Python 将 pi 打印到小数位数
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Print pi to a number of decimal places
提问by Fraser
One of the challenges on w3resources is to print pi to 'n' decimal places. Here is my code:
w3resources 的挑战之一是将 pi 打印到“n”个小数位。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)
For whatever reason, it automatically prints pi without taking into account of any inputs. Any help would be great :)
无论出于何种原因，它都会在不考虑任何输入的情况下自动打印 pi。任何帮助都会很棒:)
回答by Moses Koledoye
Why not just formatusing number_of_places:
为什么不只是format使用number_of_places：
''.format(pi)
>>> format(pi, '.4f')
'3.1416'
>>> format(pi, '.14f')
'3.14159265358979'
And more generally:
更普遍的是：
>>> number_of_places = 6
>>> '{:.{}f}'.format(pi, number_of_places)
'3.141593'
In your original approach, I guess you're trying to pick a number of digits using number_of_placesas the control variable of the loop, which is quite hacky but does not work in your case because the initial number_of_digitsentered by the user is never used. It is instead being replaced by the iteratee values from the pistring.
在您的原始方法中，我猜您正在尝试选择一些数字number_of_places作为循环的控制变量，这很棘手，但在您的情况下不起作用，因为number_of_digits从未使用用户输入的初始值。它被替换为pi字符串中的 iteratee 值。
回答by Jonas Adler
The proposed solutions using np.pi, math.pi, etc only only work to double precision (~14 digits), to get higher precision you need to use multi-precision, for example the mpmath package
使用np.pi,math.pi等的建议解决方案仅适用于双精度（~14 位），要获得更高的精度，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
Using np.pigives the wrong result
使用np.pi给出了错误的结果
>>> format(np.pi, '.20f')
3.14159265358979311600
Compare to the true value:
与真实值比较：
3.14159265358979323846264338327...
回答by cdlane
Your solution appears to be looping over the wrong thing:
您的解决方案似乎在错误的事情上循环：
for number_of_places in fraser:
For 9 places, this turns out be something like:
对于 9 个地方，结果是这样的：
for "9" in "3.141592653589793":
Which loops three times, one for each "9" found in the string. We can fix your code:
循环 3 次，每一个在字符串中找到的“9”循环一次。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)
But this still limits nto be less than the len(str(math.pi)), less than 15 in Python 2. Given a serious n, it breaks:
但这仍然限制n为小于len(str(math.pi))，在 Python 2 中小于 15。鉴于严重的n，它会中断：
> 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
>
To do better, we have to calculate PI ourselves -- using a series evaluation is one approach:
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 ([email protected])
# 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")])
Now we can take on a large value of n:
现在我们可以取一个很大的值n：
> python3 test2.py
Enter the number of decimal places you want: 100
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067
>
回答by vedang_joshi
As this question already has useful answers, I would just like to share how i created a program for the same purpose, which is very similar to the one in the question.
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)
Thanks for Reading.
谢谢阅读。
回答by Abhishek Bahukhandi
For example the mpmathpackage
例如mpmath包
from mpmath import mp
def a(n):
mp.dps=n+1
return(mp.pi)
回答by sai krishna
Why not just use:
为什么不使用：
import numpy as np
def pidecimal(round):
print(np.round(np.pi, round))
回答by SJ44
This is what I did, really elementary but works (max 15 decimal places):
这就是我所做的，非常基本但有效（最多 15 个小数位）：
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')
