Comparison of two numbers

Question

I'm trying to make a program that will check if two numbers have the same digits but in different order. For example, 232 and 223 will print "true" but 123 and 223 will print "false". Now I have no mistakes but the answer should be "true" and it's not:

my code:

a=322 
b=223

list_a = list(str(a))
list_b = list(str(b))
c=len(str(a))
d=len(str(b))

j=0

if c != d:
    print "false"
else:
    for i in range(len(list_a)):
        while j<d:
           if i == list_b[j]:
            list_b.remove(list_b[j])
            break
           j=j+1
        j=0



if list_b==[]:
    print "true"

Hint - When comparing numbers don't depend on length. You see 1 = 0001. Also you don't need a loop. Go and check if a=b. — NoChance, Oct 26 '13 at 13:36
The problem is that the numbers are not equal. 322 is not equal to 223 but it should bring me back "true" — ProductM, Oct 26 '13 at 13:38
So what do you want to do when they are not equal? What is the problem? — NoChance, Oct 26 '13 at 13:39
I need to find if they have the same digits but in a differnet order. how can I do it without the loop? — ProductM, Oct 26 '13 at 13:42
What is this project? It's just homework I have problem with. — ProductM, Oct 26 '13 at 15:06

Crowman · Answer 1 · 2013-10-26T15:12:08.040

Something like this seems like an obvious way:

#!/usr/bin/python

def same_digits(a, b):
    if sorted(str(a)) == sorted(str(b)):
        print "{0} and {1} contain the same digits".format(a, b)
    else:
        print "{0} and {1} do not contain the same digits".format(a, b)

same_digits(232, 232)
same_digits(232, 223)
same_digits(232, 233)

Output:

paul@local:~/src/python/scratch$ ./testnum.py
232 and 232 contain the same digits
232 and 223 contain the same digits
232 and 233 do not contain the same digits
paul@local:~/src/python/scratch$

If you want to match true regardless of the number of each digit, then use set to eliminate duplicates:

#!/usr/bin/python

def same_digits(a, b):
    if sorted(set(str(a))) == sorted(set(str(b))):
        print "{0} and {1} contain the same digits".format(a, b)
    else:
        print "{0} and {1} do not contain the same digits".format(a, b)

same_digits(232, 232)
same_digits(232, 223)
same_digits(232, 233)
same_digits(232, 2333332232)
same_digits(232, 2)
same_digits(232, 234)

Output:

paul@local:~/src/python/scratch$ ./testnum2.py
232 and 232 contain the same digits
232 and 223 contain the same digits
232 and 233 contain the same digits
232 and 2333332232 contain the same digits
232 and 2 do not contain the same digits
232 and 234 do not contain the same digits
paul@local:~/src/python/scratch$

If you really must do it the hard way, then this replicates the first example without using sorted():

#!/usr/bin/python

def same_digits_loop(a, b):
    a_alpha = str(a)
    b_alpha = str(b)

    if len(a_alpha) != len(b_alpha):
        return False

    for c in a_alpha:
        b_alpha = b_alpha.replace(c, "", 1)

    return False if len(b_alpha) else True


def same_digits(a, b):
    if same_digits_loop(a, b):
        print "{0} and {1} contain the same digits".format(a, b)
    else:
        print "{0} and {1} do not contain the same digits".format(a, b)


same_digits(232, 23)
same_digits(232, 232)
same_digits(232, 223)
same_digits(232, 233)
same_digits(232, 2333)

and outputs:

paul@local:~/src/python/scratch$ ./testnum3.py
232 and 23 do not contain the same digits
232 and 232 contain the same digits
232 and 223 contain the same digits
232 and 233 do not contain the same digits
232 and 2333 do not contain the same digits
paul@local:~/src/python/scratch$

For the code you have in your question in your latest edit, just change:

if i == list_b[j]:

to:

if list_a[i] == list_b[j]:

and it'll work. That being said, it won't always work, because when you do this:

while j<d:

every time you remove an element from list_b, the length of list_b will change, but d will not. You'll be going out of bounds when the digits are not the same unless you update d to equal the new length each time, and check if list_b has become empty before you've reached the end of list_a.

There is a way to do it without the sort? the mission is only with loops — ProductM, Oct 26 '13 at 14:24
You could use `return not len(b_alpha)` instead of `return False if len(b_alpha) else True` — JadedTuna, Oct 26 '13 at 14:58
@Vik2015: True, although that would be less clear and explicit. Terseness is not generally a design goal in Python. — Crowman, Oct 26 '13 at 15:00
can you please check what I did wrong? I edited the original post — ProductM, Oct 26 '13 at 15:02
@user2923032: See my latest edit, you just have one small error that's throwing you off. — Crowman, Oct 26 '13 at 15:06

tinylambda · Answer 2 · 2013-10-26T14:23:53.083

You can use the Counter object to obtain the fingerprint of each digits string, and then just compare the two fingerprint, that's all of it:

