Question

我想出了一个相当微不足道的问题，但是因为我对蟒蛇很陌生，所以我会把头撞到我的桌子上一段时间。（好痛）。虽然我认为解决问题更合乎逻辑...... 首先我要说我正在使用Python SDK for Cinema 4D，所以我不得不稍微更改以下代码。但这就是我想要做的事情并且在努力：我正在尝试对一些多边形选择进行分组，这些选择是动态生成的（基于某些规则，而不是那么重要）。以下是数学方法的工作原理：这些选择基于岛屿（意味着有几个多边形相连）。然后，必须将这些选择分组并放入我可以使用的列表中。任何多边形都有自己的索引，所以这个应该相当简单，但就像我之前说的那样，我在那里很挣扎。

主要问题很容易解释：我试图在第一个循环中访问一个不存在的索引，导致索引超出范围错误。我首先尝试评估有效性，但没有运气。对于那些熟悉Cinema 4D + Python的人，如果有人想要，我会提供一些原始代码。到目前为止，这么糟糕。这是简化和改编的代码。

编辑：忘记提到导致错误的检查实际上应该只检查重复项，因此当前选择的数字将被跳过，因为它已被处理。由于计算量很大，这是必要的。

真的希望，任何人都可以让我朝着正确的方向前进，这段代码到目前为止是有道理的。：）

def myFunc():

        sel = [0,1,5,12] # changes with every call of "myFunc", for example to [2,8,4,10,9,1], etc. - list alway differs in count of elements, can even be empty, groups are beeing built from these values
        all = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] # the whole set
        groups = [] # list to store indices-lists into
        indices = [] # list to store selected indices
        count = 0 # number of groups
        tmp = [] # temporary list to copy the indices list into before resetting

        for i in range(len(all)): # loop through values
            if i not in groups[count]: # that's the problematic one; this one actually should check whether "i" is already inside of any list inside the group list, error is simply that I'm trying to check a non existent value
                for index, selected in enumerate(sel): # loop through "sel" and return actual indices. "selected" determines, if "index" is selected. boolean.
                    if not selected: continue # pretty much self-explanatory
                    indices.append(index) # push selected indices to the list
                tmp = indices[:] # clone list
                groups.append(tmp) # push the previous generated list to another list to store groups into
                indices = [] # empty/reset indices-list
                count += 1 # increment count
        print groups    # debug
myFunc()

修改

添加第二个列表后，extend将填充，而不是append作为计数器，所有内容都按预期运行！该列表将是一个基本列表，非常简单;）

Answer 1

groups[count]

当你第一次调用它时，groups是一个空列表，count是0.你不能在组中的0点访问这个东西，因为那里什么都没有！

尝试制作 groups = []到groups = [[]]（即不是空列表，只列出空列表的列表）。

Answer 2

我不确定您为什么要将空列表添加到组中。也许这样更好

if i not in groups[count]:

到

if not groups or i not in groups[count]:

如果您不打算将其用于其他任何目的，您也无需复制该列表。所以你可以替换

            tmp = indices[:] # clone list
            groups.append(tmp) # push the previous generated list to another list to store groups into
            indices = [] # empty/reset indices-list

与

            groups.append(indices) # push the previous generated list to another list to store groups into
            indices = [] # empty/reset indices-list

您甚至可以完全放弃count（您可以随时使用len(groups)）。您还可以使用listcomprehension

替换内部循环

def myFunc():

    sel = [0,1,5,12] # changes with every call of "myFunc", for example to [2,8,4,10,9,1], etc. - list alway differs in count of elements, can even be empty, groups are beeing built from these values
    all = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] # the whole set
    groups = [] # list to store indices-lists into

    for i in range(len(all)): # loop through values
        if not groups or i not in groups[-1]: # look in the latest group
            indices = [idx for idx, selected in enumerate(sel) if selected]
            groups.append(indices) # push the previous generated list to another list to store groups into
    print groups    # debug

Answer 3

正确的第11行：

   if i not in groups[count]

为：

   if i not in groups:

嵌套for循环索引超出范围

3 个答案: