4

我想为给定列表的可能产品找到一种方法。下面，我有一个例子

到目前为止，我试过这个：

for p in itertools.product(the_list, repeat=2):
        print p

但是，这会返回可以从给定列表生成的所有可能性。您能帮我生成以下示例吗？

一个例子：

the_list=['A1','A2','A3','B1','B2','B3','C1','C2','C3','D1','D2','D3']
The results should be: 
['A1A2','A1B2','A1C2','A1D2','A2A3','A2B3','A2C3','A2D3','A3A1','A3B1','A3C1','A3D1'
 'B1A2,'B2A3'...
 'C1A2'...']

基本上，大写字母代表州，数字代表顺序。因此，如果它从 A1 开始，它只能以数字 2 的状态继续，例如先是 A1，然后是 A2 或 B2 或 C2 或 D2。它也是循环的，这意味着 A3 后面跟着 A1、B1、C1 或 D1。

4

4 回答 4

2 
l=['A1','A2','A3','B1','B2','B3','C1','C2','C3','D1','D2','D3']
result = [i + j for i in l for j in l if int(j[1]) == int(i[1]) + 1 or (j[1] == '1' and i[1] == '3')]
print(result)

这输出：

['A1A2', 'A1B2', 'A1C2', 'A1D2', 'A2A3', 'A2B3', 'A2C3', 'A2D3', 'A3A1', 'A3B1', 'A3C1', 'A3D1', 'B1A2', 'B1B2', 'B1C2', 'B1D2', 'B2A3', 'B2B3', 'B2C3', 'B2D3', 'B3A1', 'B3B1', 'B3C1', 'B3D1', 'C1A2', 'C1B2', 'C1C2', 'C1D2', 'C2A3', 'C2B3', 'C2C3', 'C2D3', 'C3A1', 'C3B1', 'C3C1', 'C3D1', 'D1A2', 'D1B2', 'D1C2', 'D1D2', 'D2A3', 'D2B3', 'D2C3', 'D2D3', 'D3A1', 'D3B1', 'D3C1', 'D3D1']

如果您更喜欢可读性而不是紧凑性，那么请打破单行

l=['A1','A2','A3','B1','B2','B3','C1','C2','C3','D1','D2','D3']
result = []
for i in l:
    for j in l:
        if int(j[1]) == int(i[1]) + 1 or (j[1] == '1' and i[1] == '3'):
            result.append(i+j)
print(result)

它输出相同的东西。

于 2013-01-05T05:12:37.357 回答
1

根据您当前的列表，如果您只想获得所有排列并在事后过滤掉不受欢迎的值，则没有太多需要消除的值。懒惰地执行此操作，您过滤的内容将远不如您保留的内容常见。（我将使用列表推导来实现透明度——在您的实际代码中将其转换为生成器表达式。）

>>> from itertools import permutations
>>> [''.join(x) for x in permutations(l, 2) if int(x[0][1]) % 3 == int(x[1][1]) - 1]
['A1A2', 'A1B2', 'A1C2', 'A1D2', 'A2A3', 'A2B3', 'A2C3', 'A2D3', 'A3A1', 'A3B1', 'A3C1', 'A3D1', 'B1A2', 'B1B2', 'B1C2', 'B1D2', 'B2A3', 'B2B3', 'B2C3', 'B2D3', 'B3A1', 'B3B1', 'B3C1', 'B3D1', 'C1A2', 'C1B2', 'C1C2', 'C1D2', 'C2A3', 'C2B3', 'C2C3', 'C2D3', 'C3A1', 'C3B1', 'C3C1', 'C3D1', 'D1A2', 'D1B2', 'D1C2', 'D1D2', 'D2A3', 'D2B3', 'D2C3', 'D2D3', 'D3A1', 'D3B1', 'D3C1', 'D3D1']

更新以回答约翰的评论：假设每个状态具有相同的序列大小：

>>> seq_size = len([x for x in l if x[0] == 'A']) # Get the length of the sequence of A
>>> [''.join(x) for x in permutations(l, 2) if int(x[0][1:]) % seq_size == int(x[1][1:]) - 1]

更新以概括 John 的下一系列评论：

#!env python

'''Get all the permutations of pieces up to a certain size
    where the sequence numbers of each piece occur in order, cyclically.
    So for a permutation_size of 5, 1->2, 2->3, 3->4, 4->5, 5->1.'''

from string import uppercase
from itertools import permutations

num_states = 5
seq_size = 13
group_size = 3

the_list = [char + str(1 + num) for char in uppercase[:num_states] for num in xrange(seq_size)]

def groupFilter(pieces):
    '''True for a list whose sequence is in order. False otherwise.'''
    ints = [int(piece.lstrip(uppercase)) for piece in pieces]
    offset = ints[0] % seq_size
    for i, piece in enumerate(ints):
        if offset + i != piece % seq_size:
            return False
    return True

[''.join(group) for group in permutations(the_list, group_size) if groupFilter(group)]
于 2013-01-05T05:08:46.467 回答
0

你可以按顺序分解它，通过

first = ['A1', 'B1', 'C1', 'D1']
second = ['A2', 'B2', 'C2', 'D2']
...

然后，取两个连续列表的乘积应该会给你你想要的（如果我正确理解你的问题......）。

于 2013-01-05T05:08:03.260 回答
0

以防万一你想要一些可以使用的东西itertools.product，你可以试试这个（如果它与你的“真实”输入不匹配，很高兴调整 - 这是一个有趣的问题:)）。这可以被包装成更少的行，但它失去了很多可读性，所以希望这会有所帮助：

from itertools import groupby, product
from operator import itemgetter


the_list = ['A1','A2','A3','B1','B2','B3','C1','C2','C3','D1','D2','D3']

# In order to work properly with groupby, we sort the list by the
# number at the end of the string
s = sorted(the_list, key=itemgetter(-1))

# Now we create a list of lists, each sub-list containing values
# with the same ending number (i.e. ['A1', 'B1', 'C1', 'D1'])
j = [list(g) for _, g in groupby(s, key=itemgetter(-1))]

# Now we create our final list
results = []

# Here we iterate through our grouped lists, using product
# similar to how you did before to create the combined strings
for index, r in enumerate(j):

    # This is the piece that lets us 'loop' the list -
    # on the first iteration, the value is -(3)+1+0 = -2,
    # which we use as our list index. This will return the item
    # 'ahead' of the current one in our main list, and when it
    # reaches the last (index=2) item, the value is -(3)+1+2 = 0 (beginning)
    inc = -len(j) + 1 + index

    # Now we just iterate through the items in our sub-list, pairing with
    # the items in the 'next' sub-list
    for val in r:
        results += [k+v for k, v in product([val], j[inc])]

print results

输出：

['A1A2', 'A1B2', 'A1C2', 'A1D2', 
 'B1A2', 'B1B2', 'B1C2', 'B1D2', 
 'C1A2', 'C1B2', 'C1C2', 'C1D2', 
 'D1A2', 'D1B2', 'D1C2', 'D1D2', '
  A2A3', 'A2B3', 'A2C3', 'A2D3', 
 'B2A3', 'B2B3', 'B2C3', 'B2D3', 
 'C2A3', 'C2B3', 'C2C3', 'C2D3', 
 'D2A3', 'D2B3', 'D2C3', 'D2D3', 
 'A3A1', 'A3B1', 'A3C1', 'A3D1', 
 'B3A1', 'B3B1', 'B3C1', 'B3D1', 
 'C3A1', 'C3B1', 'C3C1', 'C3D1', 
 'D3A1', 'D3B1', 'D3C1', 'D3D1']
于 2013-01-05T07:15:35.427 回答