Question

我不知道为什么我遇到这个问题，我是python的新手，但这个想法应该很简单。

我有一个元组列表，如：

（A，B），（B，C），（B，d），（B，A），（C，L），（d，Y），（Y，L）

给出一个起点，让我们说一个和一个终点l。

我想制作所有路径。（A，B，C，L）（A，B，d，Y，L）

我做了很多尝试，他们都在打破无益的错误消息，如：

File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths
  File "<stdin>", line 20, in findAllPaths

我认为每种方法都过于复杂。有一种简单的python方式吗？

编辑以增加一些清晰度：

我的目标是旅行元组以找到最终目标的所有不同（非圆形）路径。所以在我上面的例子中，我想从一个 - ＆gt;旅行升。

因此，该计划将检查所有＆＃39; a＆＃39;连接并创建上面的新元组。

因此，程序将采用它可以的所有路径，直到找到l，如果路径用完，并且找不到它应该返回空列表的位置。

最后，a-＆gt; b和b-＆gt; a，但这将创建一个我们可以忽略的循环。

Answer 1

您想将8个列表合并为2个列表吗？或者你想在这8个列表中查看重复的内容？我不明白(a,b,c,l) (a,b,d,y,l)你是怎么得到的？

编辑好吧，写得非常快，你可能会发现它里面有一些奇怪的名字，而且它可以更优化，但它确实起作用，也是开始的好地方。

def check_connection(Connection_DNA_List, Preview=True):
    DNA_Path = []
    for DNA_tuples in Connection_DNA_List:
        DNA_Path = check_DNA_List(newValue=DNA_tuples, Path_list=DNA_Path, Preview=False)

    if Preview:
        print DNA_Path


def check_DNA_List(newValue,Path_list, Preview=False):
    FinalList = []
    if Path_list:
        for eachpath in Path_list:
            newList, newfile = check_DNA(newValue=newValue, DNA_List=eachpath, Preview=False)
            if newfile:
                FinalList.append(eachpath)
            if "A" in newList:
                FinalList.append(newList)
    else:
        newList, newfile = check_DNA(newValue=newValue, DNA_List=Path_list, Preview=False)
        if "A" in newList:
                FinalList.append(newList)
                FinalList.append(newList)

    if Preview:
        print "Current Path List: ", Path_list
        print "New Value: %s %s" %newValue
        print "New Path List: %s" %FinalList
        print "------------------------------------"
    return FinalList

def check_DNA(newValue, DNA_List, Preview=False):
    newDNA = []
    newList = False
    if DNA_List:
        if newValue[0] in DNA_List and newValue[1] in DNA_List:
            newDNA = DNA_List
            newList = False
        elif newValue[0] in DNA_List[-1] and newValue[1] not in DNA_List:
            newDNA = DNA_List
            newDNA.append(newValue[1])
            newList = False
        else:
            newList = True
            newDNA.append(newValue[0])
            newDNA.append(newValue[1])
    else:
        newList = True
        newDNA.append(newValue[0])
        newDNA.append(newValue[1])
    if Preview:
        print "Current List: %s" %DNA_List
        print "New Value: %s %s" %newValue
        print "Created List: %s" %newDNA
        print "Is it a newList? %s" %newList
        print "------------------------------------"
    return newDNA ,newList

a = "A"
b = "B"
c = "C"
d = "D"
y = "Y"
l = "L"

DNAList = [(a,b),(b,c),(b,d),(b,a),(c,l),(d,y),(y,l)]
check_connection(Connection_DNA_List=DNAList)

注意：你发现很多预览，我经常把它们写进去调试，知道它在做什么是很好的方法，你可以把它们关掉，如果你不想看到它们的话

Answer 2

这是一个相当简洁的方法：

CREATE TRIGGER STOCKGROUP
ON [dbo].[STOCK_ITEMS]
FOR UPDATE 
AS 
    IF UPDATE (X_STOCK_GROUP3)
       set STOCKGROUP = STOCK_GROUPS.GROUPNO

    PRINT 'AFTER UPDATE Trigger fired.'
GO

...

def find_paths(segments, start, end):
    full_paths = []
    part_paths = [(start,)]
    while len(part_paths) > 0:
        new_part_paths = []
        for p in part_paths:
            for s in segments:
                if s[0] == p[-1] and s[-1] not in p:
                    # segment s should be added to path p
                    new_path = p + s[1:]
                    if new_path[-1] == end:
                        full_paths += [new_path]
                    else:
                        new_part_paths += [new_path]
        part_paths = new_part_paths
    return full_paths

a = 'a'; b = 'b'; c = 'c'; d = 'd'; l = 'l'; y = 'y'
print find_paths([(a,b),(b,c),(b,d),(b,a),(c,l),(d,y),(y,l)], a, l)

基于共同的第一个和最后一个元素合并元组

2 个答案: