如何使用过滤器、映射、lambda 求解 x^2 - y^2 = 2000^2？

Question

import numpy as np

#defining range
start = -5000000
end = abs(start)

x = np.linspace(start, end, abs(start) + end + 1)
y = np.linspace(start, end, abs(start) + end + 1)

print(x, y)

xy_dif_list1 = list(filter(lambda r: r == 2000**2, list(map(lambda a, b: a**2 - b**2, x, y))))

print(len(xy_dif_list1))

我想找出满足等式的 x 和 y 值的个数： x^2 - y^2 = 2000^2, 但是这段代码不起作用，因为，当我使用嵌套的 for 循环时，我可以看到一些结果：

xy_list = []
for x in range(-5000000, 5000000):
    for y in range(-5000000, 5000000):
        if x**2 - y**2 == (2000**2):
            xy_list.append((x, y))
            print(xy_list)

print(len(xy_list))

filter()、map() 和 lambda 函数有什么问题？

Answer 1

你的问题是 map doesn't create nested loops out of its iterable arguments. Instead, it steps through both of them sequentially in parallel, as though using the zip 函数。

因此，您将函数应用于 (x[0], y[0])，然后 (x[1], y[1])，然后 (x[2], y[2])等

您应该查看 itertools.product 以获得嵌套循环的等效功能。

Answer 2

我认为实施您正在做的事情的正确方法是

导入 itertools

# find [x, y] with x^2 - y^2 = target
def solve(target):
    # assumes target >= 1, not hard to adjust to solve for all
    xmax = (target+1)//2
    x = range(xmax + 1) # 0 to xmax
    y = range(xmax) # 0 to xmax - 1
    xy = itertools.product(x, y)
    soln = filter(lambda v: v[0]*v[0] - v[1]*v[1] == target, xy)
    return list(soln)

print(solve(1000))
#print(solve(2000*2000))

但是对于您的输入数字来说它太慢了，因为您需要构建和过滤一个 4 million-item 列表。更好的是

import math

def issquare(n):
    if (n < 0):
        return False
    m = math.isqrt(n)
    return m*m == n

# x^2 - y^2 = target
def solve2(target):
    # assumes target >= 1, not hard to adjust to solve for all
    xmax = (target+1)//2
    x = range(xmax + 1) # 0 to xmax
    soln = filter(lambda x: issquare(x*x - target), x)
    return list(map(lambda x: (x, math.isqrt(x*x - target)), soln))

print(solve2(1000))
print(solve2(2000*2000))

这让您只能迭代一个数字。（如果需要，可以进一步改进。）

如何使用过滤器、映射、lambda 求解 x^2 - y^2 = 2000^2？

How to solve x^2 - y^2 = 2000^2 using filter, map, lambda?

python

functional-programming