Python：在列表嵌套中替换项

Question

这是我的代码：

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

element = 4
x = 0
y = 0

   data[x][y] = element

我想把坐标为0,0的元素替换掉，但是当我打印数据的时候，发现这个元素并没有改变。

---

*******编辑******：好的，这里是我的完整代码：

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

z = []

#row 6
x1 = 6
for y in range(9):
  print data[x1][y]
  z.append(data[x1][y])


#column 8
y1 = 8 
for x in range(9):
  print data[x][y1]
  z.append(data[x][y1])


#finds the block coordinates
x = 6
y = 8
basex = x - x%3
basey = y - y%3
for x1 in range(basex,basex+3):
    for y1 in range(basey,basey+3):
        print x1,y1, data[x1][y1]
        z.append(data[x1][y1])

item = [1,2,3,4,5,6,7,8,9]
for element in item:
    if element not in z:
            print element

data[x][y] = element 
print data[x][y]

Answer 1

你现在用的是什么版本的Python？能不能试试在命令行里运行一下，然后把结果发过来，比如下面这样？对我来说这个是可以正常工作的。我基本上就是直接从你的帖子里复制粘贴的。

Python 2.6.4 (r264:75706, Dec  7 2009, 18:45:15) 
[GCC 4.4.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> data = [
... [5,3,0,0,7,0,0,0,0],
... [6,0,0,1,9,5,0,0,0],
... [0,9,8,0,0,0,0,6,0],
... [8,0,0,0,6,0,0,0,3],
... [4,0,0,8,0,3,0,0,1],
... [7,0,0,0,2,0,0,0,6],
... [0,6,0,0,0,0,2,8,0],
... [0,0,0,4,1,9,0,0,5],
... [0,0,0,0,8,0,0,7,9]
... ]
>>> 
>>> element = 4
>>> x = 0
>>> y = 0
>>> 
>>> data
[[5, 3, 0, 0, 7, 0, 0, 0, 0], [6, 0, 0, 1, 9, 5, 0, 0, 0], [0, 9, 8, 0, 0, 0, 0, 6, 0], [8, 0, 0, 0, 6, 0, 0, 0, 3], [4, 0, 0, 8, 0, 3, 0, 0, 1], [7, 0, 0, 0, 2, 0, 0, 0, 6], [0, 6, 0, 0, 0, 0, 2, 8, 0], [0, 0, 0, 4, 1, 9, 0, 0, 5], [0, 0, 0, 0, 8, 0, 0, 7, 9]]
>>> data[x][y] = element
>>> data
[[4, 3, 0, 0, 7, 0, 0, 0, 0], [6, 0, 0, 1, 9, 5, 0, 0, 0], [0, 9, 8, 0, 0, 0, 0, 6, 0], [8, 0, 0, 0, 6, 0, 0, 0, 3], [4, 0, 0, 8, 0, 3, 0, 0, 1], [7, 0, 0, 0, 2, 0, 0, 0, 6], [0, 6, 0, 0, 0, 0, 2, 8, 0], [0, 0, 0, 4, 1, 9, 0, 0, 5], [0, 0, 0, 0, 8, 0, 0, 7, 9]]
>>> 

Answer 2

我看到你代码里唯一的问题就是最后一行的缩进不对。把它调整到和其他代码保持一样的缩进就可以了。 :)

你可能还会对 pprint 模块感兴趣：

>>> from pprint import pprint
>>> pprint(data)
[[4, 3, 0, 0, 7, 0, 0, 0, 0],
 [6, 0, 0, 1, 9, 5, 0, 0, 0],
 [0, 9, 8, 0, 0, 0, 0, 6, 0],
 [8, 0, 0, 0, 6, 0, 0, 0, 3],
 [4, 0, 0, 8, 0, 3, 0, 0, 1],
 [7, 0, 0, 0, 2, 0, 0, 0, 6],
 [0, 6, 0, 0, 0, 0, 2, 8, 0],
 [0, 0, 0, 4, 1, 9, 0, 0, 5],
 [0, 0, 0, 0, 8, 0, 0, 7, 9]]

这样看起来会更容易读一些！

Answer 3

你需要在找到所需的元素后，立刻停止循环：

item = [1,2,3,4,5,6,7,8,9]
for element in item:
    if element not in z:
            print element
            break

data[x][y] = element 
print data[x][y]