How do I check if each digit of a number is less than 2? (python)

Question

I'm trying to create a program that checks if each digit of a given number is less than 2, possibly using

range(len(a)):

Can you elaborate? Where are you stuck/what you have tried etc. thanks. — W1ll1amvl, Apr 15 '15 at 21:22
The question doesn't make much sense... is the number an `int`, a `str`, or what? How would `range` help. Write some code and then come back with an example. — tdelaney, Apr 15 '15 at 21:22
Not sure if this is the best method, but convert to string, loop over indices, and check the re-converted values for <2 — Sterling Archer, Apr 15 '15 at 21:24
`all(c in '01' for c in str(a))` or if `a` is already a string `all(c in '01' for c in a)`. — Steven Rumbalski, Apr 15 '15 at 21:25
`all(c in '01' for c in str(a))`* or `all(int(c) < 2 for c in str(a))` — keyser, Apr 15 '15 at 21:26
Cameron will check back on this thread in a few hours and discover that people posted the same answer 6 times for some reason — keyser, Apr 15 '15 at 21:35

score 4 · Answer 1 · answered Apr 15 '15 at 21:32

4

def is_bin(number):
    try:
        int(str(number), 2)
    except ValueError:
        return False
    return True

answered Apr 15 '15 at 21:32

TigerhawkT3

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Plus one for creativity – JuniorCompressor Apr 15 '15 at 21:33

score 3 · Answer 2 · answered Apr 15 '15 at 21:29

3

You can convert number to string and check like this:

all(int(c) < 2 for c in str(n))

For example:

>>> all(int(c) < 2 for c in str(1011))
True
>>> all(int(c) < 2 for c in str(1211))
False

answered Apr 15 '15 at 21:29

JuniorCompressor

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thenaterhood · Answer 3 · 2015-04-15T21:31:59.173

Python is dynamically typed, so this is relatively easy to do. I would suggest converting your integer into a string, then iterating through the string and checking each digit.

Using that method, the code would probably look something like:

your_number_string = str(your_number)

for d in range(0, len(your_number_string)):
    i = your_number_string[d]
    if (int(i) > 2):
        raise new Exception("Digit " + str(i) + "is not less than 2")

Couple things to note with this: It's bad practice to throw bare Exceptions. If you like the exception route, then extend the exception class and make your own. This also assumes that your number is a valid integer. Finally, this also will only alert you for the first digit larger than two, it won't tell you about any subsequent ones that also are larger than 2. This will require some adjustments for negative numbers and floats.

`for d in range(0, len(your_number_string))` is quite unpythonic considering you can do `for c in your_number_string` — Steven Rumbalski, Apr 15 '15 at 21:29
I agree, but I was interpreting the "using range(len(" to mean that was what was desired. It is a good point though, would be cleaner. — thenaterhood, Apr 15 '15 at 21:30

score 1 · Answer 4 · answered Apr 15 '15 at 21:29

1

You can try this

num = 123457
>>>all(int(i)<2 for i in str(num))
False
num = 11011
>>>all(int(i)<2 for i in str(num))
True

answered Apr 15 '15 at 21:29

itzMEonTV

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score 0 · Answer 5 · edited May 23 '17 at 12:14

At the moment, all solutions depend on converting the number to a string. While that works, you could do it completely numerically:

def check_digits(nbr, max=1):
    nbr = abs(nbr) # only look at a positive numbers
    while nbr != 0:
        nbr, rem = divmod(nbr, 10)
        if rem > max: return False
    return True

The trick here is that each time the digit furthest to the right gets inspected. Once that's done, the number gets divided by 10, ignoring the remainder. So 1201 becomes 120, which becomes 12 and the cycle stops there, because divmod(12, 10) is (1, 2), and the remainder, 2, is bigger than your maximum digit.

Note that TigerhawkT3's answer and @John La Rooy's comment probably nailed it (upvoted) as it is by far the most Pythonic solution. All others, including mine, work for various programming languages, TigerhawkT3's solution uses Python exceptions to good effect.

score 0 · Answer 6 · answered Apr 15 '15 at 21:35

Could use the good ol' fashioned method :) just take your number and extract each digit and then update your number.

while yournumber != 0 :
    digit = yournumber % 10
    yournumber = younumber / 10
    if digit < 2
        do stuff

but I'm sure there are easier (maybe not as fast) ways.

How do I check if each digit of a number is less than 2? (python)

6 Answers6