python 有效且非破坏性地获取没有第 k 个元素的列表
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getting list without k'th element efficiently and non-destructively
提问by Chris B.
I have a list in python and I'd like to iterate through it, and selectively construct a list that contains all the elements except the current k'th element. one way I can do it is this:
我在 python 中有一个列表，我想遍历它，并有选择地构造一个包含除当前第 k 个元素之外的所有元素的列表。我可以做到的一种方法是：
l = [('a', 1), ('b', 2), ('c', 3)]
for num, elt in enumerate(l):
# construct list without current element
l_without_num = copy.deepcopy(l)
l_without_num.remove(elt)
but this seems inefficient and inelegant. is there an easy way to do it? note I want to get essentially a slice of the original list that excludes the current element. seems like there should be an easier way to do this.
但这似乎效率低下且不优雅。有没有简单的方法来做到这一点？请注意，我想获得排除当前元素的原始列表的一部分。似乎应该有一种更简单的方法来做到这一点。
thank you for your help.
感谢您的帮助。
回答by Chris B.
l = [('a', 1), ('b', 2), ('c', 3)]
k = 1
l_without_num = l[:k] + l[(k + 1):]
Is this what you want?
这是你想要的吗？
回答by vogonistic
It would help if you explained more how you wanted to use it. But you can do the same with list comprehension.
如果您详细解释了您想如何使用它，将会有所帮助。但是你可以对列表理解做同样的事情。
l = [('a', 1), ('b', 2), ('c', 3)]
k = 1
l_without_num = [elt for num, elt in enumerate(l) if not num == k]
This is also more memory efficient to iterate over if you don't have to store it in l_without_num.
如果您不必将其存储在 l_without_num 中，则迭代的内存效率也更高。
回答by John La Rooy
l=[('a', 1), ('b', 2), ('c', 3)]
k=1
l_without_num=l[:] # or list(l) if you prefer
l_without_num.pop(k)
回答by inspectorG4dget
new = [l[i] for i in range(len(l)) if i != k]
回答by Dan Mantyla
#!/bin/bash
`python -c "'\n'.join(mylist[:])" 2>NULL | sed '/mybadelement/d'`
lol
哈哈
回答by rajivRaja
Using difference operator on sets:
在集合上使用差分运算符：
list(set(l).difference([l[k]])
l=[('a', 1), ('b', 2), ('c', 3)]
list(set(l).difference([l[1]]))
[('a', 1), ('c', 3)]
回答by ephemient
Probably not the most efficient, but the functional programmer in me would probably write this.
可能不是最有效的，但我的函数式程序员可能会写这个。
import operator
from itertools import *
def inits(list):
for i in range(0, len(list)):
yield list[:i]
def tails(list):
for i in range(0, len(list)):
yield list[i+1:]
def withouts(list):
return imap(operator.add, inits(list), tails(list))
for elt, without in izip(l, withouts(l)):
...
import functools, operator
for elt in l:
without = filter(functools.partial(operator.ne, elt), l)
I don't think it's the rightthing to do, but it's short. :-)
我不认为这是正确的做法，但它很短。:-)
