Question

我创建了一个3D列表。现在，我想计算该3d列表中数字的出现次数。

print("Enter matrix entries column wise one matrix after another:\n")
lists=[[[int(input()) for _ in range(o)] for _ in range(a)] for _ in  range(c)] 
print(lists)

我已经完成了这段代码来创建3d列表。在这里，o = 4（行数），a = 3（列数），c = 2（矩阵数）（以前由用户提供），矩阵仅包含1,2,3,4和5。< / p>

Enter matrix entries column wise one matrix after another: 
 1 
 1 
 2 
 3 
 4 
 5
 1 
 2 
 3 
 4 
 5 
 1 
 2 
 1 
 1 
 1 
 1 
 1 
 1 
 2 
 2 
 3 
 3 
 4 
 4 
 [[[1, 2, 3, 4], [5, 1, 2, 3], [4, 5, 1, 2]], [[1, 1, 1, 1], [1, 1, 2, 2], [3, 3, 4, 4]]]

，我得到了这个输出，这是理想的情况。我们可以将生成的3d列表视为

matrix1= 1 5 4
         2 1 5
         3 2 1
         4 3 2
matrix2= 1 2 3
         1 2 3
         1 2 3
         1 2 3

现在我想要一个3d列表，上面写着“ counts”，看起来像这样：

[[[[1，1，1]，[1，1，1]，[1，1，0]，[1，0，1]，[0，1，1]]，[[4， 0，0]，[0，4，0]，[0，0，4]，[0，0，0]，[0，0，0]]] p

Here counts[0][0][0](i.e. first 1) describes how many 1's are there in first column first matrix.

counts[0][0][1](i.e. second 1) describes how many 1's are there in second column first matrix.

counts[0][0][2](i.e. third 1) describes how many 1's are there in third column first matrix.

counts[0][1][0](i.e. forth 1) describes how many 2's are there in first column first matrix.

counts[0][1][1](i.e. fifth 1) describes how many 2's are there in second column first matrix.

counts[0][1][2](i.e. sixth 1) describes how many 2's are there in third column first matrix.

counts[0][2][0](i.e. seventh 1) describes how many 3's are there in first column first matrix.

counts[0][2][1](i.e. eighth 1) describes how many 3's are there in second column first matrix.

counts[0][2][2](i.e. ninth 0) describes how many 3's are there in third column first matrix.

counts[0][3][0](i.e. tenth 1) describes how many 4's are there in first column first matrix.

counts[0][3][1](i.e. eleventh 0) describes how many 4's are there in second column first matrix.

counts[0][3][2](i.e. 12th 1) describes how many 4's are there in third column first matrix.

counts[0][4][0](i.e. 13th 0) describes how many 5's are there in first column first matrix.

counts[0][4][1](i.e. 14th 1) describes how many 5's are there in second column first matrix.

counts[0][4][2](i.e. 15th 1) describes how many 5's are there in third column first matrix.

以相同的方式[[4，0，0]，[0，4，0]，[0，0，4]，[0，0，0]，[0，0，0]]这部分还计算第二矩阵。请从“列表”列表和“ a”，“ o”，“ c”变量中计算“计数”列表。

Answer 1

我不确定您要寻找的确切输出是什么，但是可以使用for循环来实现：

lists=[[[1, 2, 3, 4], [5, 1, 2, 3], [4, 5, 1, 2]], [[1, 1, 1, 1], [1, 1, 2, 2], [3, 3, 4, 4]]]
for m_num, matrix in enumerate(lists):
    for ind, col in enumerate(matrix):
        for i in range(1, 6):  # because as you said matrix contains only 1, 2, 3, 4 and 5
            print('number of {} in column idn: {}, of {} matrix: {}'.format(i, ind+1, m_num+1, col.count(i)))

您还可以跳过枚举，仅在第一列中然后在第二列中打印数字“ 1”，依此类推：

lists=[[[1, 2, 3, 4], [5, 1, 2, 3], [4, 5, 1, 2]], [[1, 1, 1, 1], [1, 1, 2, 2], [3, 3, 4, 4]]]
for matrix in lists:
    for col in matrix:
        for i in range(1, 6):
            print(col.count(i))

在3D列表中计算特定数字

1 个答案: