How to deal with very large numbers in python (19 digits) [closed]

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Closed 8 years ago.

In python I need to be able to cycle through a 19 digit number; what is the typical way of doing this? I'm using python 2.7.x; but will use python 3.x if there is a viable solution.

I have a large number; 1000**5 (and even larger 1000**10) for example; I need to cycle through this number list in a for loop. I am aware of the time it will take; but because I cannot find a way to cycle through such a big number I'm at a loss.

I've tried with xrange, range, and itertools.islice in python 2.7 and receive Overflow errors.

Thanks

Answer 1

The following is true for python 3.x:

The int type in python does not have a limit. If you want to iterate through 0,1,2...n where n is a 19-digit number, you can just do:

for i in range(n):
    pass #whatever you like

Although that would take a very long time.

Answer 2

Not as pretty as for, but just use while() and you will overcome the problem (plus it will work with whatever version of Python):

i=0
while (i<1000**10):
    i += 1
    # do some stuff

Answer 3

In case you were trying to iterate the single digits:

long_int_string = '34234345456575657'
for digit in [int(x) for x in long_int_string]:
    # do something useful with digit        
    print digit

Answer 4

It sounds like you want to iterate through the individual digits of an integer. For situations like these, it is useful to know that x % 10 is equal to the rightmost digit of a positive integer x in base ten. With that in mind, you can get every digit by iteratively using modulus and reducing the number.

def splitIntoDigits(x):
    digits = []
    while x != 0:
        digits.insert(0, int(x%10))
        x /= 10
    return digits

print splitIntoDigits(15016023042)
[1, 5, 0, 1, 6, 0, 2, 3, 0, 4, 2]

Also, comedy one-line answer:

>>> [int(num/(10**x)%10) for x in range(int(math.log(num,10)),-1,-1)]
[1, 5, 0, 1, 6, 0, 2, 3, 0, 4, 2]