Question

继承我的代码：

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

我想替换坐标0,0处的元素，但是当我打印数据时，它没有更改元素。

*******编辑******：好了我的完整代码：**

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

对我来说效果很好......

>>> 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
>>> print data[0][0]
5
>>> data[x][y] = element
>>> print data[0][0]
4
>>>

Answer 2

你好像已经标出了你的最后一行，这在Python解释器中给了我一个错误。如果我删除该标签，它就可以。

您的数组data已更改。也许你不打印它，所以你不知道它改变了吗？

Answer 3

你在运行什么版本的python？你可以从命令行试试并发布结果，如下所示？它似乎对我有用。我基本上是直接复制并粘贴你的帖子。

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 4

我在代码中看到的唯一错误是最后一行是在不同的缩进级别。将它放在与其余代码相同的级别上工作正常。：）

您可能也对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]]

更容易阅读！

Answer 5

一旦找到必要的元素，你需要打破外观：

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]

Answer 6

啊......现在您已经发布了所有代码，它更有意义......

它没有按照你预期的方式工作的原因是因为'element'在循环的上下文之外不存在。您需要将要使用的值存储在存在于正确范围内的变量中。

Answer 7

它仍适用于编辑过的代码。我将代码的最后几行更改为：

print "before =", data[x][y]
print "element =", element
data[x][y] = element 
print "after =", data[x][y]

那些打印出来：

before = 0
element = 9
after = 9

如前所述，for循环中元素的最后一个值将是for循环完成后的值。这就是你在这里得到9的原因。

Answer 8

Python中的

for循环不会创建新范围;用于在循环中保存当前值的名称将在循环完成后保留，并将保留循环运行时的最后一个值。

Python：替换列表列表中的项目

8 个答案: