Question

尝试使用Python中的getattr和setattr函数访问/分配列表中的项目。不幸的是，似乎无法将列表索引中的位置与列表名称一起传递以下是我尝试的一些示例代码：

class Lists (object):
  def __init__(self):
    self.thelist = [0,0,0]

Ls = Lists()

# trying this only gives 't' as the second argument.  Python error results.
# Interesting that you can slice a string to in the getattr/setattr functions
# Here one could access 'thelist' with with [0:7]
print getattr(Ls, 'thelist'[0])


# tried these two as well to no avail.  
# No error message ensues but the list isn't altered. 
# Instead a new variable is created Ls.'' - printed them out to show they now exist.
setattr(Lists, 'thelist[0]', 3)
setattr(Lists, 'thelist\[0\]', 3)
print Ls.thelist
print getattr(Ls, 'thelist[0]')
print getattr(Ls, 'thelist\[0\]')

另请注意，在attr函数的第二个参数中，您无法在此函数中连接字符串和整数。

干杯

Answer 1

getattr(Ls, 'thelist')[0] = 2
getattr(Ls, 'thelist').append(3)
print getattr(Ls, 'thelist')[0]

如果您希望能够执行getattr(Ls, 'thelist[0]')之类的操作，则必须覆盖__getattr__或使用内置eval功能。

Answer 2

你可以这样做：

l = getattr(Ls, 'thelist')
l[0] = 2  # for example
l.append("bar")
l is getattr(Ls, 'thelist')  # True
# so, no need to setattr, Ls.thelist is l and will thus be changed by ops on l

getattr(Ls, 'thelist')为您提供了可以使用Ls.thelist访问的相同列表的参考。

Answer 3

正如您所发现的那样，__getattr__无法正常工作。如果您确实想要使用列表索引，请使用__getitem__和__setitem__，忘记getattr()和setattr()。像这样：

class Lists (object):

    def __init__(self):
        self.thelist = [0,0,0]

    def __getitem__(self, index):
        return self.thelist[index]

    def __setitem__(self, index, value):
        self.thelist[index] = value

    def __repr__(self):
        return repr(self.thelist)

Ls = Lists()
print Ls
print Ls[1]
Ls[2] = 9
print Ls
print Ls[2]

使用Python中的getattr / setattr访问列表项

3 个答案: