如何使用python-3.x中的字典格式化字符串?

我非常喜欢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，以及让我利用现有的字典。例如:

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()

但是我搞不懂python。X语法来做同样的事情(或者如果可能的话)。我想做以下事情

# 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)

当前回答

要将字典解压缩为关键字参数，请使用**。此外，new-style格式支持引用对象和映射项的属性:

'{0[latitude]} {0[longitude]}'.format(geopoint)
'The title is {0.title}s'.format(a) # the a from your first example

2011-05-10 15:21:42

其他回答

print("{latitude} {longitude}".format(**geopoint))

2011-05-10 15:19:52

geopoint = {'latitude':41.123,'longitude':71.091}

# working examples.
print(f'{geopoint["latitude"]} {geopoint["longitude"]}') # from above answer
print('{geopoint[latitude]} {geopoint[longitude]}'.format(geopoint=geopoint)) # alternate for format method  (including dict name in string).
print('%(latitude)s %(longitude)s'%geopoint) # thanks @tcll

2020-04-10 08:27:56

要将字典解压缩为关键字参数，请使用**。此外，new-style格式支持引用对象和映射项的属性:

'{0[latitude]} {0[longitude]}'.format(geopoint)
'The title is {0.title}s'.format(a) # the a from your first example

2011-05-10 15:21:42

这对你有好处吗?

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

2011-05-10 15:25:34

Python 2语法在Python 3中也适用:

>>> class MyClass:
...     def __init__(self):
...         self.title = 'Title'
... 
>>> a = MyClass()
>>> print('The title is %(title)s' % a.__dict__)
The title is Title
>>> 
>>> path = '/path/to/a/file'
>>> print('You put your file here: %(path)s' % locals())
You put your file here: /path/to/a/file

2011-05-11 05:11:20

如何使用python-3.x中的字典格式化字符串?

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