Question

我正在为django模型中的数据编写一些字符串表示，我的代码非常重复。什么是避开雨的最好方法？

以下是两种方法：

def get_known_citys_string(user):
    entrys = HTTPRequestData.objects.filter(user=user)
    counter = Counter()

    for entry in entrys:
        counter[entry.city] += 1

    return counter_to_string(counter)

def get_known_devices_string(user):
    entrys = HTTPRequestData.objects.filter(user=user)
    counter = Counter()

    for entry in entrys:
        counter[entry.device] += 1

    return counter_to_string(counter)

Answer 1

有很多方法可以解决这个问题。这是一个：

def get_known_fields_string(user, fieldSelector):
    entrys = HTTPRequestData.objects.filter(user=user)
    counter = Counter()

    for entry in entrys:
        counter[fieldSelector(entry)] += 1

    return counter_to_string(counter)

def get_known_citys_string(user):
    return get_known_fields_string(user, lambda entry: entry.city)

def get_known_devices_string(user):
    return get_known_fields_string(user, lambda entry: entry.device)

Answer 2

这就是发明功能的原因......

def common(user, field):
    entrys = HTTPRequestData.objects.filter(user=user)
    counter = Counter()

    for entry in entrys:
        counter[getattr(entry, field)] += 1

    return counter_to_string(counter)

def get_known_citys_string(user):
    return common(user, "city")

def get_known_devices_string(user):
    return common(user, "device")

Answer 3

另一种选择是使用functools.partial，以获得简洁的功能性风格：

from functools import partial


def getter(attrname, user):
    entrys = HTTPRequestData.objects.filter(user=user)
    counter = Counter()

    for entry in entrys:
        counter[getattr(entry, attrname)] += 1

    return counter_to_string(counter)


# Define functions as variations on getter
get_known_citys_string = partial(getter, 'city')
get_known_devices_string = partial(getter, 'device')

我怎么待干？

3 个答案: