    elif 220 >= birth_number >= 21:
        return "Tallinn"
    elif 270 >= birth_number >= 221:
        return "Kohtla-Järve"
    elif 370 >= birth_number >= 271:
        return "Tartu"
    elif 420 >= birth_number >= 371:
        return "Narva"
    elif 470 >= birth_number >= 421:
        return "Pärnu"
    elif 490 >= birth_number >= 471:
        return "Tallinn"
    elif 520 >= birth_number >= 491:
        return "Paide"
    elif 570 >= birth_number >= 521:
        return "Rakvere"
    elif 600 >= birth_number >= 571:
        return "Valga"
    elif 650 >= birth_number >= 601:
        return "Viljandi"
    elif 710 >= birth_number >= 651:
        return "Võru"
    elif 999 >= birth_number >= 711:
        return "undefined"

需要消除idcode.py:149:1: C901 'get_birth_place' is too complex (16)错误。

Answer 1

使用一个列表将每个范围的末尾映射到返回值。

if not is_valid_birth_number(birth_number):
    return "Wrong input!"

locations = [(10, "Kuressaare"), (220, "Tartu"), (270, "Tallinn"), ...]
for limit, loc in locations:
    if birth_number <= limit:
        return loc

您不需要每个范围的开头和结尾，因为它们是有序的。范围的末端就足够了。

Answer 2

使用bisect的示例。

import bisect

locations = {
    1: 'Kuressaare',
    2: 'Tartu',
    3: 'Tallinn',
    4: 'Kohtla-Järve',
    5: 'Tartu'
}

birth_number_levels = [1, 11, 21, 221, 271, 371]
position = bisect.bisect(birth_number_levels, birth_number)
return locations[position]

我更喜欢像@Barmar那样将数据保存在一起。这导致：

import bisect

locations = [
    (10, 'Kuressaare'),
    (20, 'Tartu'),
    (220, 'Tallinn'),
    (270, 'Kohtla-Järve'),
    (370, 'Tartu')
]
birth_number_levels = [location[0] for location in locations]
position = bisect.bisect_left(birth_number_levels, birth_number)
return locations[position][1]

Answer 3

我要去字典，字典包含键作为范围，值是位置的名称。然后遍历它，检查给定的birth_number是否在密钥中。

基本上：

loc_dict = {
    range(1, 11): "Kuressaare",
    range(11, 21): "Tartu",
    etc...
}

for loc_range, loc_name in loc_dict.items():
    if birth_number in loc_range:
        return loc_name

我认为这是一种非常清晰的处理方式。

Answer 4

遍历列表比必须对特定的if语句进行编码要容易得多，而且这可以是完全动态的。

def bd(birth_number):
    x = ((10,"Kuressaare"),(20,"Tartu"),(220,"Tallinn"),(270,"Tartu"),(370,"Kohtla-Järve"),(420,"Tartu"),(470,"Narva"),(490,"Pärnu"),(520,"Tallinn"),(570,"Paide"),(600,"Rakvere"))
    if not isinstance(birth_number, (int, long)):
        return "Wrong input!"
    for i in x:
        if birth_number<= i[0]:
            return i[1]
    return "undefined"

print(bd(300))

如何修复太多如果返回，则使代码过于复杂

4 个答案: