Python：大型浮点数的小数部分

Question

I'm trying to get the decimal part of (pow(10, i) - 1)/23 for 0 < i < 50 . I have tried

(pow(10, i) - 1)/23 % 1

in Python 3 but I get 0.0 for all values of i greater than 17.

How can I extract the decimal part of a large integer in Python?

Answer 1

To preserve precision, I'd probably use the fractions module:

>>> from fractions import Fraction
>>> Fraction(10)
Fraction(10, 1)
>>> Fraction(10)**50
Fraction(100000000000000000000000000000000000000000000000000, 1)
>>> Fraction(10)**50-1
Fraction(99999999999999999999999999999999999999999999999999, 1)
>>> (Fraction(10)**50-1)/23
Fraction(99999999999999999999999999999999999999999999999999, 23)
>>> ((Fraction(10)**50-1)/23) % 1
Fraction(5, 23)
>>> float(((Fraction(10)**50-1)/23) % 1)
0.21739130434782608

although using the decimal module would be another option.

update: wait, on second thought, the answer here is always going to be ((10^n-1) % 23)/23 , so the above is significant overkill (although it does scale up better to more complex problems). We can even take advantage of the three-argument pow call:

>>> pow(10, 50, 23)
6
>>> pow(10, 50, 23) - 1
5
>>> (pow(10, 50, 23) - 1) % 23 # handle possible wraparound
5
>>> ((pow(10, 50, 23) - 1) % 23) / 23.0
0.21739130434782608

Answer 2

One option that does not use other modules is

N = 10**i - 1
remainder = N % 23
print remainder / 23.0

Of course easily written on one line as well.

Using the three argument form of pow , which performs modular exponentation , you can avoid large numbers alltogether, for improved performance.

((pow(10, i, 23) + 22) % 23) / 23.0

Answer 3

You mean: "how can I extract the decimal part of a large float in Python? There is no "decimal" part anymore, because the comma is floating ... :-)

On Python 2.7, for i=20 (I divided by 23.0 in order to get a float result), the result is 4.347826086956522e+18 . There are 16 digits for a number of magnitude 10^18. Hence, there is no digit "after the comma" (this may be still incorrect für i=20 because it could be that not all digits have been printed, but at least you get the concept of what I mean).

You have to apply a library that is implementing arbitrary-precision arithmetics for large i .