Binary list from indices of ascending integer list

Question

I have an ascending list of integers e that starts from 0 and I would like to have a binary list b whose i-th element is 1 if and only if i belongs to e.

For example, if e=[0,1,3,6], then this binary list should be [1,1,0,1,0,0,1], where the first 1 is because 0 is in e, the second 1 is because 1 is in e, the third 0 is because 2 is not in e, and so on.

You can find my code for that below.

My question is: is there something built-in in python for that? If not, is my approach the most efficient?

def list2bin(e):
b=[1]
j=1
for i in range(1, e[-1]+1):
    if i==e[j]:
        b.append(1)
        j+=1
    else:
        b.append(0)     
return(b)

Ashwini Chaudhary · Accepted Answer · 2014-01-28 15:18:12Z

7

This can be done with a list comprehension, and in case e is huge then better convert it to a set first:

>>> e = [0, 1, 3, 6]
>>> [int(i in e) for i in xrange(0, e[-1]+1)]
[1, 1, 0, 1, 0, 0, 1]

The in operator returns True/False if an item is found in the list, you can convert that bool to an integer using int. Note that for lists the in is O(N) operation, so if e is large then converting it to a set will provide you much more efficiency.

answered Jan 28, 2014 at 15:18

Ashwini Chaudhary

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5 Comments

Paulo Bu Over a year ago

Just a comment, xrange doesn't exist in Python3, other than that the solution is beautiful.

geo909 Over a year ago

@Ashwini Thank you! Unfortunately I do have to use lists. e is not going to be that huge (maximum 150 elements, although later on I may try with around 400), but this will be called over and over, many times (I'm running a reccursive algorithm that will be running for weeks and what I'm asking will be in every iteration). Do you think it makes sense to turn the list into a set every time, in my case?

Ashwini Chaudhary Over a year ago

@geo909 timeit results: searching 400 in a list of size 400 takes 12 micro-seconds, while a set takes just 200 nano seconds.

geo909 Over a year ago

@Ashwini Thanks.. And although it's obviously more elegant, your solution is better in terms of effiency, right?

Ashwini Chaudhary Over a year ago

@geo909 No, the list based version is not better than your solution theoretically(as it is an O(N**2)) and will be slow for huge lists. But the set version will be almost equivalent to your solution.

Cedmundo · Accepted Answer · 2014-01-28 15:23:54Z

0

I don't think there are a built-in way to do that. But you can use List Comprehensions:

a = [ 1 if i in e else 0 for i in range(1, e[-1]+1) ]

Get fun.

answered Jan 28, 2014 at 15:23

Cedmundo

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