for k in range(0, popSiz):
for i in range(0, mn1[0]):
for j in range(0,mn1[1]):
theta1[i][j] = random.gauss(0, 1)
for i in range(0, mn2[0]):
for j in range(0,mn2[1]):
theta2[i][j] = random.gauss(0, 1)
GameLogic.globalDict["theta1-" + str(k)] = theta1
GameLogic.globalDict["theta2-" + str(k)] = theta2
theta1 = [[0 for x in range(mn1[1])] for y in range(mn1[0])] # this
theta2 = [[0 for x in range(mn2[1])] for y in range(mn2[0])] # and this
print(GameLogic.globalDict["theta1-0"]==GameLogic.globalDict["theta1-1"])
如果我评论最后两行(它显示# this
# and this
),则会使所有.globalDict["theta1-" + str(k)]
相同。如果它被评论则打印为true,如果不是则为false。我想知道为什么会这样?我不明白它的逻辑。
答案 0 :(得分:1)
theta1
和theta2
列表是对象。您的初始循环修改这些对象。然后,您可以在globalDict
。
然而,我再次修改相同的对象显然所有引用都将指向修改后的字典。您必须为每次迭代创建新对象以避免这种情况。这就是为什么添加这两行会改变行为的原因。
您的情况与更简单的情况类似:
>>> a = []
>>> d = {}
>>> d['a'] = a
>>> a.append(1)
>>> a
[1]
>>> d['a'] # modified because d is actually storing a reference to that list.
[1]
>>> a = [] # now a is a completely different object
>>> a
[]
>>> d['a'] # old reference keep original a value alive.
[1]
另请注意, nothing 与“归零”数组有关。您可以用以下内容替换最后两行:
theta1 = [el.copy() for el in theta1] # or list(el) for el in ...
theta2 = [el.copy() for el in theta2] # or el[:] for el in ...
实现相同的结果,因为下一个循环将覆盖这些值。
(唯一的区别是,这不会在最后一次迭代中将theta1
和theta2
归零......)
顺便说一下:可能最好使用numpy.random.normal
生成随机数组:
>>> numpy.random.normal(size=(5,5))
array([[ 1.16883072, -1.12038842, -0.14351093, 1.20373197, 0.79088439],
[-0.80960599, -0.56876464, 1.12962452, 0.20582962, -1.36239647],
[-1.07225523, 0.56895514, -0.07132619, 1.36478187, 0.62836829],
[ 0.69711124, -0.81957984, -1.27820196, 0.04203822, 1.68618401],
[-0.54687767, 0.34994992, -0.91724856, 0.2631614 , 0.08691433]])