CODEWITHSUNDEEP
pythonpython-3.xlist
hub_phoenix
hub_phoenix

Reputation: 21

pop from list using index list

list1 = [1,2,5,6,7,8,10,41,69,78,83,100,105,171]
index_list = [0,4,7,9,10]

how do I pop an item from list1 using indexes from index_list?

output_list = [2,5,6,8,10,69,100,105,17]

Upvotes: 1

Views: 5088

Answers (4)

Fuji Komalan
Fuji Komalan

Reputation: 2047

list1 = [1,2,5,6,7,8,10,41,69,78,83,100,105,171]
index_list = [0,4,7,9,10]


print([ t[1] for t in enumerate(list1) if t[0] not in index_list])

RESULT

[2, 5, 6, 8, 10, 69, 100, 105, 171]

enumerate will create a structure like below.

[(0, 1), (1, 2),(2, 5),(3, 6),(4, 7),(5, 8),...(13, 171)]

Where t = (0,1) (index,item)
t[0] = index
t[1] = item

Upvotes: 1

alec_djinn
alec_djinn

Reputation: 10789

Use list.remove(item)

for n in reversed(index_list):
    list1.remove(list1[n])

or list.pop(index)

for n in reversed(index_list):
    list1.pop(n)

Both methods are described here https://docs.python.org/2/tutorial/datastructures.html

Use reversed() on your index_list (assuming that the indices are always ordered like in the case you have shown), so you remove items from the end of the list and it should work fine.

Upvotes: 1

kuro
kuro

Reputation: 3226

You could try this -

for index in sorted(index_list, reverse=True):
    list1.pop(index)

print (list1)

pop() has an optional argument index. It will remove the element in index

Upvotes: 0

tobias_k
tobias_k

Reputation: 82899

How about the opposite: Retain those elements that are not in the list:

>>> list1 = [1,2,5,6,7,8,10,41,69,78,83,100,105,171]
>>> index_list = [0,4,7,9,10]
>>> index_set = set(index_list) # optional but faster    
>>> [x for i, x in enumerate(list1) if i not in index_set]
[2, 5, 6, 8, 10, 69, 100, 105, 171]

Note: This does not modify the existing list but creates a new one.

Upvotes: 1

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