生成潜在的8个字符串的所有可能的2个字符组合？

Question

我有一个生成元组元组的所有可能组合

( (base1 , position1) , (base2 , position2) )

的bases = ["U", "C", "A", "G"]和positions = [0,1,2,3,4,5,6,7,8].

要求

• 没有重复
• 基数可以相同但位置必须不同
• 必须保留订单.

例如:

( (A,1), (B,2) ) == ( (B,2) , (A,1) )并且 ( (A,1), (B,1) )应该被丢弃.

样本输出:

[ ( (U,0) , (U,1) ), ( (U,0) , (U,2) ), ( (U,0) , (U,3) ) ...]

应该有448的长度

---

例:

对于字符串长度2:

((U,0),(U,1))
((U,0),(C,1))
((U,0),(A,1))
((U,0),(G,1))

((C,0),(U,1))
((C,0),(C,1))
((C,0),(A,1))
((C,0),(G,1))

((A,0),(U,1))
((A,0),(C,1))
((A,0),(A,1))
((A,0),(G,1))

((G,0),(U,1))
((G,0),(C,1))
((G,0),(A,1))
((G,0),(G,1))

会是所有组合......我想

---

到目前为止我有这个

all_possible = []
nucleotides = ["U","C","A","G"]


for i in range(len(nucleotides)):
    for j in range(8):
        all_possible.append(((nucleotides[i],j),(nucleotides[i],j)))

Answer 1

听起来你想要(每个可能的2个基本单词)X(从范围(8)中抽取的每2个组合的笛卡尔积).

你可以得到这个

from itertools import product, combinations

def build(num_chars, length):
    bases = ["U", "C", "A", "G"]
    for letters in product(bases, repeat=num_chars):
        for positions in combinations(range(length), num_chars):
            yield list(zip(letters, positions))

这给了我们

In [4]: output = list(build(2, 8))

In [5]: len(output)
Out[5]: 448

In [6]: output[:4]
Out[6]: 
[[('U', 0), ('U', 1)],
 [('U', 0), ('U', 2)],
 [('U', 0), ('U', 3)],
 [('U', 0), ('U', 4)]]

In [7]: output[-4:]
Out[7]: 
[[('G', 4), ('G', 7)],
 [('G', 5), ('G', 6)],
 [('G', 5), ('G', 7)],
 [('G', 6), ('G', 7)]]