在 python 中生成圆周率

Question

我想知道如何修改此代码以在被问及时提供 X 位数的圆周率因为现在它只是打印一个随机数量

while True:
    print("how many digits of pi would you like")


    def make_pi():
        q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
        number = int(input())
        for j in range(int(number)):
            if 4 * q + r - t < m * t:
                yield m
                q, r, t, k, m, x = 10 * q, 10 * (r - m * t), t, k, (10 * (3 * q + r)) // t - 10 * m, x
            else:
                q, r, t, k, m, x = q * k, (2 * q + r) * x, t * x, k + 1, (q * (7 * k + 2) + r * x) // (t * x), x + 2


    digits = make_pi()
    pi_list = []
    my_array = []

    for i in make_pi():
        my_array.append(str(i))

    my_array = my_array[:1] + ['.'] + my_array[1:]
    big_string = "".join(my_array)
    print("here is the string:\n %s" % big_string)

Answer 1

print('{:.xf}'.format(math.pi))

其中x是你要在小数点后打印的圆周率的位数。

编辑：

如果你想按照你的方式去做，那么

big_string = ''.join(my_array[:x])

其中 x 是 个字符的数量，包括前导 3 和小数点。

Answer 2

完整代码如果感兴趣

虽然正确：打印（"how many digits of pi would you like"）

def make_pi():
    q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
    for j in range(1000):
        if 4 * q + r - t < m * t:
            yield m
            q, r, t, k, m, x = 10 * q, 10 * (r - m * t), t, k, (10 * (3 * q + r)) // t - 10 * m, x
        else:
            q, r, t, k, m, x = q * k, (2 * q + r) * x, t * x, k + 1, (q * (7 * k + 2) + r * x) // (t * x), x + 2


digits = make_pi()
pi_list = []
my_array = []

for i in make_pi():
    my_array.append(str(i))

number = int(input())+2
my_array = my_array[:1] + ['.'] + my_array[1:]
big_string = "".join(my_array[: number ])
print("here is the string:\n %s" % big_string)

在 python 中生成圆周率

generating pi in python

python

pi

python-3.x