Python: Checking a variable is in binary

To confirm a string contains exactly eight ones and zeros, test it for the regular expression

^[01]{8}$

Example of use:

import re
isBin = re.compile('^[01]{8}$')
s1 = "00110101"

if(s1.match(isBin)):
  print "it is a match"
else:
  print "it is not a match"

You can use a generator expression and all*:

if len(var) == 8 and all(x in "01" for x in var):
    ...

Below is a demonstration:

>>> var = "01010101"
>>> len(var) == 8 and all(x in "01" for x in var)
True
>>> var = "0101010"
>>> len(var) == 8 and all(x in "01" for x in var)
False
>>> var = "01010102"
>>> len(var) == 8 and all(x in "01" for x in var)
False
>>>

*Note: the above code assumes that var is a string.

Is the variable encoded as a string, or array of integers? You can just do a list comprehension to check each element, other solutions are possible as well.

isBinary=[x==1 or x==2 for x in variable]
isBinary=False in isBinary

You could try and parse them with int while passing a radix like this:

>>> x = int("10010", 2)
>>> print x
18

You can use the all() function to test each character:

len(var) == 8 and all(c in '01' for c in var)

or use a set:

binary_digits = set('01')
len(var) == 8 and binary_digits.issuperset(var)

or use a regular expression:

import re

binary_digits = re.compile('^[01]{8}$')
binary_digits.match(, var) is not None

Of these three choices, the regular expression option is the fastest, followed by using a set:

>>> import re
>>> import timeit
>>> def use_all(v): return len(v) == 8 and all(c in '01' for c in v)
... 
>>> def use_set(v, b=set('01')): return len(v) == 8 and b.issuperset(v)
... 
>>> def use_re(v, b=re.compile('^[01]{8}$')): return b.match(v) is not None
... 
>>> binary, nonbinary = '01010101', '01010108'
>>> timeit.timeit('f(binary); f(nonbinary)', 'from __main__ import binary, nonbinary, use_all as f')
4.871071815490723
>>> timeit.timeit('f(binary); f(nonbinary)', 'from __main__ import binary, nonbinary, use_set as f')
2.558954954147339
>>> timeit.timeit('f(binary); f(nonbinary)', 'from __main__ import binary, nonbinary, use_re as f')
2.036846160888672

I'm going to presume that you also want to convert the string into an integer at some point. If this isn't the case, please correct me.

In Python, it is usually considered better to try to do something, and handle failure if it occurs; rather than checking if something is possible, and only then doing it. This is called the EAFP principle (it is Easier to Ask for Forgiveness than Permission).

In this case, you should use a try except:

s = '01100011'
if len(s) == 8:
    try:
        n = int(s, 2)
    except ValueError:
        handle_exception_here()