Python math.sqrt loss precision

Question

I tried to get sqrt value of a long int

from math import sqrt
print sqrt(410241186411534352) 

<< 640500731.0

It returned 640500731.0, which indeed is 640500730.999999993... in precision. How to fix that?

I solved it according to @DSM and @Rob's reply, thank you so much.

from math import sqrt
from decimal import Decimal
print Decimal(410241186411534352).sqrt()

If you need an integer sqrt function, see [this question](http://stackoverflow.com/questions/15390807/integer-square-root-in-python) for several options. — DSM, Mar 29 '13 at 02:10

score 5 · Accepted Answer · answered Mar 29 '13 at 02:30

>>> import decimal
>>> decimal.getcontext().prec = 100
>>> a=decimal.Decimal(410241186411534352)**decimal.Decimal(.5)
>>> print a
640500730.9999999929742468943411712910588335443486974564849183256021518032770726219326459594197730639
>>> print a*a
410241186411534352.0000000000000000000000000000000000000000000000000000000000000000000000000000000000
>>>

score 2 · Answer 2 · answered Mar 29 '13 at 02:11

2

Python math library functions are thin wrappers around the C math library functions, so Python is converting the long integer into a double-sized floating point number before doing the computation. If you want genuine multiprecision arithmetic in Python, you could try mpmath.

answered Mar 29 '13 at 02:11

Simon

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Any change I could solve this problem using pure python, without some other packages? – 蒋艾伦 Mar 29 '13 at 02:21
2

@AaronJiang: you could use the `decimal` module. – DSM Mar 29 '13 at 02:24
The `decimal` module is nicely demonstrated by @Robᵩ in another answer. – Simon Mar 29 '13 at 02:33
@AaronJiang: Glad to help. If `decimal` works for you, you might want to tick @Robᵩ's answer as the correct one. – Simon Mar 29 '13 at 02:41

Python math.sqrt loss precision

2 Answers2