Question

我声明了一个Python字典，其中的键和值是多个数组。是否可以使用键和值索引附加数组？

这是我初始化python字典cvfoldacc的方式

a = []
b = [] 
c = [] 
d = [] 
e = [] 
f = []
classifiers = [a,b,c,d,e,f]
cvfoldacc = dict.fromkeys(range(2,11), classifiers)

此初始化的结果如下：

cvfoldacc>>

{2: [[], [], [], [], [], []],
 3: [[], [], [], [], [], []],
 4: [[], [], [], [], [], []],
 5: [[], [], [], [], [], []],
 6: [[], [], [], [], [], []],
 7: [[], [], [], [], [], []],
 8: [[], [], [], [], [], []],
 9: [[], [], [], [], [], []],
 10: [[], [], [], [], [], []]}

当我尝试将key（2）的第一个列表附加到代码中时 cvfoldacc [2] [0] .append（8），我得到的结果是：

{2: [[8], [], [], [], [], []],
 3: [[8], [], [], [], [], []],
 4: [[8], [], [], [], [], []],
 5: [[8], [], [], [], [], []],
 6: [[8], [], [], [], [], []],
 7: [[8], [], [], [], [], []],
 8: [[8], [], [], [], [], []],
 9: [[8], [], [], [], [], []],
 10: [[8], [], [], [], [], []]}

但是预期的答案应该是这样：

{2: [[8], [], [], [], [], []],
 3: [[], [], [], [], [], []],
 4: [[], [], [], [], [], []],
 5: [[], [], [], [], [], []],
 6: [[], [], [], [], [], []],
 7: [[], [], [], [], [], []],
 8: [[], [], [], [], [], []],
 9: [[], [], [], [], [], []],
 10: [[], [], [], [], [], []]}

Answer 1

这是旧的deep copy与shallow copy之类的东西。试试这个：

>>> a = []
>>> b = [] 
>>> c = [] 
>>> d = [] 
>>> e = [] 
>>> f = []
>>> classifiers = [a,b,c,d,e,f]
>>> import copy
>>> cvfoldacc = {k:copy.deepcopy(classifiers) for k in range(2,11)}
>>> cvfoldacc[2][0].append(8)
>>> cvfoldacc 
{2: [[8], [], [], [], [], []],
 3: [[], [], [], [], [], []],
 4: [[], [], [], [], [], []],
 5: [[], [], [], [], [], []],
 6: [[], [], [], [], [], []],
 7: [[], [], [], [], [], []],
 8: [[], [], [], [], [], []],
 9: [[], [], [], [], [], []],
 10: [[], [], [], [], [], []]}

您正在做

a = []
b = [] 
c = [] 
d = [] 
e = [] 
f = []
classifiers = [a,b,c,d,e,f]
cvfoldacc = dict.fromkeys(range(2,11), classifiers)

现在，要创建的字典在每个键中都有相同的列表，不仅它们看起来相同，而且是完全相同的对象，具有相同的identity（在CPython中的内存位置）。让我们看看：

>>> id(cvfoldacc[2])
171760008
>>> id(cvfoldacc[3])
171760008

这些对于其他值也将相同。因此dict.fromkeys()为字典中的所有键分配相同的值，而不创建它的副本。

现在，以另一种方式，在您进行浅表复制时，此问题将得到解决：

>>> cvfoldacc = {a:classifiers.copy() for a in range(2,11)}
>>> id(cvfoldacc[2])
171840616
>>> id(cvfoldacc[3])
171847688

解决正确了吗？

>>> cvfoldacc[2][0].append(8)
{2: [[8], [], [], [], [], []],
 3: [[8], [], [], [], [], []],
 4: [[8], [], [], [], [], []],
 5: [[8], [], [], [], [], []],
 6: [[8], [], [], [], [], []],
 7: [[8], [], [], [], [], []],
 8: [[8], [], [], [], [], []],
 9: [[8], [], [], [], [], []],
 10: [[8], [], [], [], [], []]}

显然不是！现在让我们看起来更深。让我们看一下列表中列表的id：

>>> id(cvfoldacc[2][0])
171810120
>>> id(cvfoldacc[3][0])
171810120

因此，即使list.copy()创建了外部列表的副本，内部列表也是相同的。因此，基本上，您要追加list: a中存在的keys，以便对所有内容进行修改。

Deepcopy递归复制所有对象，从而避免了该问题。

Answer 2

您需要为每个list之类的值创建新的key，

>>> a = []
>>> b = [] 
>>> c = [] 
>>> d = [] 
>>> e = [] 
>>> f = []
>>> classifiers = [a,b,c,d,e,f] 
>>> 
>>> d = {k:[x[:] for x in classifiers] for k in range(2,11)} # note i am creating a copy using the `list[:]` notation.
# either the comprehension or `k: copy.deepcopy(classifiers)` is fine
>>> d
{2: [[], [], [], [], [], []], 3: [[], [], [], [], [], []], 4: [[], [], [], [], [], []], 5: [[], [], [], [], [], []], 6: [[], [], [], [], [], []], 7: [[], [], [], [], [], []], 8: [[], [], [], [], [], []], 9: [[], [], [], [], [], []], 10: [[], [], [], [], [], []]}
>>> d[2][0].append(1)
>>> d
{2: [[1], [], [], [], [], []], 3: [[], [], [], [], [], []], 4: [[], [], [], [], [], []], 5: [[], [], [], [], [], []], 6: [[], [], [], [], [], []], 7: [[], [], [], [], [], []], 8: [[], [], [], [], [], []], 9: [[], [], [], [], [], []], 10: [[], [], [], [], [], []]}

如果值是多个数组，则使用键将值附加到python字典值

2 个答案: