CODEWITHSUNDEEP
python
user2008886
user2008886

Reputation: 111

Python trick in finding leading zeros in string

I have a binary string say '01110000', and I want to return the number of leading zeros in front without writing a forloop. Does anyone have any idea on how to do that? Preferably a way that also returns 0 if the string immediately starts with a '1'

Upvotes: 11

Views: 16551

Answers (8)

Noman Tanveer
Noman Tanveer

Reputation: 335

If you don't know which number would be next to zeros i.e. "1" in this case, and you just want to check if there are leading zeros, you can convert to int and back and compare the two.

"0012300" == str(int("0012300"))

Upvotes: 2

Neil
Neil

Reputation: 7202

I'm using has_leading_zero = re.match(r'0\d+', str(data)) as a solution that accepts any number and treats 0 as a valid number without a leading zero

Upvotes: 0

Xb74Dkjb
Xb74Dkjb

Reputation: 990

How about re module?

a = re.search('(?!0)', data)

then a.start() is the position.

Upvotes: 0

user511824
user511824

Reputation: 326

I'd use:

s = '00001010'
sum(1 for _ in itertools.takewhile('0'.__eq__, s))

Rather pythonic, works in the general case, for example on the empty string and non-binary strings, and can handle strings of any length (or even iterators).

Upvotes: 4

NPE
NPE

Reputation: 500277

Here is another way:

In [36]: s = '01110000'

In [37]: len(s) - len(s.lstrip('0'))
Out[37]: 1

It differs from the other solutions in that it actually counts the leading zeroes instead of finding the first 1. This makes it a little bit more general, although for your specific problem that doesn't matter.

Upvotes: 12

Anton Kovalenko
Anton Kovalenko

Reputation: 21507

If you're really sure it's a "binary string":

input = '01110000'
zeroes = input.index('1')

Update: it breaks when there's nothing but "leading" zeroes

An alternate form that handles the all-zeroes case.

zeroes = (input+'1').index('1')

Upvotes: 12

l4mpi
l4mpi

Reputation: 5149

A simple one-liner:

x = '01110000'
leading_zeros = len(x.split('1', 1)[0])

This partitions the string into everything up to the first '1' and the rest after it, then counts the length of the prefix. The second argument to split is just an optimization and represents the number of splits to perform, meaning the function will stop after it found the first '1' instead of splitting it on all occurences. You could just use x.split('1')[0] if performance doesn't matter.

Upvotes: 9

Foon
Foon

Reputation: 6448

If you know it's only 0 or 1:

x.find(1)

(will return -1 if all zeros; you may or may not want that behavior)

Upvotes: 3

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