将子列表中的每三个元素附加到字典中

如何解决将子列表中的每三个元素附加到字典中

假设我有一个包含列表的列表。

board = [[5,3,7,1,0],[6,9,5,[0,8,6,[8,3],[4,1],[7,2,6],4,5],9]]

我想将每个子列表中的每三个数字附加到字典中的一个键。 例如

dd = {1:[5,8]}

然后我会寻找下一个 3x3 部分。

dd = {2:[0,0]}

总共我应该有 9 个键，每个键都有 9 个元素的列表。

这显然行不通XD：

board = [[5,9]]
xBx = {}
count1 = 1
count2 = 3
count3 = 0
for row in board:
   if count1 == 4:
       count = 1
   xBx[count1] = row[count3:count2]
   count1 += 1
   count2 += 3
   count3 += 3
   
print(xBx)

解决方法

使用numpy，更容易对多维数组进行切片

import numpy as np

board = np.array(board)
result = {}
for i in range(len(board) // size):
    for j in range(len(board) // size):
        values = board[j * size:(j + 1) * size,i * size:(i + 1) * size]
        result[i * size + j + 1] = values.flatten()

然后 result 给予

{
   1: array([5,3,6,9,8]),2: array([8,4,7,0]),3: array([0,4: array([0,1,5,5: array([0,8,2,6: array([0,7: array([1,8: array([0,6]),9: array([2,9])
}

,

这就是您可能正在寻找的

size = 3
l = len(board)

dict = {}
dictIndex = 1

for x in range(0,l,size):
    for y in range(0,size):
        a = []
        for i in range(x,x + size):
            for j in range(y,y + size):
                a.append(board[i][j])
        dict[dictIndex] = a
        dictIndex += 1

print(dict)

较短的变体

size,dict,dictIndex  = 3,len(board),{},1
for x in range(0,size):
        dict[dictIndex],dictIndex = [board[i][j] for i in range(x,x + size) for j in range(y,y + size)],dictIndex + 1
print(dict)

,

这应该有效：

total = 0
for status in tweepy.Cursor(api.search,q='cricket',lang="en",geocode="24.8607,67.0011,25mi OR 40.7128,74.0060,20mi"
                           ).items(10):
    total+=1

print(total)