删除列表中的重复项 [英] delete duplicates in list
问题描述
你好，
之前一定要出现，所以我已经很抱歉要问了一下
快速谷歌搜索没有给我任何答案。
i有一个列表，我想要一个更简单的列表，没有重复项
一个简单但不知何故的设计解决方案将是
< blockquote class =post_quotes>a = [1,2,2,3]
d = {}。fromgskeys（a）
b = d.keys（）
打印b
[1,2,3]
应该有一个更简单或更直观的解决方案，可能有一个清单
理解=
类似于
b = [x代表x而不是b中的x]
print b
[]
虽然不起作用。
感谢您的帮助
chris
这个必须在之前出现，所以我已经抱歉，但是一个
快速的谷歌搜索没有给我任何答案。
我有一个列表，我想要一个更简单的列表，没有重复
一个简单但不知何故设计的解决方案将
在python 2.3中，这应该有效：
导入集
l = [1,2， 2,3]
s = sets.Set（l）
集合是一个容器，其中没有任何对象的不变量两次
持有。所以它适合你的需要。
问候，
Diez
> >>> a = [1,2,2,3]>>> d = {}。fromfromgs（a）
>>> b = d.keys（）
>>> print b
[1,2,3]
那个，或者使用Set（python 2.3+）。实际上我似乎记得
The Python CookBook仍然建议建立一个dict作为最快的
解决方案 - 如果你的元素可以被散列。请参阅
http： //aspn.activestate.com/ASPN/Coo...n/Recipe/52560
干杯，
Bernard 。
christof hoeke写道：
...我有一个列表，我从想要一个没有重复项的简单列表
规范是：
导入集
simplerlist = list（sets。设置（列表））
如果您的破坏顺序没问题，正如您的示例解决方案所示。
但是dict.fromkeys（a）.keys （）可能更快。你的断言：
应该有一个更简单或更直观的解决方案，也许有一个列表
comprehension =
不会''对我来说似乎是不言而喻的。列表理解可能是，例如：
[x代表i，x代表枚举（a）如果i == a.index（x）]
它确实具有以下优点：（a）保持秩序和（b）不需要
需要可清洗（甚至不等于可比性！）的元素 - 但是
它具有O（N * N）的非无关紧要的成本，而其他的
约为O（N）。如果你真的想要类似于你的方法：
>>> b = [x代表x而不是b中的x]
你会有，恐怖！ - ），做一个循环，所以名字b总是绑定到
到目前为止的结果列表 （在LC中，名称b仅限于末尾）：
b = []
for x in a：
如果x不在b中：
b.append（x）
但是，这也是O（N * N）。在更容易或更直观方面，
我怀疑只有后一种解决方案才有资格。
Alex
hello,
this must have come up before, so i am already sorry for asking but a
quick googling did not give me any answer.
i have a list from which i want a simpler list without the duplicates
an easy but somehow contrived solution would be
a = [1, 2, 2, 3]
d = {}.fromkeys(a)
b = d.keys()
print b [1, 2, 3]
there should be an easier or more intuitive solution, maybe with a list
comprehension=
somthing like
b = [x for x in a if x not in b]
print b
[]
does not work though.
thanks for any help
chris解决方案Hi,this must have come up before, so i am already sorry for asking but a
quick googling did not give me any answer.
i have a list from which i want a simpler list without the duplicates
an easy but somehow contrived solution would be
In python 2.3, this should work:
import sets
l = [1,2,2,3]
s = sets.Set(l)
A set is a container for which the invariant that no object is in it twice
holds. So it suits your needs.
Regards,
Diez
> >>> a = [1, 2, 2, 3]>>> d = {}.fromkeys(a)
>>> b = d.keys()
>>> print b
[1, 2, 3]
That, or using a Set (python 2.3+). Actually I seem to recall that
"The Python CookBook" still advises building a dict as the fastest
solution - if your elements can be hashed. See the details at
http://aspn.activestate.com/ASPN/Coo...n/Recipe/52560
Cheers,
Bernard.
christof hoeke wrote:
...i have a list from which i want a simpler list without the duplicates
Canonical is:
import sets
simplerlist = list(sets.Set(thelist))
if you''re allright with destroying order, as your example solution suggests.
But dict.fromkeys(a).keys() is probably faster. Your assertion:
there should be an easier or more intuitive solution, maybe with a list
comprehension=
doesn''t seem self-evident to me. A list-comprehension might be, e.g:
[ x for i, x in enumerate(a) if i==a.index(x) ]
and it does have the advantages of (a) keeping order AND (b) not
requiring hashable (nor even inequality-comparable!) elements -- BUT
it has the non-indifferent cost of being O(N*N) while the others
are about O(N). If you really want something similar to your approach:
>>> b = [x for x in a if x not in b]
you''ll have, o horrors!-), to do a loop, so name b is always bound to
"the result list so far" (in the LC, name b is only bound at the end):
b = []
for x in a:
if x not in b:
b.append(x)
However, this is O(N*N) too. In terms of "easier or more intuitive",
I suspect only this latter solution might qualify.
Alex
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