Question

我正在使用多项式并且必须执行一些操作，其中我的循环变量的总和低于某个常量d。现在我有

for i in range(0, d):
    for j in range(i, d):
        for k in range(j, d):

这对我来说似乎有点难看。是否有一些函数，可能在itertools中，允许我迭代for i, j, k in foo(d)？

Answer 1

你可以自己写。这是3个变量的强力方式：

def constant_sum(s):
    for i in range(s+1):
        for j in range(s-i+1):
            k = s - i - j
            yield i,j,k

def inferior_sum(s):
    for i in range(s+1):
        for j in range(s+1):
            if i+j >= s:
                break
            for k in range(s+1):
                if i+j+k < s:
                    yield i,j,k
                else:
                    break

for i,j,k in constant_sum(3):
    print(i,j,k)
print()
for i,j,k in inferior_sum(3):
    print(i,j,k)

输出：

以下是递归版本，可以为任何总和做任意数量的变量（n）...轻度测试：

def constant_sum(n,s):
    if n == 1:
        yield [s]
    else:
        for i in range(s+1):
            for r in constant_sum(n-1,s-i):
                yield [i] + r

def inferior_sum(n,s):
    if n == 1:
        for i in range(s):
            yield [i]
    else:
        for i in range(s):
            for r in inferior_sum(n-1,s-i):
                yield [i] + r

for x in constant_sum(3,3):
    print(*x)
print()
for x in inferior_sum(3,3):
    print(*x)

输出：

Answer 2

itertools / functional方式类似于：

from itertools import product
inferior_sum3 = filter(lambda x: sum(x)<3, product(range(4),range(4),range(4)))
for permu in inferior_sum3:
    print(permu)

输出：

(0, 0, 0)
(0, 0, 1)
(0, 0, 2)
(0, 1, 0)
(0, 1, 1)
(0, 2, 0)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(2, 0, 0)

在Python中迭代使用常量求和的索引

2 个答案: