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python - 是否可以以字节为单位转换特征图？

问题描述

蟒蛇 3.6

团结 2019

我正在尝试找到将特征图数据传输到统一的最佳解决方案。

我想以字节为单位发送数据。但是我没有找到如何将其编码为字节，然后统一解码。

基本上是一个 4d 数组，需要根据我的理解将其转换为字节

蟒蛇片

for fmap in feature_maps:
            bytes = []
            bytes.append(fmap)
            arrays_of_features.append(bytes)

        data = np.array(arrays_of_features, dtype=float) # this is not working because of the fact is multidimensional array apparently. 
        print(fmap)
        c.sendall(data.tobytes())

统一块：byte[] bytes = new byte[4000]; int idxUsedBytes = client.Receive(bytes);

    floatsReceived = new float[idxUsedBytes / 4];
    Buffer.BlockCopy(bytes, 0, floatsReceived, 0, idxUsedBytes);
    print(floatsReceived[0]);

启示：如何让 Unity 中的 c# 与 Python 通信

特征图看起来像这样：

[[[[ 0.          0.          0.         ...  0.         12.569366
 0.        ]
[ 0.          0.          0.         ...  0.          4.421044
 0.        ]
[ 0.          0.          0.         ...  0.          0.19193476
 0.        ]
...
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]]

[[ 0.          0.          0.         ...  0.         12.910363
 0.        ]
[ 0.          0.          0.         ...  0.          3.987629
 0.        ]
[ 0.          0.          0.         ...  0.          1.6041028
 0.        ]
...
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]]

[[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
 ...
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]]

...

[[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
...
[ 0.          0.          0.         ...  0.          0.
49.52598   ]
[ 0.          0.          0.         ...  0.          0.
10.050183  ]
[ 0.          0.          0.         ...  0.          9.6911745
 0.        ]]
[[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
...
[ 0.          0.          0.         ...  0.          0.
29.483086  ]
[ 0.          0.          0.         ...  0.          0.
24.422682  ]
[ 0.          0.          2.253025   ...  0.          0.
15.935954  ]]

[[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.          0.         ...  0.          0.
 0.        ]
[ 0.          0.         18.458588   ... 15.824303    0.
 0.        ]
 ...
[ 0.          0.          0.         ... 25.163502   56.87079
42.9939    ]
[ 0.          0.         11.397255   ... 36.644962   17.04247
44.108196  ]
[ 0.          0.         33.134758   ... 30.220499    8.817273
36.6427    ]]]]

标签： pythonarraysunity3ddeep-learning

解决方案

您的问题非常不清楚，我相信您对 numpy 的工作原理感到困惑。如果是这样，让我们​​解释一些事情。来自 numpy 的数组只不过是内存中的一串字节。特别是，当为您显示这些字节时，它们由 dtype 解释。dtype 不用于存储底层数据，而仅用于显示它。因此，更改 dtype 只会更改数据对您的外观，不会更改数据本身。尺寸也是一样。数据的维度只会改变数据的显示和访问方式，python 实际上并不会移动数据或改变数据本身。例如，

import numpy as np

x = np.array([[1,2,3],[4,5,6]],dtype='int64') #48 bytes, each int takes up 8 bytes.
print(x)
x.dtype = 'int32'
print(x)
x.dtype = 'float'
print(x)
x.dtype = 'int16'
print(x)

请注意，我们可以更改 dtype 并且绝对零计算由数组完成（因为基础数据已经是一个字节数组）。同样，我们可以改变形状，也可以完成绝对零计算。

x.shape = (2,2,6)
print(x)

shape 和 dtype 与内存中存储的数据无关。希望这可以清楚地说明我们现在如何将数组作为字节处理。

x = np.array([[1,2,3],[4,5,6]],dtype='int64')
print(x)
y = x.tobytes()

# Send y somewhere. Save to a file. Etc.

z = np.frombuffer(y)
z.dtype = 'int64'
z.shape = (2,3)
print(z)

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