python类可以包含自身的实例作为数据容器可能看起来像这样吗?
class A:
def __init__(self, val):
self.a = A(val)
self.val = val
aa = A(2)
#this will cause RuntimeError: maximum recursion depth exceeded
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我的目的是使用这个类作为数据容器包含一个副本,如果它被用于减少deepcopy操作.它可以用作"撤销"链,有机会在必要时获取初始值.
这样的行动有可能吗?
我需要SQL代码来解决表组合问题,如下所述:
表旧数据:表旧
name version status lastupdate ID
A 0.1 on 6/8/2010 1
B 0.1 on 6/8/2010 2
C 0.1 on 6/8/2010 3
D 0.1 on 6/8/2010 4
E 0.1 on 6/8/2010 5
F 0.1 on 6/8/2010 6
G 0.1 on 6/8/2010 7
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表新数据:表新
name version status lastupdate ID
A 0.1 on 6/18/2010
#B entry deleted
C 0.3 on 6/18/2010 #version_updated
C1 0.1 on 6/18/2010 #new_added
D 0.1 on 6/18/2010
E 0.1 off 6/18/2010 #status_updated
F 0.1 on 6/18/2010
G 0.1 …Run Code Online (Sandbox Code Playgroud) I am using twisted reactor to non-block reading sockets input. however, I want to run another loop after reactor starting running
.....
reactor.listenTCP(12345, MyFactory())
reactor.run()
# ... blah blah socket input related code
while 1:
...
...
if something:
reactor.stop()
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Problem is after reactor.run() the while loop will not working.
May I know rather than using threading in main te = Thread(target=reactor.run, args=(False,)).start(), any other way can make reactor.run() working on non-blocking?
Thank you.
一个字典
dic = {
1: 'a',
2: 'a',
3: 'b',
4: 'a',
5: 'c',
6: 'd',
7: 'd',
8: 'a',
9: 'a'}
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我想删除重复值只保留一个K/V对,关于这些重复值的"关键"选择,可以是max或min,也可以通过随机选择其中一个重复项的键.
我不想使用ak/v交换,因为它无法控制密钥选择.
以值"a"为例
1: 'a',
2: 'a',
4: 'a',
8: 'a',
9: 'a'
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最大键为{9:'a'},最小值为{1:'a'},随机将选择其中任何一个.
并且,如果键是其他类型的哈希值,例如字符串,那么如何做这样的选择呢?
谁能和我分享一个想法?
谢谢!
在python中我定义了函数:
def foo_1(p): return p + 1
def foo_2(p): return p + 1
def foo_3(p): return p + 1
def foo_4(p): return p + 1
def foo_5(p): return p + 1
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我需要执行这些功能,因为链可能是这样的:
foo_1(foo_2(foo_3(foo_4(foo_5(1)))))
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我是否可以知道是否可以将函数推入列表然后将这些函数作为链执行,也许我可以给出执行序列?
lf = [Null,foo_1,foo_2,foo_3,foo_4,foo_5] # Null is for +1 issue here
def execu(lst, seq, raw_para):
# in some way
execu(lf,(1,2,3,4,5), 1) # = foo_1(foo_2(foo_3(foo_4(foo_5(1)))))
execu(lf,(1,2,3), 1) # = foo_1(foo_2(foo_3(1)))
execu(lf,(3,3,3), 1) # = foo_3(foo_3(foo_3(1)))
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谢谢!
RGS,
KC
python ×4
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