What is the range of values a float can have in Python?

Question

What are its smallest and biggest values in python?

Answer 1

>>> import sys
>>> sys.float_info
sys.floatinfo(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308,
min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, mant_dig=53,
epsilon=2.2204460492503131e-16, radix=2, rounds=1)

The smallest is sys.float_info.min (2.2250738585072014e-308) and the biggest is sys.float_info.max (1.7976931348623157e+308). See documentation for other properties.

sys.float_info.min is the normalized min. You can usually get the denormalized min as sys.float_info.min * sys.float_info.epsilon. Note that such numbers are represented with a loss of precision. As expected, the denormalized min is less than the normalized min.

Answer 2

Technically speaking, the smallest float is -inf and the max float inf:

>>> (float('-inf')            #   negative infinity 
< -1.7976931348623157e+308    #*  smallest float that is not negative infinity 
< -4.9406564584124654e-324    #*  biggest negative float that is not zero
< 0                           #   zero duh
< 4.9406564584124654e-324     #*  smallest positive float that is not zero
< 1.7976931348623157e+308     #*  biggest float that is not positive infinity
< float('inf'))               #   positive infinity
True

numbers with * are machine-dependent and implementation-dependent.

Answer 3

Just playing around; here is an algorithmic method to find the minimum and maximum positive float, hopefully in any python implementation where float("+inf") is acceptable:

def find_float_limits():
    """Return a tuple of min, max positive numbers
    representable by the platform's float"""

    # first, make sure a float's a float
    if 1.0/10*10 == 10.0:
        raise RuntimeError("Your platform's floats aren't")

    minimum= maximum= 1.0
    infinity= float("+inf")

    # first find minimum
    last_minimum= 2*minimum
    while last_minimum > minimum > 0:
        last_minimum= minimum
        minimum*= 0.5

    # now find maximum
    operands= []
    while maximum < infinity:
        operands.append(maximum)
        try:
            maximum*= 2
        except OverflowError:
            break
    last_maximum= maximum= 0
    while operands and maximum < infinity:
        last_maximum= maximum
        maximum+= operands.pop()

    return last_minimum, last_maximum

if __name__ == "__main__":
    print (find_float_limits()) # python 2 and 3 friendly

In my case,

$ python so1835787.py
(4.9406564584124654e-324, 1.7976931348623157e+308)

so denormals are used.

Answer 4

As a kind of theoretical complement to the previous answers, I would like to mention that the "magic" value ±308 comes directly from the binary representation of floats. Double precision floats are of the form ±c*2**q with a "small" fractional value c (~1), and q an integer written with 11 binary digits (including 1 bit for its sign). The fact that 2**(2**10-1) has 308 (decimal) digits explains the appearance of 10**±308 in the extreme float values.

Calculation in Python:

>>> print len(repr(2**(2**10-1)).rstrip('L'))
308

Answer 5

Python uses double-precision floats, which can hold values from about 10 to the -308 to 10 to the 308 power.

http://en.wikipedia.org/wiki/Double_precision_floating-point_format

Try this experiment from the Python prompt:

>>> 1e308
1e+308
>>> 1e309
inf

10 to the 309 power is an overflow, but 10 to the 308 is not. QED.

Actually, you can probably get numbers smaller than 1e-308 via denormals, but there is a significant performance hit to this. I found that Python is able to handle 1e-324 but underflows on 1e-325 and returns 0.0 as the value.

Answer 6

See this post.

Relevant parts of the post:

In [2]: import kinds 
In [3]: kinds.default_float_kind.M 
kinds.default_float_kind.MAX         kinds.default_float_kind.MIN 
kinds.default_float_kind.MAX_10_EXP  kinds.default_float_kind.MIN_10_EXP 
kinds.default_float_kind.MAX_EXP     kinds.default_float_kind.MIN_EXP 
In [3]: kinds.default_float_kind.MIN 
Out[3]: 2.2250738585072014e-308