w3resources的挑战之一是将pi打印到小数点后n位。这是我的代码：
from math import pi
fraser = str(pi)
length_of_pi = []
number_of_places = raw_input("Enter the number of decimal places you want to
see: ")
for number_of_places in fraser:
length_of_pi.append(str(number_of_places))
print "".join(length_of_pi)
无论出于何种原因，它都会自动打印 pi，而不考虑任何输入。任何帮助都会很棒:)
使用np.pi、math.pi等的提议解决方案仅适用于双精度(~14 位(，为了获得更高的精度，您需要使用多精度，例如 mpmath 包
>>> from mpmath import mp
>>> mp.dps = 20 # set number of digits
>>> print(mp.pi)
3.1415926535897932385
使用np.pi会给出错误的结果
>>> format(np.pi, '.20f')
3.14159265358979311600
与真实值进行比较：
3.14159265358979323846264338327...
为什么不直接使用number_of_placesformat：
''.format(pi)
>>> format(pi, '.4f')
'3.1416'
>>> format(pi, '.14f')
'3.14159265358979'
更一般地说：
>>> number_of_places = 6
>>> '{:.{}f}'.format(pi, number_of_places)
'3.141593'
在您最初的方法中，我猜您正在尝试使用number_of_places作为循环的控制变量来选择一些数字，这非常笨拙，但在您的情况下不起作用，因为用户输入的初始number_of_digits从未使用过。相反，它被替换为pi字符串中的迭代值。
例如mpmath包
from mpmath import mp
def a(n):
mp.dps=n+1
return(mp.pi)
很好的答案！ 有很多方法可以实现这一点。看看我在下面使用的这种方法，它可以工作任意数量的小数位直到无穷大：
#import multp-precision module
from mpmath import mp
#define PI function
def pi_func():
while True:
#request input from user
try:
entry = input("Please enter an number of decimal places to which the value of PI should be calculatednEnter 'quit' to cancel: ")
#condition for quit
if entry == 'quit':
break
#modify input for computation
mp.dps = int(entry) +1
#condition for input error
except:
print("Looks like you did not enter an integer!")
continue
#execute and print result
else:
print(mp.pi)
continue
祝你好运，朋友！
您的解决方案似乎循环了错误的内容：
for number_of_places in fraser:
对于 9 个地方，结果是这样的：
for "9" in "3.141592653589793":
循环三次，字符串中找到的每个"9"对应一个。 我们可以修复您的代码：
from math import pi
fraser = str(pi)
length_of_pi = []
number_of_places = int(raw_input("Enter the number of decimal places you want: "))
for places in range(number_of_places + 1): # +1 for decimal point
length_of_pi.append(str(fraser[places]))
print "".join(length_of_pi)
但这仍然限制n小于len(str(math.pi))，在 Python 2 中小于 15。 给定一个严重的n，它中断：
> python test.py
Enter the number of decimal places you want to see: 100
Traceback (most recent call last):
File "test.py", line 10, in <module>
length_of_pi.append(str(fraser[places]))
IndexError: string index out of range
>
为了做得更好，我们必须自己计算PI - 使用系列评估是一种方法：
# Rewrite of Henrik Johansson's (Henrik.Johansson@Nexus.Comm.SE)
# pi.c example from his bignum package for Python 3
#
# Terms based on Gauss' refinement of Machin's formula:
#
# arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + ...
from decimal import Decimal, getcontext
TERMS = [(12, 18), (8, 57), (-5, 239)] # ala Gauss
def arctan(talj, kvot):
"""Compute arctangent using a series approximation"""
summation = 0
talj *= product
qfactor = 1
while talj:
talj //= kvot
summation += (talj // qfactor)
qfactor += 2
return summation
number_of_places = int(input("Enter the number of decimal places you want: "))
getcontext().prec = number_of_places
product = 10 ** number_of_places
result = 0
for multiplier, denominator in TERMS:
denominator = Decimal(denominator)
result += arctan(- denominator * multiplier, - (denominator ** 2))
result *= 4 # pi == atan(1) * 4
string = str(result)
# 3.14159265358979E+15 => 3.14159265358979
print(string[0:string.index("E")])
现在我们可以取一个很大的n值：
> python3 test2.py
Enter the number of decimal places you want: 100
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067
>
这就是我所做的，真的很基本但有效(最多 15 位小数(：
pi = 22/7
while True:
n = int(input('Please enter how many decimals you want to print: '))
if n<=15:
print('The output with {} decimal places is: '.format(n))
x = str(pi)
print(x[0:n+2])
break
else:
print('Please enter a number between 0 and 15')
由于这个问题已经有了有用的答案，我只想分享我如何为相同目的创建一个程序，这与问题中的程序非常相似。
from math import pi
i = int(input("Enter the number of decimal places: "))
h = 0
b = list()
for x in str(pi):
h += 1
b.append(x)
if h == i+2:
break
h = ''.join(b)
print(h)
感谢您的阅读。
为什么不直接使用：
import numpy as np
def pidecimal(round):
print(np.round(np.pi, round))