How to check if a input is in binary format(1 and 0)?

Question

I have made a program however I wanted to add an exception if the user inputs is not in a binary format. I have tried many times adding exceptions but I can't seem to get it to work. The below is the program code. I would appreciate if someone could help.

import time
error=True
n=0
while n!=1:
    print"***Welcome to the Bin2Dec Converter.***\n"
    while error:
        try:
            bin2dec =raw_input("Please enter a binary number: ")
            error=False
        except NameError: 
            print"Enter a Binary number. Please try again.\n"
            time.sleep(0.5)
        except SyntaxError: 
            print"Enter a Binary number. Please try again.\n"
            time.sleep(0.5)


        #converts bin2dec
        decnum = 0 
        for i in bin2dec: 
            decnum = decnum * 2 + int(i)
            time.sleep(0.25)
        print decnum, "<<This is your answer.\n" #prints output

score 9 · Answer 1 · edited Sep 15 '13 at 11:59

9

Better to ask for forgiveness. Try to convert it to integer using int(value, 2):

while True:
    try:
        decnum = int(raw_input("Please enter a binary number: "), 2)
    except ValueError:
        print "Enter a Binary number. Please try again.\n"
    else:
        break

print decnum

edited Sep 15 '13 at 11:59

ThiefMaster

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answered Sep 15 '13 at 11:55

alecxe

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Yup - much more filled out than my answer :) – Jon Clements Sep 15 '13 at 11:55
1

I think it would be prettier to get rid of the useless `continue` and maybe move `break` into an `else` block. – ThiefMaster Sep 15 '13 at 11:57
This clearly demonstrates how to convert a binary number to decimal in Python (+1), but it's not what the OP asked... – Alex Chamberlain Sep 15 '13 at 12:03
2

@AlexChamberlain, it's an implicit check (converting only works for binary input), so this is kind of what the OP asked + a better way to handle the problem IMO – tamasgal Sep 15 '13 at 12:04
1

Note that this function allows more valid inputs than the OP's version. You should check whether the user input contains a `b` before calling `int(..., 2)`, otherwise inputs starting with the `0b` prefix will be allowed(which I believe is not something the OP wants, and anyway he should be aware of this change). – Bakuriu Sep 15 '13 at 12:15
Thanks for all your suggestions, if i could request someone to put the above in the context of my code. I'm still a beginer at python coding. Thanks – DaCruzR Sep 15 '13 at 14:12

score 6 · Answer 2 · answered Sep 15 '13 at 11:54

6

int(bin2dec, 2) will throw a ValueError if the input isn't in binary format. But of course that solves the whole problem for you.

answered Sep 15 '13 at 11:54

Imre Kerr

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tamasgal · Answer 3 · 2013-09-15T12:02:05.427

5

Using set():

def is_binary(x):
    return set(input_string) <= set('01')

input_string = "0110110101"
print(is_binary(input_string))

input_string = "00220102"
print(is_binary(input_string))

edited Sep 15 '13 at 12:02

answered Sep 15 '13 at 11:56

tamasgal

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1

Note, this won't work for "00000" or "111111" (or any other pattern containing just one of the two characters). – Sylvain Defresne Sep 15 '13 at 11:58
1

I think you could write `set(input_string) <= set('01')` http://docs.python.org/2/library/sets.html#set-objects – Stuart Sep 15 '13 at 12:01
@Septi if you really wanted to use a set for this, then another way is `not set('01').symmetric_difference(input_string)` – Jon Clements Sep 15 '13 at 12:03
Another option: `is_binary = set('01').issuperset`. This works on strings. – Stuart Sep 15 '13 at 12:19

score 2 · Answer 4 · answered Sep 15 '13 at 11:53

2

Using all:

>>> b = '01011'
>>> all(c in '01' for c in b) # OR  c in ('0', '1')
True
>>> b = '21011'
>>> all(c in '01' for c in b) # OR  c in ('0', '1')
False

answered Sep 15 '13 at 11:53

falsetru

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is it faster than others? – Grijesh Chauhan Sep 15 '13 at 12:02
I mean because [`any()` and `all()` functions also support short-circuiting](http://stackoverflow.com/questions/2580136/does-python-support-short-circuiting/14892812#14892812) – Grijesh Chauhan Sep 15 '13 at 12:06
1

@GrijeshChauhan, I'm not sure. But I think `int` will also stop if it found some invalid character/digit inside loop. – falsetru Sep 15 '13 at 12:10
Ok `int(your_string, 2)` looks more proper way too...thanks to add a technique – Grijesh Chauhan Sep 15 '13 at 12:16
1

@GrijeshChauhan, See [source code](http://hg.python.org/cpython/file/tip/Objects/longobject.c) (esp. `PyLong_FromString`) – falsetru Sep 15 '13 at 12:19

ThiefMaster · Answer 5 · 2013-09-15T11:59:44.730

2

The proper way to do this (i.e. if it's not a stupid homework exercise) is using int(your_string, 2) and catching ValueError which is raised if the string contains an invalid character.

http://docs.python.org/2/library/functions.html#int

edited Sep 15 '13 at 11:59

answered Sep 15 '13 at 11:56

ThiefMaster

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score 1 · Answer 6 · answered Sep 15 '13 at 11:58

1

>>> b = '01011'
>>> not(b.translate(None, '01'))
True
>>> b = '21011'
>>> not(b.translate(None, '01'))
False

answered Sep 15 '13 at 11:58

John La Rooy

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`not` is a unary operator, so you don't need to add parenthesis as for a function call. – Yann Vernier Sep 15 '13 at 12:20

Alex Chamberlain · Answer 7 · 2013-09-15T12:04:15.423

1

Using re:

>>> import re
>>> matches = re.match('[01]*$', bin2dec)
>>> if matches:
...    process(bin2dec)

edited Sep 15 '13 at 12:04

answered Sep 15 '13 at 11:58

Alex Chamberlain

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@JonClements Thanks, I agree with the `$`; I always forget `match` has an implicit `^`, but not a `$`. `*` vs `+` is a question for the OP really. – Alex Chamberlain Sep 15 '13 at 12:06

Stuart · Accepted Answer · 2013-09-15T12:49:30.333

If you are avoiding Python's built in way of doing this (int(..., 2)), as a learning exercise, then a logical and Pythonic approach would be to make your own error class and build the error checking in to your conversion function.

class BinaryError(Exception):
    def __str__(self):
        return "Not a valid binary number"

def bin2dec(input_string):
    r = 0
    for character in input_string:
        if character == '0':
            r = r * 2
        elif character == '1':
            r = r * 2 + 1
        else:
            raise BinaryError()
    return r

while True:
    try:
        print bin2dec(raw_input("Please enter a binary number: "))
    except BinaryError:
        print "Enter a Binary number. Please try again.\n"
    else:
        break

How to check if a input is in binary format(1 and 0)?

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