如何在 python-3.x 中使用字典格式化字符串?

问题描述

我非常喜欢使用字典来格式化字符串.它帮助我阅读我正在使用的字符串格式，并让我利用现有的词典.例如:

class MyClass:def __init__(self):self.title = '标题'a = MyClass()打印 '标题是 %(title)s' % a.__dict__path = '/path/to/a/file'打印 '你把你的文件放在这里:%(path)s' % locals()

但是，我无法弄清楚执行相同操作的 python 3.x 语法(或者甚至可能).我想做以下事情

# 失败，KeyError '纬度'geopoint = {'纬度':41.123，'经度':71.091}打印 '{latitude} {longitude}'.format(geopoint)# 成功打印 '{latitude} {longitude}'.format(latitude=41.123,longitude=71.091)

解决方案

由于 Python 3.0 和 3.1 已经停产并且没有人使用它们，您可以并且应该使用 str.format_map(mapping) (Python 3.2+):

类似于str.format(**mapping)，除了直接使用映射而不是复制到dict.例如，如果映射是 dict 子类，这很有用.

这意味着您可以使用例如 defaultdict 来设置(并返回)丢失的键的默认值:

>>>从集合导入 defaultdict>>>vals = defaultdict(lambda: '', {'bar': 'baz'})>>>'foo 是 {foo} 而 bar 是 {bar}'.format_map(vals)'foo 是 <unset>酒吧是 baz'

即使提供的映射是 dict，而不是子类，这可能仍然会稍微快一些.

虽然差别不大，但

>>>d = dict(foo='x', bar='y', baz='z')

然后

>>>'foo 是 {foo}，bar 是 {bar}，baz 是 {baz}'.format_map(d)

比

快约 10 ns (2 %)>>>'foo 是 {foo}，bar 是 {bar}，baz 是 {baz}'.format(**d)

在我的 Python 3.4.3 上.随着字典中的键越多，差异可能会更大，并且

请注意，格式语言比这灵活得多；它们可以包含索引表达式、属性访问等，因此您可以格式化整个对象，或其中的 2 个:

>>>p1 = {'纬度':41.123，'经度':71.091}>>>p2 = {'纬度':56.456，'经度':23.456}>>>'{0[纬度]} {0[经度]} - {1[纬度]} {1[经度]}'.format(p1, p2)'41.123 71.091 - 56.456 23.456'

从 3.6 开始，您也可以使用内插字符串:

>>>f'lat:{p1["latitude"]} lng:{p1["longitude"]}''纬度:41.123 lng:71.091'

您只需要记住在嵌套引号中使用 other 引号字符.这种方法的另一个优点是它比调用格式化方法要快得多.

I am a big fan of using dictionaries to format strings. It helps me read the string format I am using as well as let me take advantage of existing dictionaries. For example:

class MyClass:
    def __init__(self):
        self.title = 'Title'

a = MyClass()
print 'The title is %(title)s' % a.__dict__

path = '/path/to/a/file'
print 'You put your file here: %(path)s' % locals()

However I cannot figure out the python 3.x syntax for doing the same (or if that is even possible). I would like to do the following

# Fails, KeyError 'latitude'
geopoint = {'latitude':41.123,'longitude':71.091}
print '{latitude} {longitude}'.format(geopoint)

# Succeeds
print '{latitude} {longitude}'.format(latitude=41.123,longitude=71.091)

解决方案

As Python 3.0 and 3.1 are EOL'ed and no one uses them, you can and should use str.format_map(mapping) (Python 3.2+):

What this means is that you can use for example a defaultdict that would set (and return) a default value for keys that are missing:

>>> from collections import defaultdict
>>> vals = defaultdict(lambda: '<unset>', {'bar': 'baz'})
>>> 'foo is {foo} and bar is {bar}'.format_map(vals)
'foo is <unset> and bar is baz'

Even if the mapping provided is a dict, not a subclass, this would probably still be slightly faster.

The difference is not big though, given

>>> d = dict(foo='x', bar='y', baz='z')

then

>>> 'foo is {foo}, bar is {bar} and baz is {baz}'.format_map(d)

is about 10 ns (2 %) faster than

>>> 'foo is {foo}, bar is {bar} and baz is {baz}'.format(**d)

on my Python 3.4.3. The difference would probably be larger as more keys are in the dictionary, and

---

Note that the format language is much more flexible than that though; they can contain indexed expressions, attribute accesses and so on, so you can format a whole object, or 2 of them:

>>> p1 = {'latitude':41.123,'longitude':71.091}
>>> p2 = {'latitude':56.456,'longitude':23.456}
>>> '{0[latitude]} {0[longitude]} - {1[latitude]} {1[longitude]}'.format(p1, p2)
'41.123 71.091 - 56.456 23.456'

---

Starting from 3.6 you can use the interpolated strings too:

>>> f'lat:{p1["latitude"]} lng:{p1["longitude"]}'
'lat:41.123 lng:71.091'

You just need to remember to use the other quote characters within the nested quotes. Another upside of this approach is that it is much faster than calling a formatting method.

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