我有一个具有这种格式的目标值numpy数组，可以说10个维度（24,10）
|-- app
|-- [+] configs
|
|
|-- core
|-- [+] authentication
|-- [+] guards
|-- [+] http
|-- [+] interceptors
|-- [+] layout
|-- [+] mocks
|-- [+] services
|-- [+] strategies
|-- core.module.ts
|-- router.animations.ts
|-- template-core.module.ts
|-- theme.module.ts
|-- ensureModuleLoadedOnceGuard.ts
|-- logger.service.ts
|
|
|-- modules
|-- client
|-- [+] components
|-- client-routing.module.ts
|-- client.module.ts
|--- [+] other modules
|
|-- shared
|-- [+] components
|-- [+] directives
|-- [+] pipes
|-- [+] models
|-- [+] module
|
|-- app-routing.module.ts
|-- app.module.ts
|-- etc ...
|
|-- assets
|-- images
|-- icons
|-- css
|-- styles.scss
所以我应用了这张地图
Target = [[ 2, 0, 2, 0, 0, 3, 0, 0, 1, 0, 0, -2, 4, -2, 0, 0,
-3, -3, -5, 1, 0, 0, 0, 2],...]
目标值现在是（10,24,5）：
checker_presence = {
5: np.array([1,1,1,1,1], dtype=int),
4: np.array([1,1,1,1,0], dtype=int),
3: np.array([1,1,1,0,0], dtype=int),
2: np.array([1,1,0,0,0], dtype=int),
1: np.array([1,0,0,0,0], dtype=int),
0: np.array([0,0,0,0,0], dtype=int),
-1: np.array([-1,0,0,0,0], dtype=int),
-2: np.array([-1,-1,0,0,0], dtype=int),
-3: np.array([-1,-1,-1,0,0], dtype=int),
-4: np.array([-1,-1,-1,-1,0], dtype=int),
-5: np.array([-1,-1,-1,-1,-1], dtype=int)}
labels_ = np.array([list(map(checker_presence.__getitem__, row)) for row in target])
此后，我将flatten函数应用为一个数组（10,120）：
[[ 1, 1, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 1, 1, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 1, 1, 1, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 1, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[-1, -1, 0, 0, 0],
[ 1, 1, 1, 1, 0],
[-1, -1, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[-1, -1, -1, 0, 0],
[-1, -1, -1, 0, 0],
[-1, -1, -1, -1, -1],
[ 1, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0],
[ 1, 1, 0, 0, 0]],...]
我的问题是如何将这一过程恢复为我的目标值
labels = np.array([i.flatten() for i in labels_])
[ 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, -1, -1, 0, 0, 0, 1, 1, 1, 1, 0, -1, -1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, -1, -1, 0, 0,
-1, -1, -1, 0, 0, -1, -1, -1, -1, -1, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0,
0],...]
在此示例中，我仅使用一行来更好地理解。问题是我需要对神经网络进行此转换，但是要分析预测，则需要逆向过程。
答案 0 :(得分：3)
虽然flatten会折叠尺寸，但reshape可以进行任意尺寸转换。它以函数和方法形式存在。快速示例：
import numpy as np
a = np.empty((10, 24, 5))
b = a.reshape(10, 120) # or b = np.reshape(a, (10, 120))
print(b.shape) # shows (10, 120)
c = b.reshape(10, 24, 5) # or c = np.reshape(b, (10, 24, 5))
print(c.shape) # shows (10, 24, 5)
答案 1 :(得分：2)
您可以使用嵌套列表理解：
import numpy as np
labels = [ 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, -1, -1, 0, 0, 0, 1, 1, 1, 1, 0, -1, -1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, -1, -1, 0, 0,
-1, -1, -1, 0, 0, -1, -1, -1, -1, -1, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0,
0]
checker_presence = {
5: np.array([1,1,1,1,1], dtype=int),
4: np.array([1,1,1,1,0], dtype=int),
3: np.array([1,1,1,0,0], dtype=int),
2: np.array([1,1,0,0,0], dtype=int),
1: np.array([1,0,0,0,0], dtype=int),
0: np.array([0,0,0,0,0], dtype=int),
-1: np.array([-1,0,0,0,0], dtype=int),
-2: np.array([-1,-1,0,0,0], dtype=int),
-3: np.array([-1,-1,-1,0,0], dtype=int),
-4: np.array([-1,-1,-1,-1,0], dtype=int),
-5: np.array([-1,-1,-1,-1,-1], dtype=int)}
chunked = [labels[i:i + 5] for i in range(0, len(labels), 5)]
target = [key for chunk in chunked for key, value in checker_presence.items() if list(value)==chunk]
收益：
[2, 0, 2, 0, 0, 3, 0, 0, 1, 0, 0, -2, 4, -2, 0, 0, -3, -3, -5, 1, 0, 0, 0, 2]
答案 2 :(得分：1)
这是一个使用高级索引进行查找的解决方案，它应该比任何基于字典的方法都快很多。我还使用此方法添加了更快版本的正向地图。顺便说一句。如果您知道您的网络仅产生合法模式，则可以重塑形状，然后对最后一个轴求和以恢复原始表示形式。
import numpy as np
checker_presence = {
5: np.array([1,1,1,1,1], dtype=int),
4: np.array([1,1,1,1,0], dtype=int),
3: np.array([1,1,1,0,0], dtype=int),
2: np.array([1,1,0,0,0], dtype=int),
1: np.array([1,0,0,0,0], dtype=int),
0: np.array([0,0,0,0,0], dtype=int),
-1: np.array([-1,0,0,0,0], dtype=int),
-2: np.array([-1,-1,0,0,0], dtype=int),
-3: np.array([-1,-1,-1,0,0], dtype=int),
-4: np.array([-1,-1,-1,-1,0], dtype=int),
-5: np.array([-1,-1,-1,-1,-1], dtype=int)}
forward_lookup = np.empty((11, 5), int)
for k, v in checker_presence.items():
forward_lookup[k] = v
reverse_lookup = np.full((3, 3, 3, 3, 3), 999999)
for k, v in checker_presence.items():
reverse_lookup[(*v,)] = k
def forward(data, flatten=True):
mapped = forward_lookup[data]
return mapped.ravel() if flatten else mapped
def reverse(mapped, shape=(-1,), assume_all_legal=False):
if assume_all_legal:
return mapped.reshape(*shape, 5).sum(-1)
return reverse_lookup[(*np.moveaxis(mapped.reshape(*shape, 5), -1, 0),)]
# small example
A = np.random.randint(-5, 6, (1, 5))
B = forward(A)
C = reverse(B, A.shape)
D = reverse(B, A.shape, True)
print(A)
print(B)
print(C)
print(D)
# large example
A = np.random.randint(-5, 6, (200, 1000))
B = forward(A)
C = reverse(B, A.shape)
D = reverse(B, A.shape, True)
assert np.all(A==C) and np.all(A==D)