I have this string: 11000000101010000000010000000000
I would like to count the 0s starting at the back until I hit 1 and stop there, determining the total number of 0s at the end. In this particular case it would give me 10 as an answer.
Any help greatly appreciated.
You can use rindex() to get the index of the last 1 and then subtract that from the maximum index (len(s) - 1):
>>> s = '11000000101010000000010000000000'
>>> len(s) - s.rindex('1') - 1
10
rindex will do) seems most appropriate to me in such cases.Use str.rsplit() and str.count()
>>> s = '11000000101010000000010000000000'
>>> len(s.rsplit('1', 1)[-1])
10
rsplit (instead of just split), use the maxsplit argument: s.rsplit('1', 1). Also, no need for count once it's been split. Just use len(s.rsplit('1',1)[-1])len(s) - len(s.rstrip('0'))1s.Here is how to do it in regex, because why not!
>>> s = '11000000101010000000010000000000'
>>> match = re.search('0*$', s)
>>> match.end() - match.start()
10
I know the question was answered already, but I thought I would add yet another way that you could do this.
You could use itertools.takewhile on the reverse of the string and takewhile the digit is not '1'. Apply a sum to all the 1s that were generated and you get the answer.
>>> test = "11000000101010000000010000000000"
>>> sum(1 for x in takewhile(lambda i: i != '1', test[::-1]))
10
You could either :
perform successive divisions by 10 on the corresponding integer taken as if it was in base 10.
use strings operations to find the last 1 and take everything after it
use regular expressions to get the 0s at the end and count them
look for operations to convert to binary and perform successive divisions by two.
binary_str = "11000000101010000000010000000000"
import re
match_obj = re.search(r"0*$", binary_str)
if match_obj:
print len(match_obj.group())
