Get top-6 elements, subject to the condition that element >= 3

Question

I'm trying to make a list of 11 elements

I want to pop out the maximum element only from l[0] to l[5]. under one of the two conditions:

l[5] >= 3 or l[5] is no longer exist or None.

l = [2,8,6,2,8,7,9,8,6,7,4]
max = 0
maxIndex = 0
while (l[5] >= 3 or l[5] is None):
    for x in range(6):
        if l[x] > max:
            max = l[x]
            maxIndex = x
    l.pop(maxIndex)
print(l)

I am getting the error:

IndexError: list index out of range

I know at some point l[5] will no longer exist, but how can I create this code.

Add a condition in the while statement which should go first. It should be: `len(l) > 5` — sshashank124, Apr 20 '15 at 20:34
Your function is rather complex; all you really need is [`heapq.nsmallest()`](https://docs.python.org/2/library/heapq.html#heapq.nsmallest) here. — Martijn Pieters, Apr 20 '15 at 20:50
You want the **top-n (here, n=6), subject to the condition that x >= 3**. As @MartijnPieters said, don't start from a list in the first place, use a self-sorting data structure like `heapq` — smci, Apr 20 '15 at 20:58
Don't say *"pop... from a list"*. Say *"Get top-n"*. State your problem without nailing it down to specific data structures or operations. — smci, Apr 20 '15 at 21:00

score 2 · Answer 1 · answered Apr 20 '15 at 20:36

Minimal reproducible code:

l = [1, 2, 3]
l[3] is None # raises IndexError
assert len(l) == 3 # len(l) returns 3

As you can see, accessing non-existing item raises an IndexError, so correct condition should contain length test.

while (len(l) >= 6 and l[5] >= 3):

score 0 · Answer 2 · edited Apr 20 '15 at 20:47

0

Reverse the conditions:

l = [2,8,6,2,8,7,9,8,6,7,4]
max = 0
maxIndex = 0
while len(l) > 5 or l[5] >= 3:
    for x in range(6):
        if l[x] > max:
            max = l[x]
            maxIndex = x
    l.pop(maxIndex)
print(l)

edited Apr 20 '15 at 20:47

sshashank124

31,495
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answered Apr 20 '15 at 20:35

hd1

33,938
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score 0 · Answer 3 · answered Apr 20 '15 at 20:56

I think this does all you need - returns maximum if there is a 5th element and it's >=3 or None

>>> l =  [2,8,6,2,8,7,9,8,6,7,4]

>>> max(l[0:5]) if len(l)>5 and (l[5]>=3 or l[5] is None) else None
8

>>> l = [1,2,3]
>>> max(l[0:5]) if len(l)>5 and (l[5]>=3 or l[5] is None) else None
>>>

Get top-6 elements, subject to the condition that element >= 3

3 Answers3