In [1]: n1 = 123

In [2]: n2 = 312

In [3]: n1, n2 = map(str, [n1, n2])

In [4]: n1,n2
Out[4]: ('123', '312')   

In [5]: from collections import Counter

In [6]: c1 = Counter(n1)

In [7]: c2 = Counter(n2)

In [8]: c1 == c2
Out[8]: True

In [9]: c1
Out[9]: Counter({'1': 1, '3': 1, '2': 1})

In [10]: c2
Out[10]: Counter({'1': 1, '3': 1, '2': 1})

If you are not care about the number of digits in the string, you can use the set builtin type to obtain the fingerprint:

In [13]: n1 = 112

In [14]: n2 = 212

In [15]: n1, n2 = map(str, [n1, n2])

In [16]: s1 = set(n1)

In [17]: s2 = set(n2)

In [18]: s1
Out[18]: set(['1', '2'])

In [19]: s2
Out[19]: set(['1', '2'])

In [20]: s1 == s2
Out[20]: True

The only work you should do is just find some kind of Fingerprint!

aga · Answer 3 · 2013-10-26T14:38:16.523

0

i variable in your snippet is string, not integer, that's why you have TypeError. Change

if list_a[i] == list_b[j]:

to

if i == list_b[j]:

If you wan to use i like index, change for loop like so:

for i in range(len(list_a)):

Also, indices in python start from 0, not from 1, so change

j = 1

to

j = 0

One more correction:

while j <= d:

to

while j < d:

So you won't have index-out-of-range.

Also this one:

list_a = list(str(a))
list_b = list(str(b))

edited Oct 26 '13 at 14:38

answered Oct 26 '13 at 14:23

aga

27,954
13
86
121

AttributeError: 'str' object has no attribute 'remove' – ProductM Oct 26 '13 at 14:32
I still have the problem with the "remove" – ProductM Oct 26 '13 at 14:37
@user2923032 that's because your `list_b` is string, not a list. Take a look at my last correction. – aga Oct 26 '13 at 14:39
Right now I have no mistakes but the answer should be true and it's not. I edited my original post. what is wrong? thank you!` – ProductM Oct 26 '13 at 14:48

score 0 · Answer 4 · answered Jan 07 '19 at 18:24

Comparing two numbers with different digits using sort():

 def check_different_numbers_having_same_digits(num1,num2):
            num1=[int(i) for i in str(num1)] #converting to int to list
            num1.sort()                     #sorting the list
            num2=[int(i) for i in str(num2)]
            num2.sort()
            if(num1==num2):
                return True
            else:
                return False

        print(check_different_numbers_having_same_digits(234,324))

score 0 · Answer 5 · answered Nov 01 '22 at 22:44

Python Comparison Operators In fact, this is the first and easiest practice about the Python Comparison Operators. Let's Get Code !

first_Number = input('Enter firstNumber: ') #Prompt first number from user
second_Number = input('Enter secondNumber: ') #Prompt second number from user
# Pay Attention that the numbers we got from users are integers in the string Format
first_Number_int = int(first_Number) # Then we need to convert the string to integer ( with int() Function)
second_Number_int = int(second_Number) # Same conversion for second number 
#We could write in this way to optimize and legiblely write code instead of the top 4 lines of code in summary. 
# first_Number = int(input('Enter firstNumber: '))
# second_Number = int(input('Enter secondNumber: '))
# We need conditional statment to measure and compare between two numbers. 
# We'll have three different results. 
if (first_Number_int > second_Number_int): # First : first number is bigger than second number
    print('firstNumber is Bigger than secondNumber') 
elif (first_Number_int < second_Number_int)) : # Second : second number is bigger than first number 
    print('secondNumber is Bigger than firstNumber')
else: # Third: Equal ( Some people only take two conclusions that only numbers are bigger than each other )  
    print('firstNumber and secondNumber are Euqal to Eachother')

I hope I've been able to help you, my dear friends.

The problem that my mind is busy with, is there another way to compare two numbers in python? If you, my dear friends, find another solution, I'd be happy to learn.

Thank you so much ,

Sepehr Tajbakhsh

2 November , 2022

But this is entirely off topic. Please be sure to understand the question. I am reviewing this, and, frankly, it is so off-topic, that I am not entirely sure that this is not a strange form of vandalism. — chrslg, Nov 06 '22 at 16:55

score -1 · Answer 6 · answered Oct 26 '13 at 14:08

-1

I think it helps

    a = 323
    b = 233

    al = [c for c in str(a)]
    bl = [c for c in str(b)]

    print sorted(al) == sorted(bl)

answered Oct 26 '13 at 14:08

fryday

387
2
14

There is a way to do it without the sort? the mission is only with loops – ProductM Oct 26 '13 at 14:22
The list comparison is not needed, just takes time and memory; `sorted()` can sort string too, and using `sorted()` with list is a bad idea if you don't want to save a copy; you `list.sort()` to sort in-place. – Chayim Friedman Sep 01 '19 at 04:01

Comparison of two numbers

6 Answers6