d = dict()

or

d = {}

or

import types
d = types.DictType.__new__(types.DictType, (), {})

知道如何编写预先设定的字典也很有用:

cmap =  {'US':'USA','GB':'Great Britain'}

# Explicitly:
# -----------
def cxlate(country):
    try:
        ret = cmap[country]
    except KeyError:
        ret = '?'
    return ret

present = 'US' # this one is in the dict
missing = 'RU' # this one is not

print cxlate(present) # == USA
print cxlate(missing) # == ?

# or, much more simply as suggested below:

print cmap.get(present,'?') # == USA
print cmap.get(missing,'?') # == ?

# with country codes, you might prefer to return the original on failure:

print cmap.get(present,present) # == USA
print cmap.get(missing,missing) # == RU

你可以这样做

x = {}
x['a'] = 1

我还没有足够的声誉来评论，所以我分享这个作为答案。

@David Wheaton在他的评论中分享的接受答案的链接不再有效，因为Doug Hellmann已经迁移了他的网站(来源:https://doughellmann.com/posts/wordpress-to-hugo/)。

这是关于“在CPython 2.7中使用dict()而不是{}对性能的影响”的更新链接:https://doughellmann.com/posts/the-performance-impact-of-using-dict-instead-of-in-cpython-2-7-2/

所以有两种方法来创建字典:

My_dict = dict() My_dict = {}

但在这两个选项中，{}比dict()加上它的可读性更有效。 检查在这里

>>> dict.fromkeys(['a','b','c'],[1,2,3])


{'a': [1, 2, 3], 'b': [1, 2, 3], 'c': [1, 2, 3]}