Question

我正在尝试创建转换数据的方法的列表（或字典）。例如，我有如下数据：

data = [
{'Result': 1, 'Reason1': False, 'Reason2': 1},
{'Result': 0, 'Reason1': False, 'Reason2':'haha'},
{'Result': 0, 'Reason1': True, 'Reason2': 'hehe'},
{'Result': 0, 'Reason1': True, 'Reason2': 0},
]


def rule_1(datum):
    modified_datum = datum
    if datum['Reason1']:
        modified_datum['Result'] = 1 # always set 'Result' to 1 whenever 'Reason1' is True
    else:
        modified_datum['Result'] = 1 # always set 'Result' to 0 whenever 'Reason1' is False
    return modified_datum


def rule_2(datum):
    modified_datum = datum
    if type(datum['Reason2']) is str:
        modified_datum['Result'] = 1 # always set 'Result' to 1 whenever 'Reason2' is of type 'str'
    elif type(datum['Reason2']) is int:
        modified_datum['Result'] = 2 # always set 'Result' to 2 whenever 'Reason2' is of type 'int'
    else:
        modified_datum['Result'] = 0
    return modified_datum


# There can be 'rule_3', 'rule_4' and so on... Also, these rules may have different method signatures (that is, they may take in more than one input parameter)
rule_book = [rule_2, rule_1] # I want to apply rule_2 first and then rule_1

processed_data = []
for datum in data:
    for rule in rule_book:
        # Like someone mentioned here, the line below works, but what if I want to have different number of input parameters for rule_3, rule_4 etc.?
        # processed_data.append(rule(datum))

我认为关于Stack Overflow的this answer与我正在尝试的工作非常接近，但是我想向有Python经验的人学习如何最好地处理它。我用“ dispatch”标记了此帖子，我认为这是我要达到的目标的称呼（？）谢谢您的帮助和建议！

Answer 1

如前所述，您非常接近。您需要做的就是在迭代过程中致电rule。

关于处理各种长度的参数，您可能会选择在规则中使用*args和**kwargs。这是一个简单的示例：

def rule1(*args, **kwargs):
    # Handling of non-keyword params passed in, if any
    if args:
        for arg in args:
            print(f'{arg} is type {type(arg)}')
    # if kwargs is not necessary if you don't intend to handle keyword params

def rule2(*args, **kwargs):
    # if args is not necessary if you don't intend to handle non-keyword params

    # handling of keyword params passed in, if any
    if kwargs:
        for k, v in kwargs.items():
            print(f'Keyword arg {k} has value {v}')

rule_book = [rule2, rule1]
for rule in rule_book:
    # iterate through the rule_book with the same amount of args and kwargs
    rule('I am a string', 123, ('This', 'is', 'Tuple'), my_list=[0, 1, 2], my_dict={'A': 0, 'B': 1})

结果：

Keyword arg my_list has value [0, 1, 2]
Keyword arg my_dict has value {'A': 0, 'B': 1}
I am a string is type <class 'str'>
123 is type <class 'int'>
('This', 'is', 'Tuple') is type <class 'tuple'>

关键是要使规则之间的参数保持一致，一旦所有内容都传入，只需获取相关对象并加以利用即可：

def rule3(*args, **kwargs):
    if args:
        for arg in args:
            if isinstance(arg, tuple):
                # if there's a tuple presented, reverse each of the inner items
                print([a[::-1] for a in arg])

 # ['sihT', 'si', 'elpuT']

以您的代码结构方式，我相信您应该能够理解并将其应用于您自己的代码。

如何使用字典调用Python方法

1 个答案: