Question

我是一名蟒蛇初学者。我想从一个列表中获取两个组合列表。

例如，我有一个列表：

c = [1, 2, 3, 4]

我想使用每四个项目来填充两个列表。会有((2^4)/2)-1种可能性。

c1 = [1]  c2 = [2, 3, 4]
c1 = [2]  c2 = [1, 3, 4]
c1 = [3]  c2 = [2, 3, 4]
c1 = [4]  c2 = [1, 2, 3]
c1 = [1, 2]  c2 = [3, 4]
c1 = [1, 3]  c2 = [2, 4]
c1 = [1, 4]  c2 = [2, 3]

此功能通常适用于此类任务itertools，但我无法选择itertools.combination生成的列表数。

该功能只允许我选择每个列表应该有多少项。

例如，如果我尝试跟随功能，

print list(itertools.combinations(c, 2))

我只能得到这样的结果。

[(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)]

我很难找到这个，但我找不到任何东西。

更新

哦，我可怜的英语造成了这样的混乱！我完全改变了我的榜样。我想将4个项目分配到2个列表。抱歉混乱！

Answer 1

我不确定您对 10选择2 的理解是什么。您从list(itertools.combinations(c, 2))收到的输出在数学上定义为 10C2 。

修改

从编辑到您的问题，您似乎想要一种完全不同的组合。结果的数量仍然不是45，而是： 10C1 + 10C2 + 10C3 + 10C4 + 10C5 。

我希望以下内容可以帮助您前进：

for i in range(1, 6):
    for c1 in itertools.combinations(c, i):
            c1 = set(c1)
            c2 = set(c) - c1
            print c1, c2

以上代码的灵感来自于this (deleted) answer CSZ {/ 3}}。

使用range(1, 3)时收到的输出：

[1] [2, 3, 4, 5, 6, 7, 8, 9, 10]
[2] [1, 3, 4, 5, 6, 7, 8, 9, 10]
[3] [1, 2, 4, 5, 6, 7, 8, 9, 10]
[4] [1, 2, 3, 5, 6, 7, 8, 9, 10]
[5] [1, 2, 3, 4, 6, 7, 8, 9, 10]
[6] [1, 2, 3, 4, 5, 7, 8, 9, 10]
[7] [1, 2, 3, 4, 5, 6, 8, 9, 10]
[8] [1, 2, 3, 4, 5, 6, 7, 9, 10]
[9] [1, 2, 3, 4, 5, 6, 7, 8, 10]
[10] [1, 2, 3, 4, 5, 6, 7, 8, 9]
[1, 2] [3, 4, 5, 6, 7, 8, 9, 10]
[1, 3] [2, 4, 5, 6, 7, 8, 9, 10]
[1, 4] [2, 3, 5, 6, 7, 8, 9, 10]
[1, 5] [2, 3, 4, 6, 7, 8, 9, 10]
[1, 6] [2, 3, 4, 5, 7, 8, 9, 10]
[1, 7] [2, 3, 4, 5, 6, 8, 9, 10]
[8, 1] [2, 3, 4, 5, 6, 7, 9, 10]
[1, 9] [2, 3, 4, 5, 6, 7, 8, 10]
[1, 10] [2, 3, 4, 5, 6, 7, 8, 9]
[2, 3] [1, 4, 5, 6, 7, 8, 9, 10]
[2, 4] [1, 3, 5, 6, 7, 8, 9, 10]
[2, 5] [1, 3, 4, 6, 7, 8, 9, 10]
[2, 6] [1, 3, 4, 5, 7, 8, 9, 10]
[2, 7] [1, 3, 4, 5, 6, 8, 9, 10]
[8, 2] [1, 3, 4, 5, 6, 7, 9, 10]
[9, 2] [1, 3, 4, 5, 6, 7, 8, 10]
[2, 10] [1, 3, 4, 5, 6, 7, 8, 9]
[3, 4] [1, 2, 5, 6, 7, 8, 9, 10]
[3, 5] [1, 2, 4, 6, 7, 8, 9, 10]
[3, 6] [1, 2, 4, 5, 7, 8, 9, 10]
[3, 7] [1, 2, 4, 5, 6, 8, 9, 10]
[8, 3] [1, 2, 4, 5, 6, 7, 9, 10]
[9, 3] [1, 2, 4, 5, 6, 7, 8, 10]
[10, 3] [1, 2, 4, 5, 6, 7, 8, 9]
[4, 5] [1, 2, 3, 6, 7, 8, 9, 10]
[4, 6] [1, 2, 3, 5, 7, 8, 9, 10]
[4, 7] [1, 2, 3, 5, 6, 8, 9, 10]
[8, 4] [1, 2, 3, 5, 6, 7, 9, 10]
[9, 4] [1, 2, 3, 5, 6, 7, 8, 10]
[10, 4] [1, 2, 3, 5, 6, 7, 8, 9]
[5, 6] [1, 2, 3, 4, 7, 8, 9, 10]
[5, 7] [1, 2, 3, 4, 6, 8, 9, 10]
[8, 5] [1, 2, 3, 4, 6, 7, 9, 10]
[9, 5] [1, 2, 3, 4, 6, 7, 8, 10]
[10, 5] [1, 2, 3, 4, 6, 7, 8, 9]
[6, 7] [1, 2, 3, 4, 5, 8, 9, 10]
[8, 6] [1, 2, 3, 4, 5, 7, 9, 10]
[9, 6] [1, 2, 3, 4, 5, 7, 8, 10]
[10, 6] [1, 2, 3, 4, 5, 7, 8, 9]
[8, 7] [1, 2, 3, 4, 5, 6, 9, 10]
[9, 7] [1, 2, 3, 4, 5, 6, 8, 10]
[10, 7] [1, 2, 3, 4, 5, 6, 8, 9]
[8, 9] [1, 2, 3, 4, 5, 6, 7, 10]
[8, 10] [1, 2, 3, 4, 5, 6, 7, 9]
[9, 10] [1, 2, 3, 4, 5, 6, 7, 8]

Answer 2

     <!DOCTYPE html>
      <html ng-app>
        <head>
             <link rel="stylesheet" href="style.css">
             <!--<link rel="stylesheet" href="bootstrap.css">--> 
            <!-- Latest compiled and minified CSS -->
            <link rel="stylesheet" href="bootstrap.css">
            <!-- Latest compiled and minified JavaScript -->
             <script src="angular.min.js"></script> 
              <script src="cordova.js"></script> 
             <script src="angular.animate.min.js"></script>
             <script src="angular.route.min.js"></script>     
             <link rel="stylesheet" href="font-awesome.css">
             <meta name="viewport" content="width=device-width, initial-scale=1.0"/>
        </head>
        <body>
</body>

     <script type="text/javascript" src="pro.js"></script>
    <script type="text/javascript" src="bootstrap.js"></script>
     <script type="text/javascript" src="script.js"></script>
  </html>

如果你想要所有的组合。你可以这样做。

l = [1,2,3,4, 5,  6, 7, 8]

print [[l[:i], l[i:]] for i in range(1, len(l))]

或

print [l[i:i+n] for i in range(len(l)) for n in range(1, len(l)-i+1)]

一个列表中的两个组合列表

2 个答案:

修改