Question

我有这个：

array1 = [[1,2,3],[1,2,3],[2,1,3],[2,1,3],[1,-2,3]]
array2 = [[1,2,3],[1,2,3],[1,2,3],[1,2,3],[0,2,3],[2,1,3]]

并想要创建：

multiArray1 = {[1,2,3]:2, [2,1,3]:2}
multiArray2 = {[1,2,3]:4, [2,1,3]:1}

问题：我正在尝试将multiArray1和multiArray2作为包含相同值的字典，但键分别给出了这些值在array1和array2中出现的次数。

我不确定在我的代码中要更改什么。任何帮助将不胜感激。感谢。

from collections import defaultdict

array1 = [[1,2,3],[1,2,3],[2,1,3],[2,1,3],[1,-2,3]]
array2 = [[1,2,3],[1,2,3],[1,2,3],[1,2,3],[0,2,3],[2,1,3]]

def f(arrA,arrB):
    multiArray1 = {}
    multiArray2 = {}

    intersect = set(map(tuple,arrA)).intersection(map(tuple,arrB))
    print(set(map(tuple,arrA)).intersection(map(tuple,arrB)))

    for i in intersect:
        multiArray1.update({i:0})
        multiArray2.update({i:0})
    print(multiArray1)
    print(multiArray2)

    multipleArray1 = {}
    multipleArray2 = {}

    for i in intersect:
        for j in range(len(arrA)):
            if str(tuple(arrA[j])) in set(intersect):
                multiArray1[tuple(arrA[j])].append(j)
                print(multiArray1)

                multipleArray1 = defaultdict(list)
                for key, value in multipleArray1:
                    multipleArray1[i].append(j)
                    print(multipleArray1)

    for j in range(len(arrB)):
        if str(tuple(arrB[j])) in set(intersect):
            multiArray2[tuple(arrB[j])].append(j)

            multipleArray2 = defaultdict(list)
            for key, value in multipleArray2:
                multipleArray2[i].append(j)
                print(multipleArray2)

    print(multiArray1)
    print(multiArray2)

f(array1,array2)

您从上面的代码中获得的输出是：

{(2, 1, 3), (1, 2, 3)}
{(2, 1, 3): 0, (1, 2, 3): 0}
{(2, 1, 3): 0, (1, 2, 3): 0}
{(2, 1, 3): 0, (1, 2, 3): 0}
{(2, 1, 3): 0, (1, 2, 3): 0}

Answer 1

有人指出，你不能使用列表。在我的方法中，您需要将子列表转换为元组，然后更新字典。

In [48]: array1 = [[1,2,3],[1,2,3],[2,1,3],[2,1,3],[1,-2,3]]

In [49]: array1 = list(map(lambda x: tuple(x), array1))

In [50]: array1
Out[50]: [(1, 2, 3), (1, 2, 3), (2, 1, 3), (2, 1, 3), (1, -2, 3)]

In [51]: res = dict()

In [52]: for i in array1:
    if i not in res:
        res[i] = 1
    else:
        res[i] += 1
   ....:         

In [53]: res
Out[53]: {(1, -2, 3): 1, (1, 2, 3): 2, (2, 1, 3): 2}

Answer 2

诀窍是，在将它们作为dict的键之前将它们转换为字符串，因为你不能将列表作为字典的键。

PopupWindow pw = new PopupWindow(layout, (int)mContext.getResources().getDimension(R.dimen.popup_width),
        (int)mContext.getResources().getDimension(R.dimen.popup_height), true); // notice that they have to be casted as int

输出

dct1 = {}
dct2 = {}
array1 = [[1,2,3],[1,2,3],[2,1,3],[2,1,3],[1,-2,3]]
array2 = [[1,2,3],[1,2,3],[1,2,3],[1,2,3],[0,2,3],[2,1,3]]

for x in array1:
    cnt = array1.count(x)
    dct1[str(x)] = cnt #here str(x) convert the list to string
for x in array2:
    cnt = array2.count(x)
    dct2[str(x)] = cnt #again here


print (dct1)
print (dct2)

创建一个字典，其键给出了数组数组中出现的值的多重性

2 个答案: