问题描述
给定以下代码:
class Gridspace:
space = "e"
class GridRow:
space1 = Gridspace()
space2 = Gridspace()
space3 = Gridspace()
space4 = Gridspace()
space5 = Gridspace()
space6 = Gridspace()
space7 = Gridspace()
space8 = Gridspace()
space9 = Gridspace()
space10 = Gridspace()
spaceList = [space1,space2,space3,space4,space5,space6,space7,space8,space9,space10]
class Grid:
gridRow1 = GridRow()
gridRow2 = GridRow()
gridRow3 = GridRow()
gridRow4 = GridRow()
gridRow5 = GridRow()
gridRow6 = GridRow()
gridRow7 = GridRow()
gridRow8 = GridRow()
gridRow9 = GridRow()
gridRow10 = GridRow()
rowList = [gridRow1,gridRow2,gridRow3,gridRow4,gridRow5,gridRow6,gridRow7,gridRow8,gridRow9,gridRow10]
grid = Grid()
grid.rowList[0].spaceList[0].space = "s"
for x in grid.rowList:
rowWord = ""
for y in x.spaceList:
rowWord = rowWord + y.space + " "
print(rowWord)
我想输出一个 e 字符的 10x10 网格,除了左上角的应该是 s 字符的字符。相反,我正在更改每个行列表的第一个元素的类变量,而不是仅更改第一行列表的第一个元素的实例变量。如何使它只将第一个字符更改为 s,而将其他 99 个字符保留为 e?
解决方法
如果您希望变量成为实例变量,则必须如此定义它们(例如,在 __init__
函数中使用 self
参数):
class GridSpace:
def __init__(self):
self.space = "e"
class GridRow:
def __init__(self):
self.spaceList = [GridSpace() for i in range(10)]
class Grid:
def __init__(self):
self.rowList = [GridRow() for i in range(10)]
PS:我使用了列表理解来简化代码。
,发生这种情况是因为以下语句是 True
grid = Grid()
print(Grid.rowList[0].spaceList[0] is Grid.rowList[1].spaceList[0]) # outputs True
在这里您可以看到第一行的 spaceList[0]
与第二行的 spaceList[0]
完全相同(它们指的是同一个对象,当一个对象发生变化时,另一个也是如此)。
正如您已经猜到的那样,您需要将这些转换为实例变量。但是您必须执行此 2 级深度,这意味着您必须将其应用于 GridSpace
和 GridRow
class GridSpace:
def __init__(self):
self.space = "e" # Now "e" is unique to every column
以及
class GridRow:
def __init__(self):
self.spaceList = [GridSpace() for _ in range(10)]
现在语句 print(Grid.rowList[0].spaceList[0] is Grid.rowList[1].spaceList[0])
确实是 False
并且您的代码运行良好
s e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e e