我的清单是这样的,
['"third:4"', '"first:7"', '"second:8"']
我想把它转换成这样的字典......
{"third": 4, "first": 7, "second": 8}
我如何在Python中执行此操作?
答案 0 :(得分:6)
以下是两种可能的解决方案,一种为您提供字符串值,另一种为您提供int值:
>>> lst = ['"third:4"', '"first:7"', '"second:8"']
>>> dict(x[1:-1].split(':', 1) for x in lst)
{'second': '8', 'third': '4', 'first': '7'}
>>> dict((y[0], int(y[1])) for y in (x[1:-1].split(':', 1) for x in lst))
{'second': 8, 'third': 4, 'first': 7}
但是,为了便于阅读,您可以将转换分为两个步骤:
>>> lst = ['"third:4"', '"first:7"', '"second:8"']
>>> dct = dict(x[1:-1].split(':', 1) for x in lst)
>>> {k: int(v) for k, v in dct.iteritems()}
当然,由于你创建了一个dict两次,这有一些开销 - 但是对于一个小列表来说并不重要。
答案 1 :(得分:3)
>>> data
['"third:4"', '"first:7"', '"second:8"']
>>> dict((k,int(v)) for k,v in (el.strip('\'"').split(':') for el in data))
{'second': 8, 'third': 4, 'first': 7}
或
>>> data = ['"third:4"', '"first:7"', '"second:8"']
>>> def convert(d):
for el in d:
key, num = el.strip('\'"').split(':')
yield key, int(num)
>>> dict(convert(data))
{'second': 8, 'third': 4, 'first': 7}
答案 2 :(得分:1)
def listToDic(lis):
dic = {}
for item in lis:
temp = item.strip('"').split(':')
dic[temp[0]] = int(temp[1])
return dic
答案 3 :(得分:0)
使用map and dict
非常简单。
map(func, iterables)
to_dict
将返回key, value
dict
def to_dict(item):
return item[0].replace('"',''), int(item[1].replace('"', ''))
>>> items
6: ['"third:4"', '"first:7"', '"second:8"']
>>> dict(map(to_dict, [item.split(':') for item in items]))
7: {'first': 7, 'second': 8, 'third': 4}
help(dict)
class dict(object)
dict() -> new empty dictionary
dict(mapping) -> new dictionary initialized from a mapping object's
(key, value) pairs
dict(iterable) -> new dictionary initialized as if via:
d = {}
for k, v in iterable:
d[k] = v
dict(**kwargs) -> new dictionary initialized with the name=value pairs
in the keyword argument list. For example: dict(one=1, two=2)
>>> dict(map(to_dict, [item.split(':') for item in items]))