如何转换元组的元组列在一行(pythonic)? [英] How do I convert tuple of tuples to list in one line (pythonic)?
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问题描述
query = 'select mydata from mytable'
cursor.execute(query)
myoutput = cursor.fetchall()
print myoutput
(('aa',), ('bb',), ('cc',))
b $ b
为什么(cursor.fetchall)返回一个元组的元组而不是一个元组,因为我的查询只需要一列数据?
Why is it (cursor.fetchall) returning a tuple of tuples instead of a tuple since my query is asking for only one column of data?
将其转换为 ['aa','bb','cc']
?
我可以这样做:
mylist = []
myoutput = list(myoutput)
for each in myoutput:
mylist.append(each[0])
这不是最好的做法。请告诉我!
I am sure this isn't the best way of doing it. Please enlighten me!
推荐答案
这也很好:
>>> tu = (('aa',), ('bb',), ('cc',))
>>> import itertools
>>> list(itertools.chain(*tu))
['aa', 'bb', 'cc']
b $ b
编辑 您能否评论成本折衷? (for循环和itertools)
Itertools显着更快:
Itertools is significantly faster:
>>> t = timeit.Timer(stmt="itertools.chain(*(('aa',), ('bb',), ('cc',)))")
>>> print t.timeit()
0.341422080994
>>> t = timeit.Timer(stmt="[a[0] for a in (('aa',), ('bb',), ('cc',))]")
>>> print t.timeit()
0.575773954391
编辑2 code>你可以解释itertools.chain(*)
Edit 2 Could you pl explain itertools.chain(*)
*
将序列解包成位置参数,在这种情况下是元组的嵌套元组。
That *
unpacks the sequence into positional arguments, in this case a nested tuple of tuples.
示例:
>>> def f(*args):
... print "len args:",len(args)
... for a in args:
... print a
...
>>> tu = (('aa',), ('bb',), ('cc',))
>>> f(tu)
len args: 1
(('aa',), ('bb',), ('cc',))
>>> f(*tu)
len args: 3
('aa',)
('bb',)
('cc',)
另一个例子:
>>> f('abcde')
len args: 1
abcde
>>> f(*'abcde')
len args: 5
a
b
c
d
e
请参阅文档开箱。
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