How to Get the Largest Integer One Can Use in Python

Maximum and Minimum values for ints

Python 3

In Python 3, this question doesn't apply. The plain int type is unbound.

However, you might actually be looking for information about the current interpreter's word size, which will be the same as the machine's word size in most cases. That information is still available in Python 3 as sys.maxsize, which is the maximum value representable by a signed word. Equivalently, it's the size of the largest possible list or in-memory sequence.

Generally, the maximum value representable by an unsigned word will be sys.maxsize * 2 + 1, and the number of bits in a word will be math.log2(sys.maxsize * 2 + 2). See this answer for more information.

Python 2

In Python 2, the maximum value for plain int values is available as sys.maxint:

>>> sys.maxint
9223372036854775807

You can calculate the minimum value with -sys.maxint - 1 as shown here.

Python seamlessly switches from plain to long integers once you exceed this value. So most of the time, you won't need to know it.

Maximum value for long integer

Long integers:

There is no explicitly defined limit. The amount of available address space forms a practical limit.

(Taken from this site). See the docs on Numeric Types where you'll see that Long integers have unlimited precision. In Python 2, Integers will automatically switch to longs when they grow beyond their limit:

>>> import sys
>>> type(sys.maxsize)
<type 'int'>
>>> type(sys.maxsize+1)
<type 'long'>

for integers we have

maxint and maxsize:

The maximum value of an int can be found in Python 2.x with sys.maxint. It was removed in Python 3, but sys.maxsize can often be used instead. From the changelog:

The sys.maxint constant was removed, since there is no longer a limit
to the value of integers. However, sys.maxsize can be used as an
integer larger than any practical list or string index. It conforms to
the implementation’s “natural” integer size and is typically the same
as sys.maxint in previous releases on the same platform (assuming the
same build options).

and, for anyone interested in the difference (Python 2.x):

sys.maxint The largest positive integer supported by Python’s regular
integer type. This is at least 2**31-1. The largest negative integer
is -maxint-1 — the asymmetry results from the use of 2’s complement
binary arithmetic.

sys.maxsize The largest positive integer supported by the platform’s
Py_ssize_t type, and thus the maximum size lists, strings, dicts, and
many other containers can have.

and for completeness, here's the Python 3 version:

sys.maxsize
An integer giving the maximum value a variable of type Py_ssize_t can take. It’s usually 2^31 - 1 on a 32-bit platform and
2^63 - 1 on a 64-bit platform.

floats:

There's float("inf") and float("-inf"). These can be compared to other numeric types:

>>> import sys
>>> float("inf") > sys.maxsize
True

What is sys.maxint in Python 3?

The sys.maxint constant was removed, since there is no longer a limit
to the value of integers. However, sys.maxsize can be used as an
integer larger than any practical list or string index. It conforms to
the implementation’s “natural” integer size and is typically the same
as sys.maxint in previous releases on the same platform (assuming the
same build options).

http://docs.python.org/3.1/whatsnew/3.0.html#integers

Python: Choosing a large number as initial max

In python, you can do something like that

from math import inf
'initialize maximum with very small value'
maximum=float('-inf')
'traverse the array'
for v in array:
    maximum=max(maximum,v)
print(maximum)

Or You can use max() function directly on the given array

print(max(array))

How do I find the largest integer less than x?

math.ceil(x)-1 is correct and here is the proof.

if x is in Z (the set of integers), then math.ceil(x) = x. Therefore math.ceil(x)-1=x-1, the largest integer smaller than x.

Else we have x in R \ Z and math.ceil(x) is the smallest integer y such that x ≤ y. But then y-1 is an integer smaller than the smallest integer such that x ≤ y, therefore x > y-1 and by construction y-1 is the largest such integer smaller than x.

It's simple enough that I wouldn't bother with those if-else. But to avoid computation errors with floats I would do the -1 outside the int conversion.

int(math.ceil(x))-1

Why can't I use max with Python to find maximum one-digit integer in array?

Here is a solution that ticks more points in my book.

def solution(integers):
    """Return max single digit integer in sequence `integers`"""
    try:
        return max(i for i in integers if -9 <= i <= 9)
    except ValueError:
        return ValueError("Could not find single digit integer in list")

print(solution([-6, -91, 1011, -100, 84, -22, 0, 1, 473]))

It uses a descriptive parameter name, has a comment, does a simple comparison for digit size and uses a generator with the built in function max instead of the needless list. It also reraises the most expected fault with a more topical error message. This function is easier to maintain and use.

Max integer value of python3

In Python3, value of an integer is not restricted by the number of bits and can expand to the limit of the available memory

print(100**100)

output:
100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

How do i find the biggest number that are not in a list?

Let me share you a solution for your problem. (Take a look at the if clause after the check of the number being "-999"). Feel free to ask if you have any questions! Hope it helps you

num=0 
total=0 
average=0 
count=0
biggest=0 

while True: 
  num=input("enter a number:")
  num=int(num) 
  if num==-999: 
    break
  if num > biggest:
    biggest = num
  total=total + num
  count=count+1 

average=total/count 

#print the results
print("the total is:", total)
print("the biggest number is:", biggest)
print("the average is:", average)

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