有没有更好的方法来做到这一点？ [英] Is there a nicer way to do this?
问题描述
是否有更好的方法来执行以下操作？
attributes = [''foo''，''bar'']
attributeNames = {}
n = 1
属性中的属性：
attributeNames [" AttributeName ％d" ％n] =属性
n = n + 1
它有效，但我想知道是否有更多的pythonic方式
这样做。
S.
Is there a better way to do the following?
attributes = [''foo'', ''bar'']
attributeNames = {}
n = 1
for attribute in attributes:
attributeNames["AttributeName.%d" % n] = attribute
n = n + 1
It works, but I am wondering if there is a more pythonic way to
do this.
S.
推荐答案
你好，
Stefan Arentzaécrit：
Hello,
Stefan Arentz a écrit :
有更好的方法可以做到以下几点吗？
attributes = [''foo' '，''bar'']
attributeNames = {}
n = 1
属性中的属性：
attributeNames [" AttributeName。％d" ％n] =属性
n = n + 1
它有效，但我想知道是否有更多的pythonic方式
这样做。
S.
Is there a better way to do the following?
attributes = [''foo'', ''bar'']
attributeNames = {}
n = 1
for attribute in attributes:
attributeNames["AttributeName.%d" % n] = attribute
n = n + 1
It works, but I am wondering if there is a more pythonic way to
do this.
S.
您可以使用enumerate（）对项目进行编号（小心它以0）：
attributes = [''foo''，''bar'']
attributeNames = {}
对于n，枚举（属性）中的属性：
attributeNames [" AttributeName。％d" ％（n + 1）] =属性
然后使用生成器表达式来提供dict：
attributes = [''foo''，'' bar'']
attributeNames = dict（（" AttributeName。％d"％（n + 1），attribute）
代表n，属性为枚举（属性） ）
希望这会有所帮助，
-
Amaury
You could use enumerate() to number the items (careful it starts with 0):
attributes = [''foo'', ''bar'']
attributeNames = {}
for n, attribute in enumerate(attributes):
attributeNames["AttributeName.%d" % (n+1)] = attribute
Then use a generator expression to feed the dict:
attributes = [''foo'', ''bar'']
attributeNames = dict(("AttributeName.%d" % (n+1), attribute)
for n, attribute in enumerate(attributes))
Hope this helps,
--
Amaury
attributes = [''foo''，''bar'']
attributes = [''foo'', ''bar'']
>
attributeNames = {}
n = 1
属性中的属性：
attributeNames [" AttributeName。％d" ％n] =属性
n = n + 1
它有效，但我想知道是否有更多的pythonic方式
这样做。
>
attributeNames = {}
n = 1
for attribute in attributes:
attributeNames["AttributeName.%d" % n] = attribute
n = n + 1
It works, but I am wondering if there is a more pythonic way to
do this.
更好可能是主观的，但你可以做类似的事情
attributes = [''foo''，''bar'']
attributeNames = dict（
（" AttributeName。％d"％（i + 1），attrib）
for i，attrib
in enumerate（attributes）
）
HTH，
-tkc
"Better" may be subjective, but you could do something like
attributes = [''foo'', ''bar'']
attributeNames = dict(
("AttributeName.%d" % (i+1), attrib)
for i, attrib
in enumerate(attributes)
)
HTH,
-tkc
不确定这是否真的更好，甚至更加pythonic，但是如果你像朋友一样运行这种语言的
：
attributeNames = dict（[（" AttributeName。％d"％（n + 1），属性）
n，枚举（属性）中的属性]）
它的作用是创建一个列表（使用列表推导和
枚举函数）（AttributeName.x，属性）元组，然后是
用于初始化字典。
Not sure if this is really better or even more pythonic, but if you
like one-liners that exercise the language:
attributeNames = dict( [("AttributeName.%d" % (n+1), attribute) for
n,attribute in enumerate(attributes)] )
What this does is create a list (using a list comprehension and the
enumerate function) of ("AttributeName.x", attribute) tuples which is
then be used to initialize a dictionary.
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