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Length of an integer in Python

Questions:

In Python, how do you find the number of digits in an integer?

If you want the length of an integer as in the number of digits in the integer, you can always convert it to string like str(133) and find its length like len(str(123)).

Questions:

Without conversion to string

import math
digits = int(math.log10(n))+1

To also handle zero and negative numbers

import math
if n > 0:
digits = int(math.log10(n))+1
elif n == 0:
digits = 1
else:
digits = int(math.log10(-n))+2 # +1 if you don't count the '-'

You’d probably want to put that in a function 🙂

Here are some benchmarks. The len(str()) is already behind for even quite small numbers

timeit math.log10(2**8)
1000000 loops, best of 3: 746 ns per loop
timeit len(str(2**8))
1000000 loops, best of 3: 1.1 µs per loop

timeit math.log10(2**100)
1000000 loops, best of 3: 775 ns per loop
timeit len(str(2**100))
100000 loops, best of 3: 3.2 µs per loop

timeit math.log10(2**10000)
1000000 loops, best of 3: 844 ns per loop
timeit len(str(2**10000))
100 loops, best of 3: 10.3 ms per loop

Questions:

Python 2.* ints take either 4 or 8 bytes (32 or 64 bits), depending on your Python build. sys.maxint (2**31-1 for 32-bit ints, 2**63-1 for 64-bit ints) will tell you which of the two possibilities obtains.

In Python 3, ints (like longs in Python 2) can take arbitrary sizes up to the amount of available memory; sys.getsizeof gives you a good indication for any given value, although it does also count some fixed overhead:

>>> import sys
>>> sys.getsizeof(0)
12
>>> sys.getsizeof(2**99)
28

If, as other answers suggests, you’re thinking about some string representation of the integer value, then just take the len of that representation, be it in base 10 or otherwise!

Questions:

All math.log10 solutions will give you problems.

math.log10 is fast but gives problem when your number is greater than 999999999999997. This is because the float have too many .9s, causing the result to round up.

The solution is to use a while counter method for numbers above that threshold.

To make this even faster, create 10^16, 10^17 so on so forth and store as variables in a list. That way, it is like a table lookup.

def getIntegerPlaces(theNumber):
if theNumber <= 999999999999997:
return int(math.log10(theNumber)) + 1
else:
counter = 15
while theNumber >= 10**counter:
counter += 1
return counter

Questions:

Let the number be n then the number of digits in n is given by:

math.floor(math.log10(n))+1

Note that this will give correct answers for +ve integers < 10e15. Beyond that the precision limits of the return type of math.log10 kicks in and the answer may be off by 1. I would simply use len(str(n)) beyond that; this requires O(log(n)) time which is same as iterating over powers of 10.

Thanks to @SetiVolkylany for bringing my attenstion to this limitation. Its amazing how seemingly correct solutions have caveats in implementation details.

Questions:

Well, without converting to string I would do something like:

def lenDigits(x):
"""
Assumes int(x)
"""

x = abs(x)

if x < 10:
return 1

return 1 + lenDigits(x / 10)

Minimalist recursion FTW

Questions:

For posterity, no doubt by far the slowest solution to this problem:

def num_digits(num, number_of_calls=1):
"Returns the number of digits of an integer num."
if num == 0 or num == -1:
return 1 if number_of_calls == 1 else 0
else:
return 1 + num_digits(num/10, number_of_calls+1)

Questions:
from math import log10
digits = lambda n: ((n==0) and 1) or int(log10(abs(n)))+1

Questions:

As mentioned the dear user @Calvintwr, the function math.log10 has problem in a number outside of a range [-999999999999997, 999999999999997], where we get floating point errors. I had this problem with the JavaScript (the Google V8 and the NodeJS) and the C (the GNU GCC compiler), so a 'purely mathematically' solution is impossible here.

Based on this gist and the answer the dear user @Calvintwr

import math

def get_count_digits(number: int):
"""Return number of digits in a number."""

if number == 0:
return 1

number = abs(number)

if number <= 999999999999997:
return math.floor(math.log10(number)) + 1

count = 0
while number:
count += 1
number //= 10
return count

I tested it on numbers with length up to 20 (inclusive) and all right. It must be enough, because the length max integer number on a 64-bit system is 19 (len(str(sys.maxsize)) == 19).

assert get_count_digits(-99999999999999999999) == 20
assert get_count_digits(-10000000000000000000) == 20
assert get_count_digits(-9999999999999999999) == 19
assert get_count_digits(-1000000000000000000) == 19
assert get_count_digits(-999999999999999999) == 18
assert get_count_digits(-100000000000000000) == 18
assert get_count_digits(-99999999999999999) == 17
assert get_count_digits(-10000000000000000) == 17
assert get_count_digits(-9999999999999999) == 16
assert get_count_digits(-1000000000000000) == 16
assert get_count_digits(-999999999999999) == 15
assert get_count_digits(-100000000000000) == 15
assert get_count_digits(-99999999999999) == 14
assert get_count_digits(-10000000000000) == 14
assert get_count_digits(-9999999999999) == 13
assert get_count_digits(-1000000000000) == 13
assert get_count_digits(-999999999999) == 12
assert get_count_digits(-100000000000) == 12
assert get_count_digits(-99999999999) == 11
assert get_count_digits(-10000000000) == 11
assert get_count_digits(-9999999999) == 10
assert get_count_digits(-1000000000) == 10
assert get_count_digits(-999999999) == 9
assert get_count_digits(-100000000) == 9
assert get_count_digits(-99999999) == 8
assert get_count_digits(-10000000) == 8
assert get_count_digits(-9999999) == 7
assert get_count_digits(-1000000) == 7
assert get_count_digits(-999999) == 6
assert get_count_digits(-100000) == 6
assert get_count_digits(-99999) == 5
assert get_count_digits(-10000) == 5
assert get_count_digits(-9999) == 4
assert get_count_digits(-1000) == 4
assert get_count_digits(-999) == 3
assert get_count_digits(-100) == 3
assert get_count_digits(-99) == 2
assert get_count_digits(-10) == 2
assert get_count_digits(-9) == 1
assert get_count_digits(-1) == 1
assert get_count_digits(0) == 1
assert get_count_digits(1) == 1
assert get_count_digits(9) == 1
assert get_count_digits(10) == 2
assert get_count_digits(99) == 2
assert get_count_digits(100) == 3
assert get_count_digits(999) == 3
assert get_count_digits(1000) == 4
assert get_count_digits(9999) == 4
assert get_count_digits(10000) == 5
assert get_count_digits(99999) == 5
assert get_count_digits(100000) == 6
assert get_count_digits(999999) == 6
assert get_count_digits(1000000) == 7
assert get_count_digits(9999999) == 7
assert get_count_digits(10000000) == 8
assert get_count_digits(99999999) == 8
assert get_count_digits(100000000) == 9
assert get_count_digits(999999999) == 9
assert get_count_digits(1000000000) == 10
assert get_count_digits(9999999999) == 10
assert get_count_digits(10000000000) == 11
assert get_count_digits(99999999999) == 11
assert get_count_digits(100000000000) == 12
assert get_count_digits(999999999999) == 12
assert get_count_digits(1000000000000) == 13
assert get_count_digits(9999999999999) == 13
assert get_count_digits(10000000000000) == 14
assert get_count_digits(99999999999999) == 14
assert get_count_digits(100000000000000) == 15
assert get_count_digits(999999999999999) == 15
assert get_count_digits(1000000000000000) == 16
assert get_count_digits(9999999999999999) == 16
assert get_count_digits(10000000000000000) == 17
assert get_count_digits(99999999999999999) == 17
assert get_count_digits(100000000000000000) == 18
assert get_count_digits(999999999999999999) == 18
assert get_count_digits(1000000000000000000) == 19
assert get_count_digits(9999999999999999999) == 19
assert get_count_digits(10000000000000000000) == 20
assert get_count_digits(99999999999999999999) == 20

All example of codes tested with the Python 3.5

Questions:

Assuming you are asking for the largest number you can store in an integer, the value is implementation dependent. I suggest that you don’t think in that way when using python. In any case, quite a large value can be stored in a python ‘integer’. Remember, Python uses duck typing!

Edit:
I gave my answer before the clarification that the asker wanted the number of digits. For that, I agree with the method suggested by the accepted answer. Nothing more to add!

Questions:
def length(i):
return len(str(i))

Questions:

It can be done for integers quickly by using:

len(str(abs(1234567890)))

Which gets the length of the string of the absolute value of “1234567890”

abs returns the number WITHOUT any negatives (only the magnitude of the number), str casts/converts it to a string and len returns the string length of that string.

If you want it to work for floats, you can use either of the following:

# Ignore all after decimal place
len(str(abs(0.1234567890)).split("."))

# Ignore just the decimal place
len(str(abs(0.1234567890)))-1

For future reference.

Questions: