Question

给定一个列表，我可以得到列表中每个项目的产品：

['apple apple',
 'apple orange',
 'apple pair',
 'apple None',
 'orange apple',
 'orange orange',
 'orange pair',
 'orange None',
 'pair apple',
 'pair orange',
 'pair pair',
 'pair None',
 'None apple',
 'None orange',
 'None pair',
 'None None']

[OUT]：

product(list, list)

如果我想将from itertools import product x = 'apple orange pair None'.split() for i, j in product(x, x): i = '' if i == 'None' else i j = '' if j == 'None' else j y = i + ' ' + j y = y.strip() print(y) for k, l in product(x, [y]): k = '' if k == 'None' else k l = '' if l == 'None' else l z = k + ' ' + l z = z.strip() print(z)的输出与初始列表嵌套，我可以这样做：

apple apple
apple apple apple
orange apple apple
pair apple apple
apple apple
apple orange
apple apple orange
orange apple orange
pair apple orange
apple orange
apple pair
apple apple pair
orange apple pair
pair apple pair
apple pair
apple
apple apple
orange apple
pair apple
apple
orange apple
apple orange apple
orange orange apple
pair orange apple
orange apple
orange orange
apple orange orange
orange orange orange
pair orange orange
orange orange
orange pair
apple orange pair
orange orange pair
pair orange pair
orange pair
orange
apple orange
orange orange
pair orange
orange
pair apple
apple pair apple
orange pair apple
pair pair apple
pair apple
pair orange
apple pair orange
orange pair orange
pair pair orange
pair orange
pair pair
apple pair pair
orange pair pair
pair pair pair
pair pair
pair
apple pair
orange pair
pair pair
pair
apple
apple apple
orange apple
pair apple
apple
orange
apple orange
orange orange
pair orange
orange
pair
apple pair
orange pair
pair pair
pair

apple
orange
pair

[OUT]：

from itertools import product
x = 'apple orange pair None'.split()
for i, j in product(x, x):
    i = '' if i == 'None' else i
    j = '' if j == 'None' else j
    y = i + ' ' + j
    y = y.strip()
    print(y)
    for k, l in product(x, [y]):
        k = '' if k == 'None' else k
        l = '' if l == 'None' else l
        z = k + ' ' + l
        z = z.strip()
        print(z)
        for m, n in product(x, [z]):
            m = '' if m == 'None' else m
            n = '' if n == 'None' else n
            zz = m + ' ' + n
            zz = zz.strip()
            print(zz)

如果我想进入另一级别的嵌套，我可以对其进行硬编码：

{{1}}

但是有没有其他方法可以在没有硬编码的情况下实现相同的输出？

Answer 1

关键在于递归过程自然形成递归模式。

为了说明这个想法，'None'由于简单性而未被''取代。在进一步的解决方案中，它是针对嵌套模式完成的。

def product_combine(a, b):
  return [i + ' ' + j for i, j in product(a, b)]

# for n times of nesting
def products_combine(x, n):
  if n == 0:
    return x
  else:
    return product_combine(x, products_combine(x, n-1)) + products_combine(x, n-1)

x = 'apple orange pair None'.split()    
print(products_combine(x, 3))

如果有以下情况，您需要不同的数据类型来保存结果。更通用的解决方案允许更灵活地选择输出数据类型：

# for different types of combination
def products_combine(combine):
  def products(x, n):
    if n == 0:
      return x
    else:
      return combine(x, products(x, n-1)) + products(x, n-1)
  return products

# combine to string ('None' is replaced for nested patterns, not for initial)
def tostr(a, b):
  NoneToEmpty = lambda x: '' if x == 'None' else x
  return [' '.join(map(NoneToEmpty, (i, j))).strip() for i, j in product(a, b)]

# combine to iterator (list/tuple)
def toiter(iter_type):
  def to_thatiter(a, b):
    return [iter_type((i,))+j if isinstance(j, iter_type) else iter_type((i, j)) for i, j in product(a, b)]
  return to_thatiter

tolist=toiter(list)
totuple=toiter(tuple)

products_str=products_combine(tostr)
products_list=products_combine(tolist)
products_tuple=products_combine(totuple)

x = 'apple orange pair None'.split()
print(products_str(x, 3))
print(products_list(x, 3))
print(products_tuple(x, 3))

一般表单适用于任何binary function f(x, x) ，将其嵌套到 f(x, f(x, f(x,...f(x, f(x, x)) 。不仅是产品而且是任意二进制操作的通用方法是：

def nest(f_binary, n):
  def g(x):
    if n == 1:
      return x
    else:
      return f_binary(x, nest(f_binary, n-1)(x))
  return g

add = lambda x, y: x + y
power = lambda x,y: x**y
concatenate = lambda l1, l2: l1 + l2

x = 'apple orange pair None'.split()
print(list(nest(product, 3)(x)))
print(nest(add, 3)(5))
print(nest(power,3)(5))
print(nest(concatenate, 3)(['a','b']))

另一个想法是使用多个参数而不是显式整数N来表示嵌套级别。它看起来很奇怪，但它确实有效。

def nest(f):
  def expand(x, *args):
    return x if not args else f(x, expand(*args))
  return expand

products = nest(product)

x = 'apple orange pair None'.split()

# instead of giving N, you call products with number n of argument x
# indicating n levels of nesting (here: 3 x, product(x, product(x, x))
print(list(products(x, x, x)))

如何嵌套itertools产品？

1 个答案: