在循环中的指定索引处将字符串插入二维数组

Question

我想我应该用矩阵来解释我的问题。

我想从这里收到：

A=[[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

这个：

 A=[[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
       [ 0, 0, 0, 0, 0, **h**, 0, 0, 0, 0, 0], 
       [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
       [ 0, 0, 0, **h**, 0, 0, 0, 0, 0, 0, 0]]

粗体字母仅突出问题。我试过了：

def r(i):
    for k in i[-1:-4:-2]:
        for l in k[3:6:2]:
            k[l]='h'
    print (i)
r(A)

但它会返回：

[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
['**h**', 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 
['**h**', 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

这绝对不符合我的期望。我想从列表列表的最后一个元素（最后一个列表）开始在指定索引处插入字符串，从 rhe 行的某个索引开始。你能帮帮我吗？

如你所问，我再添加一个例子： 基本数组：

B=[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

输出：

B=[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, !, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, !, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, !, 0, 0, 0, 0, 0, 0, 0, 0]]

Answer 1

你去吧：

def r(i, substitution, col_start, row_stop):
    for row, R in enumerate(i):
        found = False
        for col, el in enumerate(R):
            if not found and col >= (col_start + (len(i)-1) - row) and \
                    (col % 2 == col_start % 2) and row > row_stop and \
                    (row % 2 == (len(i)-1) % 2):
                found = True
                i[row][col] = substitution
    return i
A=[[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
B=[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
   [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
r(A, 'h', 3, 1)
r(B, '!', 1, 2)

[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 'h', 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 'h', 0, 0, 0, 0, 0, 0, 0]]
[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, '!', 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, '!', 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, '!', 0, 0, 0, 0, 0, 0, 0, 0]]

提醒您，如果您需要对代码进行此类修改，或者您玩得很开心，或者您的设计必须进行修改。保持简单！