使用牛顿方法查找平方根(错误!)

Question

我正在努力完成一个数学问题,它使用Newton的猜测和检查方法来近似数字的平方根.用户应该输入一个数字,该数字的初始猜测,以及他们想要在返回之前检查他们的答案的次数.为了让事情变得更容易并且了解Python(我几个月前才开始学习这门语言),我把它分解成了许多小函数; 但现在的问题是,我无法调用每个函数并传递数字.

这是我的代码,有帮助的注释(每个函数按使用顺序):

# This program approximates the square root of a number (entered by the user)
# using Newton's method (guess-and-check). I started with one long function,
# but after research, have attempted to apply smaller functions on top of each
# other.
# * NEED TO: call functions properly; implement a counting loop so the
# goodGuess function can only be accessed the certain # of times the user
# specifies. Even if the - .001 range isn't reached, it should return.

# sqrtNewt is basically the main, which initiates user input.

def sqrtNewt():
    # c equals a running count initiated at the beginning of the program, to
    # use variable count.
    print("This will approximate the square root of a number, using a guess-and-check process.")
    x = eval(input("Please type in a positive number to find the square root of: "))
    guess = eval(input("Please type in a guess for the square root of the number you entered: "))
    count = eval(input("Please enter how many times would you like this program to improve your initial guess: ")) 
    avg = average(guess, x)
    g, avg = improveG(guess, x)
    final = goodGuess(avg, x)
    guess = square_root(guess, x, count)
    compare(guess, x)


# Average function is called; is the first step that gives an initial average,
# which implements through smaller layers of simple functions stacked on each
# other.
def average(guess, x) :
    return ((guess + x) / 2)

# An improvement function which builds upon the original average function.
def improveG(guess, x) :
    return average(guess, x/guess)

# A function which determines if the difference between guess X guess minus the
# original number results in an absolute vale less than 0.001. Not taking
# absolute values (like if guess times guess was greater than x) might result
# in errors
from math import *
def goodGuess(avg, x) :
    num = abs(avg * avg - x)
    return (num < 0.001)

# A function that, if not satisfied, continues to "tap" other functions for
# better guess outputs. i.e. as long as the guess is not good enough, keep
# improving the guess.
def square_root(guess, x, count) :
    while(not goodGuess(avg, x)):
        c = 0
        c = c + 1
        if (c < count):
            guess = improveG(guess, x)
        elif (c == count):
            return guess
        else :
            pass

# Function is used to check the difference between guess and the sqrt method
# applied to the user input.
import math
def compare(guess, x):
    diff = math.sqrt(x) - guess
    print("The following is the difference between the approximation") 
    print("and the Math.sqrt method, not rounded:", diff)

sqrtNewt()

目前,我得到这个错误:g, avg = improveG(guess, x) TypeError: 'float' object is not iterable. 最终函数使用猜测的最后一次迭代从数学平方根方法中减去,并返回整体差异.我甚至这样做了吗？如果您能提供建议,我们将非常感谢工作代码.再一次,我是一个新手,所以我为误解或盲目的明显错误道歉.

Answer 1

牛顿法的实现:

在需要时添加一些调整应该相当容易.试试,告诉我们什么时候你卡住了.

from math import *
def average(a, b):
    return (a + b) / 2.0
def improve(guess, x):
    return average(guess, x/guess)
def good_enough(guess, x):
    d = abs(guess*guess - x)
    return (d < 0.001)
def square_root(guess, x):
    while(not good_enough(guess, x)):
        guess = improve(guess, x)
    return guess
def my_sqrt(x):
    r = square_root(1, x)
    return r

>>> my_sqrt(16)
4.0000006366929393

注意:你会发现在SO或google搜索时如何使用原始输入的足够的例子,但是,如果你在计算循环,c=0必须在循环之外,或者你将陷入无限循环.

Quiqk和脏,很多方法来改善:

from math import *
def average(a, b):
    return (a + b) / 2.0
def improve(guess, x):
    return average(guess, x/guess)
def square_root(guess, x, c):
    guesscount=0
    while guesscount < c :
        guesscount+=1
        guess = improve(guess, x)
    return guess
def my_sqrt(x,c):
    r = square_root(1, x, c)
    return r

number=int(raw_input('Enter a positive number'))
i_guess=int(raw_input('Enter an initial guess'))
times=int(raw_input('How many times would you like this program to improve your initial guess:'))    
answer=my_sqrt(number,times)

print 'sqrt is approximately ' + str(answer)
print 'difference between your guess and sqrt is ' + str(abs(i_guess-answer))

Answer 2

选择的答案有点令人费解......没有对OP的不尊重.

对于将来使用谷歌搜索的人来说,这是我的解决方案:

def check(x, guess):
    return (abs(guess*guess - x) < 0.001)

def newton(x, guess):
    while not check(x, guess):
        guess = (guess + (x/guess)) / 2.0
    return guess

print newton(16, 1)