Python：检查变量是否为二进制

Question

How would I make an if/else statement where the if checks whether a variable 8 digit variable is in binary? I know how to do the 8 digit part (len()) but I can't work out how to restrict it to 1s and 0s.

Answer 1

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"

Answer 2

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.

Answer 3

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

Answer 4

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

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

Answer 5

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

Answer 6

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()