我有一个元组列表:
现在我想把这组元组变成字典的列表,类似的效果是这样:
meat = [
{
"地区词":"深圳福田区",
"品牌词":"TTM全身体检",
"疑问词":"地址怎么走",
"价格词":"价格要多少钱"
},
{
"地区词":"深圳保安区",
"品牌词":"TTM全身体检",
"疑问词":"地址怎么走",
"价格词":"要做多少项目"
},
.....
]
有什么比较方便的方法?
怪我咯2017-04-18 09:08:51
@dokelung’s method can be more concise
names = 'area brand question price'.split()
lst = [dict(zip(names, t)) for t in tlst]
大家讲道理2017-04-18 09:08:51
names = 'area brand question price'.split()
lst = [{name:value for name, value in zip(names, t)} for t in tlst]
Test:
tlst = [('a1','b1','q1','p1'),
('a2','b2','q2','p2'),
('a3','b3','q3','p3')]
names = 'area brand question price'.split()
lst = [{name:value for name, value in zip(names, t)} for t in tlst]
print(lst)
Result:
[{'brand': 'b1', 'area': 'a1', 'question': 'q1', 'price': 'p1'}, {'brand': 'b2', 'area': 'a2', 'question': 'q2', 'price': 'p2'}, {'brand': 'b3', 'area': 'a3', 'question': 'q3', 'price': 'p3'}]
Questions I answered: Python-QA
迷茫2017-04-18 09:08:51
Simple way of writing: (a is an array of tuples)
meat = [{'地区词':t[0], '品牌词':t[1], '疑问词':t[2], '价格词':t[3]} for t in a]
Of course you can also try the mapping method, which is not covered here
天蓬老师2017-04-18 09:08:51
In order to forcefully show off, use $lambda$ expression to write:
list(map(lambda x: {'地区词':x[0], '品牌词':x[1], '疑问词':x[2], '价格词':x[3]}, a))
Please ignore my answer. . .
迷茫2017-04-18 09:08:51
The correct solution upstairs is to form a separate dictionary, and then call the append of the array and it will be ok