Question

我有两个像这样的列表：

[[0,3,5],['cat','dog','chicken'], [15,14,11]]
[[1,2,4],['whale', 'rabbit', 'zebra'], [10,9,22]]

我如何创建一个列表列表：

[[0,1,2,3,4,5], ['cat', 'whale', 'rabbit','dog','zebra','chicken'], [15,10,9,14,22,11]]

修改只是为了澄清一下，我需要什么输出 - 按第一个列表排序的列表列表。

Answer 1

这是怎么回事。按照你想要的方式工作。逻辑是，在第一次迭代中，项目从“l1”然后l2中获取。在另外两个中，l2后跟l1：

l1=[[0,3,5],['cat','dog','chicken'], [15,14,11]]
l2=[[1,2,4],['whale', 'rabbit', 'zebra'], [10,9,22]]
l3=[[],[],[]]
for i in range (3):
    for j in range(3):
        if(j==0):
            l3[i].append(l1[i][j])
            l3[i].append(l2[i][j])
        else:
            l3[i].append(l2[i][j])
            l3[i].append(l1[i][j])
print(l3)

输出：[[0,1,2,3,4,5], ['cat', 'whale', 'rabbit','dog','zebra','chicken'], [15,10,9,14,22,11]]

行动准则：http://ideone.com/jIlAeK

Answer 2

这应该可以解决问题：

  // getting the image file from the stream and saving to server
                HttpPostedFile f1 = (HttpPostedFile)Session["f1"];
                int nFileLen = f1.ContentLength;
                byte[] myData = new byte[nFileLen];
                f1.InputStream.Read(myData, 0, nFileLen);

                if (f1.ContentLength != 0)
                {
                    string filename = f1.FileName;
                    string extention = Path.GetExtension(filename);
                    if (extention.ToLower() == ".jpg" | extention.ToLower().ToLower() == ".png" | extention.ToLower().ToLower() == ".jpeg")
                    {
                        filename = txtMobile.Text;
                        int width = Convert.ToInt32(ConfigurationManager.AppSettings["ProfilePicWidth"]);
                        string Destination = Convert.ToString(Server.MapPath("../Images/Members/" + filename + extention));
                        CommanFunctions.ResizeImage(width, f1.InputStream, Destination);
                        regi.ProfilePic = Destination;
                    }
                    else
                    {
                        lblError.Text = "Please Upload a valid Image";
                        multiview.ActiveViewIndex = 1;
                    }
                }

说明：

In [30]: l1 = [[0,3,5],['cat','dog','chicken'], [15,14,11]] In [31]: l2 = [[1,2,4],['whale', 'rabbit', 'zebra'], [10,9,22]] In [32]: from operator import itemgetter In [33]: import itertools In [37]: list(zip(*sorted(itertools.chain(zip(*l1), zip(*l2)), key=itemgetter(0)))) Out[37]: [(0, 1, 2, 3, 4, 5), ('cat', 'whale', 'rabbit', 'dog', 'zebra', 'chicken'), (15, 10, 9, 14, 22, 11)] In [39]: list(map(list, zip(*sorted(itertools.chain(zip(*l1), zip(*l2)), key=itemgetter(0))))) Out[39]: [[0, 1, 2, 3, 4, 5], ['cat', 'whale', 'rabbit', 'dog', 'zebra', 'chicken'], [15, 10, 9, 14, 22, 11]]将您的列表收集到元组列表中。每个元组包含三个元素：zip(*l)。例如：
```
(id, name, some_integer_field)
```
元组应按第一个元素排序 - 这就是In [35]: list(itertools.chain(zip(*l1), zip(*l2))) Out[35]: [(0, 'cat', 15), (3, 'dog', 14), (5, 'chicken', 11), (1, 'whale', 10), (2, 'rabbit', 9), (4, 'zebra', 22)]
然后它应该被收集到单个列表中。另一次致电sorted(..., key=operator.itemgetter(0))。
不幸的是，zip会返回一个元组列表。当您需要列表列表时，您应该将zip的结果打包到zip

Answer 3

使用列表压缩，您只需要1行代码

aa = [[0,3,5]，['cat'，'dog'，'chicken']，[15,14,11]]

bb = [[1,2,4]，['whale'，'rabbit'，'zebra']，[10,9,22]]

new_list = [a + b代表a，b代表zip（aa，bb）]

打印new_list

[[0,3,5,1,2,4]，['猫'，'狗'，'鸡'，'鲸'，'兔子'，'斑马']，[15,14,11] ，10,9,22]]

Answer 4

使用列表中的第一个条目作为排序索引，以下脚本将为您提供预期的输出：

list_1 = [[0, 3, 5], ['cat','dog','chicken'], [15, 14, 11]]
list_2 = [[1, 2, 4], ['whale', 'rabbit', 'zebra'], [10, 9, 22]]

sort_order = list_1[0] + list_2[0]
output = []

for i1, i2 in zip(list_1, list_2):
    source = i1 + i2
    entry = [None] * len(source)

    for i, v in enumerate(source):
        entry[sort_order[i]] = v

    output.append(entry)

print output

这将显示以下内容：

[[0, 1, 2, 3, 4, 5], ['cat', 'whale', 'rabbit', 'dog', 'zebra', 'chicken'], [15, 10, 9, 14, 22, 11]]

Answer 5

只需要几行：

>>> a = [[0,3,5],['cat','dog','chicken'], [15,14,11]]
>>> b = [[1,2,4],['whale', 'rabbit', 'zebra'], [10,9,22]]
>>> the_list = [c+d for c, d, in zip(a,b)]
>>> print the_list
[[0, 3, 5, 1, 2, 4], ['cat', 'dog', 'chicken', 'whale', 'rabbit', 'zebra'], [15, 14, 11, 10, 9, 22]]
>>> the_list[0].sort()
>>> print the_list
[[0, 1, 2, 3, 4, 5], ['cat', 'dog', 'chicken', 'whale', 'rabbit', 'zebra'], [15, 14, 11, 10, 9, 22]]

Answer 6

此解决方案适用于顶级列表中未指定数量的列表。合并两个输入列表后，它们将根据第一个列表值进行排序：

a = [[0,3,5],['cat','dog','chicken'], [15,14,11]]
b = [[1,2,4],['whale', 'rabbit', 'zebra'], [10,9,22]]
combined = list(zip(*a)) + list(zip(*b))  # create tuples with an element from each sublist then merge them
ordered_tuples = sorted(combined)  # or  more explicitly: sorted(combined, key=lambda x: x[0])
output = list(zip(*ordered_tuples))  # this reverses the sorted tuples into lists as requested

Answer 7

查看此脚本，它对我有用

list_1 = [[0,3,5],['cat','dog','chicken'], [15,14,11]]
list_2 = [[1,2,4],['whale', 'rabbit', 'zebra'], [10,9,22]]

list_3 = []

for i in range(len(list_1)):
    list_3.append(list_1[i] + list_2[i])

print(list_3)

从两个列表列表中创建列表列表

7 个答案: