CODEWITHSUNDEEP
pythonregex
Martin
Martin

Reputation: 1243

Regex filter numbers divisible by 3

I have a list of comma-separated ids(digits) . And I need to get only these which are divisible by 3.

Example: i = "3454353, 4354353, 345352, 2343242, 2343242 ..."

Upvotes: 3

Views: 10874

Answers (4)

merger
merger

Reputation: 41

A hopefully complete version, from reduction of DEA[1]:

^([0369]|[147][0369]*[258]|(([258]|[147][0369]*[147])([0369]|[258][0369]*[147])*([147]|[258][0369]*[258])))+$

[1:] Converting Deterministic Finite Automata to Regular Expressions', C. Neumann 2005
NOTE: There is a typo in Fig.4: the transition from q_j to itself should read ce*b instead of ce*d.

Upvotes: 4

Tim Pietzcker
Tim Pietzcker

Reputation: 336128

Just for the heck of it:

reobj = re.compile(
    r"""\b            # Start of number
    (?:               # Either match...
     [0369]+          # a string of digits 0369
    |                 # or
     [147]            # 1, 4 or 7
     (?:              # followed by
      [0369]*[147]    # optional 0369s and one 1, 4 or 7
      [0369]*[258]    # optional 0369s and one 2, 4 or 8
     )*               # zero or more times,
     (?:              # followed by
      [0369]*[258]    # optional 0369s and exactly one 2, 5 or 8
     |                # or
      [0369]*[147]    # two more 1s, 4s or 7s, with optional 0369s in-between.
      [0369]*[147]
     )
    |                 # or the same thing, just the other way around,
     [258]            # this time starting with a 2, 5 or 8
     (?:
      [0369]*[258]
      [0369]*[147]
     )*
     (?:
      [0369]*[147]
     |
      [0369]*[258]
      [0369]*[258]
     )
    )+                # Repeat this as needed
    \b                # until the end of the number.""", 
    re.VERBOSE)
result = reobj.findall(subject)

will find all numbers in a string that are divisible by 3.

Upvotes: 18

Matthias
Matthias

Reputation: 13222

Using the idea from this question i get:

i = "1, 2, 3, 4, 5, 6, 60, 61, 3454353, 4354353, 345352, 2343241, 2343243"

for value in i.split(','):
    result = re.search('^(1(01*0)*1|0)+$', bin(int(value))[2:])
    if result:
        print '{} is divisible by 3'.format(value)

But you don't want to use regular expressions for this task.

Upvotes: 2

georg
georg

Reputation: 214949

If you really mean digits (not numbers), this is as easy as

 re.findall(r'[369]', my_str)

For a list of numbers, it's quite easy without regular expressions:

lst = "55,62,12,72,55"
print [x for x in lst.split(',') if int(x) % 3 == 0]

Upvotes: 10

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