for循环，如果语句

-3

我真的很新的python。我试图让这个工作。for循环，如果语句

import math 
number, times = eval(input('Hello please enter the value and number of times to improve the guess followed by comma:')) 
guess=number/2 
sq_math= math.sqrt(number) 
if times>1: 
    for i in range(2,times+1): 
     guess=(guess+times/guess)/2 
     if round(guess,1) == round(sq_math,1): 
     break 

else: 
    pass 

print('Newtons method guessed {0}, square root was {1}'.format(guess, sq_math))

那么他最好的办法是什么？感谢你们！

来源

2017-07-24 Artem Dergach

嗨，欢迎来到堆栈溢出。请回顾[问]并帮助我们解释您想要发生的事情，您遇到的错误以及您不了解的内容。 –

它是做什么的？它应该做什么？任何错误？你期望输出什么？你会得到什么输出？ – jacoblaw

请不要这样做：'number，times = eval（input（...））' –

你想要做的布尔值不等于比较round(guess,1) != round(sq_math,1)在一个单独的if条款，就像你已为相等比较==完成：

if times>1: 
    # break this next line up in to two lines, `for` and `if` 
    # for i in range(2,times+1) and round(guess,1) != round(sq_math,1): 
    for i in range(2,times+1):     # for check 
     if round(guess,1) != round(sq_math,1): # if check 
      guess=(guess+times/guess)/2 
     if round(guess,1) == round(sq_math,1): 
      break 
     times-=1 #decrement times until we reach 0

演示：

Hello please enter the value and number of times to improve the guess followed by comma:9,56 
Newtons method guessed 3.0043528214, square root was 3.0

来源

2017-07-25 01:18:31 davedwards

“您好，请输入值和次数以改善猜测，然后加上逗号：9,56 牛顿方法猜测7.483314773547883，平方根3.0“由于某种原因，它不想给我正确的答案。 –

对不起，我不知道正确的答案是什么。你想得到什么正确的答案？ https://stackoverflow.com/questions/45291577/for-loop-if-statement/45292363?noredirect=1#comment77545467_45291577 – davedwards

你可以在公式中看到。当round（猜测）将是3时，它应该打破并打印猜测。 –

我相信主要问题是这个公式不正确：

guess = (guess + times/guess)/2

它应该是：

guess = (guess + number/guess)/2

我看不出有任何问题与您if声明也不是你for循环。完整的解决方案：

import math 

number = int(input('Please enter the value: ')) 
times = int(input('Please enter the number of times to improve the guess: ')) 

answer = math.sqrt(number) 

guess = number/2 

if times > 1: 
    for _ in range(times - 1): 
     guess = (guess + number/guess)/2 
     if round(guess, 1) == round(answer, 1): 
      break 

print("Newton's method guessed {0}; square root was {1}".format(guess, answer))

用法

% python3 test.py 
Please enter the value: 169 
Please enter the number of times to improve the guess: 6 
Newton's method guessed 13.001272448567825; square root was 13.0 
%

虽然我相信我真的实施寻找平方根巴比伦的方法。

来源

2017-07-25 23:27:23 cdlane

谢谢是的公式不正确！ –

for循环，如果语句

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