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 ...'

Solution 1:

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.

Solution 2:

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 [xforx in lst.split(',') ifint(x) % 3 == 0]

Solution 3:

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.

Solution 4:

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_i to itself should read ce*b instead of ce*d.