Python：如何按列遍历列表列表，就好像它是常规2D数组（矩阵）一样？

Question

I have a list of lists like this: list=[[0,1,2,3],[4,5,6,7],[8,9,10,11],[12,13,14,15]] .

I want to use this list in a function that "slices" my list into 9 groups of four values each, that are dumped into a dict of lists . If len(list)=n , the function should create [sqrt(n)-1]*[sqrt(n)-1] slices. In this case, given len(list)=16 , slices=[4-1]*[4-1]=9 .

This is the function:

def dictionarize1(array):
    dict1 = {}
    count = 0
    for x in range(len(array[0]) - 1) :
        for y in range(len(array[0]) - 1):
            dict1[count] = [array[x][y], array[x][y+1], array[x+1][y], array[x + 1][y+1]]
            count = count + 1
    return dict1 


dictionarize1(list) #Calling the function

I think of my list as a 2D array - a 4x4 matrix in this case - which must be traversed as if it was a matrix: the row is fixed and the column is incremented:

Columns ->
Rows v    0    1    2    3
          4    5    6    7
          8    9    10   11
          12   13   14   15  

Therefore, my slices must be [0,1,4,5] , [1,2,5,6] , [2,3,6,7] , [4,5,8,9] etc., without wrapping the matrix up.

The final result should be dict={'0':[0,1,4,5],'1':[1,2,5,6],'2':[2,3,6,7],'3':[4,5,8,9],'4':[5,6,9,10],'5':[6,7,10,11],'6':[8,9,12,13],'7':[9,10,13,14],'8':[10,11,14,15]} .

My question: is this the correct way for my for loops to traverse the list by columns (which means: the row is fixed, the column is incremented)? I cannot judge as if I print my list I only get a regular list so it confuses me. Thanks!

Answer 1

I do not know if I understand what you want. If you want to show correct results do this

list = dictionarize1(list) #Calling the function
print(list) #your result is ok