Python中的小数位数

Question

I am trying to put all the decimal places of 1/n into a list in Python.

def dec(n):

    result = float(1) / n
    while (result >= 1):
        result = result - 1

    while (result != 0):
        result = result * 10
        decimals = int(result)
        yield(decimals)
        result = result - decimals

However, when I tried list(dec(3)) , it doesn't give me a list full of 3s but instead something like [3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 0, 3, 7, 2]

How should I fix this?

Answer 1

Since Python floats are represented in binary, float representations of fractions are not just inexact - decimal approximation are often just as inexact - but inexact in a counter-intuitive way. We are used to approximating a fraction like 1/3 as something like 0.3333333333, or 33333333333/10**10 (assuming 10 significant digits). However, the binary floating-point used by Python represents numbers as fractions with a power-of-two denominator. In this representation, 1/3 is approximated as 6004799503160661/2**54, and the digits in your output come from that fraction.

To calculate the digits of a decimal approximation of the fraction 1/3, import the decimal module and replace float(1) with decimal.Decimal(1) . The Decimal type was designed with the express goal of supporting what you're attempting to do here, ie getting results that work the same way as pencil-and-paper calculations with algorithms taught in school.

The Decimal instance will of course contain a limited number of digits, which are for 1/3 repeated infinitely. To be able to access digits without limit, import fractions and use fractions.Fraction() . In that case, the sequence produced by the generator will be infinite and you won't be able to convert it to a list, but you will still be able to iterate over it and analyze as much of it as you need.

Answer 2

I'm assuming you're using Python 3, since you've got only a single "/". You'll note that float(1)/3*100000000 = 33333333.333333332. This is likely due to the vagaries of binary representation of decimal fractions, because you run out of bits for precision in the width of the float.

So, you'd be better off handling it as a string, and converting that to a list:

list(str(float(1)/3))[2:]

['3', '3', '3', '3', '3', '3', '3', '3', '3', '3', '3', '3', '3', '3', '3', '3']

Answer 3

You are trying to present the rational number as a repeating decimal . And you need to know that:

Every rational number is either a terminating or repeating decimal.

So you do not need to store all the digits there (non repeating part and repeating part). Another important point is that if you have m/n then your repeating part is at maximum n-1 length.

Because you are basically asking for solution for project Euler 26 task, I will not post it, but rather will give you a link with some explanation .

Answer 4

Floating point arithmetic is necessarily inaccurate. One way to compute the digits exactly is to use the division algorithm you probably learnt at school.

Here's an implementation:

def dec(n):
    v = 1
    while v:
        v *= 10
        yield v // n
        v %= n

You can test it like this, which shows the first 40 digits for 1/3, 1/5, 1/7 and 1/11.

import itertools

for i in [3, 5, 7, 11]:
    print '1/%d = 0.%s' % (i, ''.join(map(str, itertools.islice(dec(i), 40))))

The output is:

1/3 = 0.3333333333333333333333333333333333333333
1/5 = 0.2
1/7 = 0.1428571428571428571428571428571428571428
1/11 = 0.0909090909090909090909090909090909090909