使用列表理解使项目不在第二列表中

Question

I am trying to use list comprehension to remove a number of items from a list by just keeping those not specified.

For example if I have 2 lists a = [1,3,5,7,10] and b = [2,4] I want to keep all items from a that are not at an index corresponding to a number in b .

Now, I tried to use y = [a[x] for x not in b] but this produces a SyntaxError.

y = [a[x] for x in b] works fine and keeps just exact the elements that i want removed.

So how do I achieve this? And on a side note, is this a good way to do it or should I use del ?

Answer 1

You can use enumerate() and look up indexes in b :

>>> a = [1, 3, 5, 7, 10]
>>> b = [2, 4]
>>> [item for index, item in enumerate(a) if index not in b]
[1, 3, 7]

Note that to improve the lookup time, better have the b as a set instead of a list. Lookups into sets are O(1) on average while in a list - O(n) where n is the length of the list.

Answer 2

Guess you're looking for somthing like :

[ x  for x  in a if a.index(x) not in b  ] 

Or, using filter:

filter(lambda x : a.index(x) not in b , a)

Answer 3

试试这个会起作用

   [j for i,j in enumerate(a) if i not in b ]

Answer 4

after this:

y = [a[x] for x in b]

just add:

for x in y:
    a.remove(x)

then you end up with a stripped down list in a