Python bisect: pass value instead of returning insertion index

Question

I'm using the binary search function mentioned here: When are bisect_left and bisect_right not equal?, but instead of returning False, I want to just skip values that aren't in the list e.

from bisect import bisect_left

def binsearch(l,e):
    index = bisect_left(l,e)
    if index == len(l) or l[index] != e:
        return False
    return index

l = [1, 2, 3, 6, 7, 8, 9]
e = [7, 9, 2, 4, 7]
index = []

for i in e:
   index.append(binsearch(l,i))

print index # [4, 6, 1, False, 4]

I tried replacing return False with pass but am getting the index of where the value not in the list would be placed. Is there a way to simply pass a value if it is not in l and output [4, 6, 1, 4]?

Mark Ransom · Accepted Answer · 2015-11-05 05:20:47Z

1

If you replace the return with a pass in the if statement, it's as if the if didn't exist. That's why you get an index returned.

What you can do instead is return either a sentinel value or a tuple combining the index with a True/False indicator of whether the item was found.

Sentinel style:

if index == len(l) or l[index] != e:
    return -1
return index

Tuple style:

if index == len(l) or l[index] != e:
    return False, None
return True, index

To complete the picture you need to put some logic into the spot where you're building the final list. Example for the tuple form:

for i in e:
    found, ind = binsearch(l,i)
    if found:
        index.append(ind)

edited Nov 5, 2015 at 5:20

answered Nov 5, 2015 at 4:44

Mark Ransom

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