如何写float - >字符串并获得与"%。f"相同的数字。

时间:2018-02-10 00:23:05

标签: python floating-point

今天我写了一个天真的函数,通过反复修改10并除以10将float转换为字符串。

def to_string(x):
    r = ""
        while x >= 1.0:
            r = str(int(x%10)) + r
            x /= 10.0
    return r

然后我根据Python的内置功能测试了我的函数,将float转换为字符串。不出所料,我的功能大不相同。

>>> to_string(1e30)
'1000000000000000424684240284426'

>>> "%.f"%1e30
'1000000000000000019884624838656'

所以,我的问题是:我需要做什么计算来获取Python得到的数字?

1 个答案:

答案 0 :(得分:1)

这是一个非常简单的解决方案。它无意提高效率或处理正整数以外的任何值。

#!/usr/bin/python

import math

# This code represents a hexadecimal number using a list in which each item
# is a single hexadecimal digit (an integer from 0 to 15).

# Divide the base-16 digit-list in x by digit d.  This uses grade-school
# division, for a single-digit divisor.  It returns the quotient in a new
# digit-list and the remainder as a plain integer.
def Divide(x, d):

    # If the first digit is smaller than d, start an empty quotient with
    # a remainder being carried.
    if x[0] < d:
        q = []
        r = x[0]
    # Otherwise, start the quotient by dividing the first digit by d.
    else:
        q = [x[0] / d]
        r = x[0] % d

    # For the remaining digits, multiply the remainder by the base and
    # add the new digit.  Then divide by d, calculating a new digit of the
    # quotient and a new remainder to carry.
    for i in x[1:]:
        r = r*16 + i
        q.append(r/d)
        r = r % d

    return (q, r)

# Convert a positive integer floating-point value to decimal and print it.
def to_string(x):

    # Convert input to base 16.  This has no rounding errors since the
    # floating-point format uses base two, and 16 is a power of two.
    t= []
    while 0 < x:
        # Use remainder modulo 16 to calculate the next digit, then insert it.
        t.insert(0, int(x % 16))
        # Remove the digit from x.
        x = math.floor(x/16)

    # Start an empty output string.
    output = ""

    # Divide the number by 10 until it is zero (is an empty digit list).
    # Each time we divide, the remainder is a new digit of the answer.
    while t != []:
        (t, r) = Divide(t, 10)
        output = str(r) + output

    print output


to_string(1e30)