slice (unsorted) array at value in python

Question

Given the array a = [1,1,12,3,5,8,13,21] I can slice off the first 3 elements like a[:3] giving [1,1,2]. What I want is to slice off up to the element of vlaue i (e.g. if i=8 I want [1,1,12,3,5,8] or [1,1,12,3,5] (I can work with either)).

This works:

return a[:a.index(i)]

but only if I give it a value that's in the array.

Is there a clean built in way to do this that does something sane on the missing value case?

Answer 1

Assuming the array is sorted, use a binary search. The function is in the bisect module.

from bisect import bisect_right
a[:bisect_right(a, value)]

Answer 2

Create a generator, and use that:

a = [1,1,2,3,5,8,13,21]

def _gen(listing, cutoff):
    for i in listing:
        if i == cutoff:
            return

        yield i

new_a = list(_gen(a, 5))

... or, if you really want a slice ...

for i, val in enumerate(a):
    if val == cutoff:
        break

new_a = a[:i]

Answer 3

1. That's a list.

2. Try

   >>> a = [1,1,2,3,5,8,13,21]
   >>> import itertools
   >>> for x in itertools.takewhile(lambda val: val != 8, a):
   ...     print x
   ...
   1
   1
   2
   3
   5