如何使用关键字作为变量名? [英] How do I use a keyword as a variable name?

问题描述

我有带有变量from，to和rate的以下类. from是关键字.如果要在下面的init方法中使用它，编写它的正确方法是什么?

I have the following class with the variables from, to and rate. from is a keyword. If I want to use it in the init method below, what's the correct way to write it?

更多上下文:该类明确需要from变量，因为它是另一位开发人员用另一种语言编写的POST端点所需的json的一部分.因此，更改变量名称是不可能的.

More context: The class needs the from variable explicitly as it's part of a json required by a POST endpoint written up by another developer in a different language. So changing the variable name is out of the question.

class ExchangeRates(JsonAware):
    def __init__(self, from, to, rate):
        self.from = from
        self.to = to
        self.rate = rate

JsonAware代码:

JsonAware code:

class PropertyEquality(object):
    def __eq__(self, other):
        return (isinstance(other, self.__class__) and self.__dict__ == other.__dict__)

    def __ne__(self, other):
        return not self.__eq__(other)

    def __repr__(self):
        return '%s(%s)' % (self.__class__.__name__, ', '.join(['%s=%s' % (k, v) for (k, v) in self.__dict__.items()]))

class JsonAware(PropertyEquality):
    def json(self):
        return json.dumps(self, cls=GenericEncoder)

    @classmethod
    def from_json(cls, json):
        return cls(**json)

GenericEncoder代码:

GenericEncoder code:

class GenericEncoder(json.JSONEncoder):
    def default(self, obj):
        return obj.__dict__

推荐答案

如注释中所述，from是Python关键字，因此不能将其用作变量名或属性名.因此，您需要使用备用名称，并在读取或写入JSON数据时进行转换.

As mentioned in the comments, from is a Python keyword so you can't use it as a variable name, or an attribute name. So you need to use an alternative name, and do a conversion when reading or writing the JSON data.

要进行输出转换，可以为json.dumps提供一个新的编码器.您可以通过覆盖ExchangeRates.json方法来实现.要进行输入转换，请覆盖ExchangeRates.from_json.

To do the output conversion you can supply a new encoder for json.dumps; you can do that by overriding the ExchangeRates.json method. To do the input conversion, override ExchangeRates.from_json.

两种情况下的策略相似:我们创建字典的副本(因此我们不会对原始字典进行变异)，然后创建具有所需名称和值的新键，然后删除旧键.

The strategy is similar in both cases: we create a copy of the dictionary (so we don't mutate the original), then we create a new key with the desired name and value, then delete the old key.

这是一个快速的演示，已在Python 2.6和3.6上进行了测试:

Here's a quick demo, tested on Python 2.6 and 3.6:

import json

class PropertyEquality(object):
    def __eq__(self, other):
        return (isinstance(other, self.__class__) and self.__dict__ == other.__dict__)

    def __ne__(self, other):
        return not self.__eq__(other)

    def __repr__(self):
        return '%s(%s)' % (self.__class__.__name__, ', '.join(['%s=%s' % (k, v) for (k, v) in self.__dict__.items()]))

class JsonAware(PropertyEquality):
    def json(self):
        return json.dumps(self, cls=GenericEncoder)

    @classmethod
    def from_json(cls, json):
        return cls(**json)

class ExchangeRatesEncoder(json.JSONEncoder):
    def default(self, obj):
        d = obj.__dict__.copy()
        d['from'] = d['frm']
        del d['frm']
        return d

class ExchangeRates(JsonAware):
    def __init__(self, frm, to, rate):
        self.frm = frm
        self.to = to
        self.rate = rate

    def json(self):
        return json.dumps(self, cls=ExchangeRatesEncoder)

    @classmethod
    def from_json(cls, json):
        d = json.copy()
        d['frm'] = d['from']
        del d['from']
        return cls(**d)

# Test

a = ExchangeRates('a', 'b', 1.23)
print(a.json())

jdict = {"from": "z", "to": "y", "rate": 4.56, }

b = ExchangeRates.from_json(jdict)
print(b.json())    

典型输出

{"from": "a", "to": "b", "rate": 1.23}
{"from": "z", "to": "y", "rate": 4.56}

这篇关于如何使用关键字作为变量名?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！