How do I use a keyword as a variable name?

问题:

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?

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 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 code:

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

回答1:

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.

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.

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.

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())    

typical output

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

回答2:

Add a single underscore to your preferred names: from_ and to_

(see PEP 8)

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

回答3:

Use a synonym. Try "origin" or "source" instead.