是否有一个函数可以返回给定令牌集的所有可能的完成组合？

Question

是否有python函数返回给定标记集的所有可能的完成组合：

-------------------
|col_1|col_2|col_3|
------+-----+------
| c11 | c12 | c13 |
| c21 | c22 | c23 |  
| c 31| c 32| c 33|
-------------------

应该返回：

tokens = ["Afoo","fooB","Abar","Bbar","barA","barBX","barBY"]

complete(tokens,"foo","barB")

Answer 1

据我所知，没有这样的功能，但使用循环这是一件简单的事情：

def complete(tokens, foo, bar):
    return [(i, j) for i in tokens if foo in i for j in tokens if bar in j]

但绝对不是超级高效！您可以使用生成器执行相同的操作：

def complete(tokens, foo, bar):
    return ((i, j) for i in tokens if foo in i for j in tokens if bar in j)

Answer 2

将其分解为搜索步骤并找到组合＆＃39;使用itertools步骤

from itertools import product
def complete(tokens, searches):
    # search
    matches = [[t for t in tokens if s in t] for s in searches]

    # find all combinations https://docs.python.org/3/library/itertools.html
    return product(*matches)

tokens = ["Afoo", "fooB", "Abar", "Bbar", "barA", "barBX", "barBY"]
search = ["foo", "barB"]

combinations = list(complete(tokens, search)

Answer 3

我想到了这个解决方案

我得到了所有的foo和barB 然后我结合两个列表

tokens = ["Afoo", "fooB", "Abar", "Bbar", "barA", "barBX", "barBY"]

foo = [a for a in tokens if "foo" in a]

barB = [a for a in tokens if "barB" in a]

x = [(a, b) for a in foo for b in barB]

print(x)

  

输出：

[('Afoo', 'barBX'), ('Afoo', 'barBY'), ('fooB', 'barBX'), ('fooB', 'barBY')]

更快

tokens = ["Afoo", "fooB", "Abar", "Bbar", "barA", "barBX", "barBY"]

x = [(a, b) for a in tokens if "foo" in a for b in tokens if "barB" in b]

print(x)

  

输出：

[('Afoo', 'barBX'), ('Afoo', 'barBY'), ('fooB', 'barBX'), ('fooB', 'barBY')]