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,似乎没有办法将列表索引位置与列表名称一起传递。
以下是我的一些尝试和示例代码:
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\[0\]')
请注意,在attr函数的第二个参数中,您不能在此函数中连接字符串和整数。
祝好