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How do I trim a sorted list in Python?

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How do I trim a sorted list in Python, so that I only keep the elements greater than a certain value?

For example, if I have this list:

l = [1, 3, 5, 6, 7, 8]

How can I efficiently get a list of numbers greater than 4?

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You can use the functions in the bisect module with a list slicing operator. For example bisect_left finds the index of the insertion point in a sorted list, and if the values are equal, it will find the position to its left.

>>> l = [1, 3, 5, 6, 7, 8]
>>> from bisect import bisect_left
>>> bisect_left(l, 4)
2
>>> bisect_left(l, 5)
2
>>> l[bisect_left(l, 4):]
[5, 6, 7, 8]
>>> l[bisect_left(l, 5):]
[5, 6, 7, 8]
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Python has inbuilt list filtering syntax:

l = [1,3,5,6,7,8]
trimmed_list = [item for item in l if item > 4]
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The library module itertools provides a function called dropwhile().

The example they give is:

dropwhile(lambda x: x<5, [1,4,6,3,8]) 
→ 6 3 8

So, to get a list of numbers greater than 4, drop items while they are less than or equal to 4:

import itertools

l = [1, 3, 5, 6, 7, 8]
result = itertools.dropwhile(lambda x: x<=4, l)

Note that result at this point will be a lazy iterator object of some kind, not an actual list. So you might have to do: print(*result) to confirm it works. But you can iterate it, or call list(result) to make it a true list. (Remember, also, that iterators get consumed.)

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Not the most efficient solution (1 comment)

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