关于列表的查询 [英] A query about list
问题描述
大家好，
我对python很新。我在python中查询了
列表。
假设我有一个列表
a = [1，[2,3 ，4]，5,6,7，[8,9,10]，11,12]
我想知道是否有任何简单的python
设施可用这将扩大上面的列表
给予
a = [1,2,3,4,5,6,7,8,9,10,11,12]
我知道我可以用type（）和for / while循环来实现，
但是有没有更简单的方法？
问候，
Santanu
Hello everybody,
I am very new to python. I have a query about
list in python.
Suppose I have a list
a = [1,[2,3,4],5,6,7,[8,9,10],11,12]
I want to know if there is any simple python
facility available that would expand the above list
to give
a = [1,2,3,4,5,6,7,8,9,10,11,12]
I know I can do that with a type() and a for/while loop,
but is there any simpler way?
Regards,
Santanu
推荐答案
文章< pa ****** **********************@softhome.net> ;, Santanu Chatterjee写道：
In article <pa****************************@softhome.net>, Santanu Chatterjee wrote:
我对python很新。我在python中查询了
列表。
假设我有一个列表
a = [1，[2,3,4]，5,6,7，[ 8,9,10]，11,12]
我想知道是否有任何简单的python
设施可以扩展上面的列表
给
a = [ 1,2,3,4,5,6,7,8,9,10,11,12]
我知道我可以用type（）和for / while循环来做到这一点，
但是有没有更简单的方法？
I am very new to python. I have a query about
list in python.
Suppose I have a list
a = [1,[2,3,4],5,6,7,[8,9,10],11,12]
I want to know if there is any simple python
facility available that would expand the above list
to give
a = [1,2,3,4,5,6,7,8,9,10,11,12]
I know I can do that with a type() and a for/while loop,
but is there any simpler way?
通常情况下，我建议使用reduce（operator.add，...）。压扁一个清单，但是因为你有一些赤裸裸的条目，那不行......
我不知道你是否认为这个更简单，但你可以定义一个减少
的功能检查第二个参数的类型，并使用它如下：
Normally, I''d suggest "reduce(operator.add, ...)" to flatten a list, but
since you''ve got some "naked" entries, that won''t work...
I don''t know if you consider this simpler, but you could define a reduction
function that checks the type of the second argument, and use it like this:
def merge（x，y） ：
....如果type（y）是type（[]）：return x + y
.... return x + [y]
.... reduce（merge，a，[]）
def merge(x, y): .... if type(y) is type([]): return x + y
.... return x + [y]
.... reduce(merge, a, [])
[1,2,3,4,5,6,7,8 ，9,10,11,12]
戴夫
-
..：[dave benjamin （ramenboy） - ： - www.ramenfest.com - ： - www.3dex.com ]：
：drinklifeoutofthec ontainer：
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
Dave
--
..:[ dave benjamin (ramenboy) -:- www.ramenfest.com -:- www.3dex.com ]:.
: d r i n k i n g l i f e o u t o f t h e c o n t a i n e r :
Santanu Chatterjee写道：
Santanu Chatterjee wrote:
Hello eve rybody，
我是python的新手。我在python中查询了
列表。
假设我有一个列表
a = [1，[2,3,4]，5,6,7，[ 8,9,10]，11,12]
我想知道是否有任何简单的python
设施可以扩展上面的列表
给
a = [ 1,2,3,4,5,6,7,8,9,10,11,12]
我知道我可以用type（）和for / while循环来做到这一点，
但是有没有更简单的方法？
Hello everybody,
I am very new to python. I have a query about
list in python.
Suppose I have a list
a = [1,[2,3,4],5,6,7,[8,9,10],11,12]
I want to know if there is any simple python
facility available that would expand the above list
to give
a = [1,2,3,4,5,6,7,8,9,10,11,12]
I know I can do that with a type() and a for/while loop,
but is there any simpler way?
为什么你首先有这样的数据结构？难道不是吗？b
$ b b b b b b b b b b b b b b b b b b b b b b b b b b b b b >
然后你可以扩展（）它。
- 格哈德
Why do you have such a data structure in the first place? Can''t it be
avoided?
If for example you made the mistake of .append()-ing lists to a list,
then you can .extend() it instead.
-- Gerhard
开星期四，2003年10月9日20:05:08 +0000，Dave Benjamin写道：
On Thu, 09 Oct 2003 20:05:08 +0000, Dave Benjamin wrote:
假设我有一个清单
a = [1 ，[2,3,4]，5,6,7，[8,9,10]，11,12]
我想知道是否有任何简单的python工具可以扩展以上列表给出了
a = [1,2,3,4,5,6,7,8,9,10,11,12]
我知道我能做到这一点使用type（）和for / while循环，但有没有更简单的方法？
Suppose I have a list
a = [1,[2,3,4],5,6,7,[8,9,10],11,12]
I want to know if there is any simple python facility available that
would expand the above list to give
a = [1,2,3,4,5,6,7,8,9,10,11,12]
I know I can do that with a type() and a for/while loop, but is there
any simpler way?
我不知道你是否认为这个更简单，但你可以定义一个
缩减函数来检查第二个参数的类型，并使用它
像这样：
def merge（x，y）：...如果type（y）是type（[]）：return x + y
... return x + [y]
... reduce（merge，a，[]）
I don''t know if you consider this simpler, but you could define a
reduction function that checks the type of the second argument, and use it
like this:
def merge(x, y): ... if type(y) is type([]): return x + y
... return x + [y]
... reduce(merge, a, [])
[1,2,3,4,5,6,7,8,9,10,11,12 ]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
感谢您的快速回复。是的，这类似于我将要做的事情
使用for循环。
但是更简单，我真正的意思是操纵
列表''a''本身没有创建新的列表对象。
我读到创建新对象非常耗时（而且它是
更好地使用填充列表然后''join''来创建一个字符串，而不是使用+ =来增加一个字符串
。对不起，我以前忘记提了。
我正在尝试类似a.insert（1，a.pop（1））的东西，但我需要
以某种方式修改a.pop（1）以便括号消失......你知道
我的意思。这可能吗？
问候，
Santanu
Thanks for the quick reply. Yes, this is similar to what I would do
using for loop.
But by ''simpler'' what I really meant was that manipulating
the list ''a'' itself without creating a new list object.
I read that creating new objects is time consuming (and that it is
better to use fill a list and then ''join'' to create a string, than to
increase a string using += ). Sorry I forgot to mention it before.
I was trying something like a.insert(1,a.pop(1)) but I need to
modify a.pop(1) somehow so that the brackets vanish ...you know
what I mean. Is that possible ?
Regards,
Santanu
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