Python:cerberus check_与函数

Question

我想验证一个dict，其中的值遵循以下规则：

• 值必须是单个float或List(float)
• 如果是单个float，则该值必须为1
• 如果它是List(float)，则每个浮点必须为正

以下是我的代码和一些正常工作的测试断言：

import cerberus

v = cerberus.Validator()

schema1 = {
    "key1": {
        "type": ["float", "list"],
        "min": 1,
        "max": 1,
        "schema": {"type": "float", "min": 0},
    }
}
document1 = {"key1": 1}
document2 = {"key1": 5}
document3 = {"key1": "5"}
document4 = {"key1": [0.5, 0.3]}
document5 = {"key1": ["0.5", 0.3]}

assert v.validate(document1, schema1)
assert not v.validate(document2, schema1)
assert not v.validate(document3, schema1)
assert v.validate(document4, schema1)
assert not v.validate(document5, schema1)

现在，我必须执行另一个条件：

• 如果它是List(float)，则float的sum必须等于1

因此，我编写了一个check_with函数，如文档（https://docs.python-cerberus.org/en/stable/validation-rules.html）中所述

from cerberus import Validator

class MyValidator(Validator):
    def _check_with_sum_eq_one(self, field, value):
        """Checks if sum equals 1"""

        if sum(value) != 1:
            self._error(field, f"Sum of '{field}' must exactly equal 1")

调整后的架构和测试文档如下所示：

v = MyValidator()

schema2 = {
    "key1": {
        "type": ["float", "list"],
        "min": 1,
        "max": 1,
        "schema": {"type": "float", "min": 0, "max": 1, "check_with": "sum_eq_one"},
    }
}

document1 = {"key1": 1}
document2 = {"key1": 5}
document3 = {"key1": "5"}
document4 = {"key1": [0.5, 0.3]}  # error
document5 = {"key1": ["0.5", 0.3]}  # error
document6 = {"key1": [0.5, 0.5]}  # error

现在，每当值是List(float)时，只有list的第一个元素将被注入到我的函数中，从而产生一个TypeError: 'float' object is not iterable。
验证document4时field将是int=0和value=0.5。因此，错误消息是有意义的

我想知道，为什么整个列表没有传递给我的函数？我错过了什么

Answer 1

如果您试图捕获错误并仅在错误发生时继续执行您的函数，该怎么办？例如，以这种方式：

class MyValidator(Validator):

def _check_with_sum_eq_one(self, field, value):
    """ Checks whether value is a list and its sum equals 1.0. """
    if isinstance(value, list) and sum(value) != 1.0:
        self._error(str(value), f"Sum of '{field}' must exactly equal 1")


schema2 = {
    "key1": {
        "type": ["list", "float"],
        "min": 1,
        "max": 1,
        "schema": {"type": "float", "min": 0, "max": 1},
        "check_with": "sum_eq_one",
    }
}

v = MyValidator(schema2)

document1 = {"key1": 1}
document2 = {"key1": 5}
document3 = {"key1": "5"}
document4 = {"key1": [0.3, 0.5]}  # error
document5 = {"key1": ["0.5", 0.3]}  # error
#document6 = {"key1": [0.5, 0.5]}  # error
assert v.validate(document1)
assert not v.validate(document2)
assert not v.validate(document3)
assert v.validate(document4)
assert not v.validate(document5)

Answer 2

下面的答案工作正常。然而，在我看来，这太复杂了

首先，调整schema2如下：

schema2 = {
    "key1": {
        "type": ["float", "list"],
        "min": 0,
        "max": 1,
        "check_with": "sum_eq_one"
    }
}

接下来，调整_check_with_sum_eq_one，如下所示：

class MyValidator(Validator):
    def _check_with_sum_eq_one(self, field, value):
        """Checks if sum equals 1"""

        if (isinstance(value, float) or isinstance(value, int)) and value != 1:
            self._error(field, f"Sum of '{field}' must exactly equal 1")
        if isinstance(value, list):
            if all([isinstance(x, float) for x in value]):
                if sum(value) != 1:
                    self._error(field, f"Sum of '{field}' must exactly equal 1")
            else:
                self._error(field, f"All list members must be of type ['float']")

最后，断言一切都按预期进行

v = MyValidator()

document1 = {"key1": 1}
document2 = {"key1": 5}
document3 = {"key1": "5"}
document4 = {"key1": [0.5, 0.3]}
document5 = {"key1": ["0.5", 0.3]}
document6 = {"key1": [0.5, 0.5]}

assert v.validate(document1, schema2)
assert not v.validate(document2, schema2)
assert not v.validate(document3, schema2)
assert not v.validate(document4, schema2)
assert not v.validate(document5, schema2)
assert v.validate(document6, schema2)

我不喜欢这里的事实是，如果所有列表成员都是float（if all([isinstance(x, float) for x in value])）类型，我需要“手动”检查。在我看来，这个测试属于schema2。然而，在某种程度上，我没有成功地调整schema2，即float类型的测试先于check_with验证

任何进一步简化此任务的提示将不胜感激