英文:
Use data matrix as a fiducial to obtain angle of rotation
问题
<img src="https://i.stack.imgur.com/ubm2b.png" width="200">
我有一堆像上面那样的图像。它们每个都包含一个数据矩阵，但不能保证它们是朝向轴的。尽管如此，我可以使用`libdmtx`相当可靠地读取这些矩阵，而不管它们的旋转如何。但是，我还需要旋转图像，以便标签朝上。我的思路是我需要获取数据矩阵的旋转角度，以便我可以使用PIL将图像旋转到正确的方向。`pylibdmtx.decode`返回矩阵包含的数据，以及一个矩形，最初我以为它是数据矩阵的边界框。为了测试这一点，我使用上面的图像运行了以下代码：
```python
from PIL import Image
from pylibdmtx.pylibdmtx import decode
def segment_qr_code(image: Image.Image):
data = decode(image)[0]
print(data.rect)
if __name__ == "__main__":
segment_qr_code(Image.open('<path to image>'))
不幸的是，这段代码返回了Rect(left=208, top=112, width=94, height=-9)。由于高度是负数，我认为这不是数据矩阵的边界框，如果是的话，我不知道如何使用它来获取旋转角度。
我的问题是，**获取数据矩阵的旋转角度的最佳方法是什么？**我最初以为我可以裁剪图像，使用边界框获得仅数据矩阵的分割图像。然后，我可以使用图像阈值化或轮廓来获取旋转角度。但是，我不确定如何获取正确的边界框，即使我这样做了，我也不知道如何使用阈值化。我也希望尽量不使用阈值化，因为它不总是准确的。数据矩阵的底部和左侧始终有一个实心边框，因此我认为可能可以将其用作基准来对齐图像，但我找不到任何能够返回数据矩阵旋转角度的库。
我对任何建议都持开放态度。提前感谢。
<details>
<summary>英文:</summary>
<img src="https://i.stack.imgur.com/ubm2b.png" width="200">
I have a bunch of images such as the one above. They each contain a data matrix, but do not guarantee that it is oriented to an axis. Nevertheless, I can read these matrices with `libdmtx` pretty reliably regardless of their rotation. However, I also need to rotate the image so that the label is oriented right-side-up. My thought process is that I need to get the angle of rotation of the data matrix so that I can rotate the image with PIL to orient it correctly. `pylibdmtx.decode` returns the data that the matrix contains, as well as a rectangle which I originally thought was the bounding box of the data matrix. To test this, I ran the following code with the image above:
```python
from PIL import Image
from pylibdmtx.pylibdmtx import decode
def segment_qr_code(image: Image.Image):
data = decode(image)[0]
print(data.rect)
if __name__ == "__main__":
segment_qr_code(Image.open('<path to image>'))
Unfortunately, this code returned Rect(left=208, top=112, width=94, height=-9). Because the height is negative, I don't think it is the bounding box to the data matrix, and if it is, I don't know how to use it to get the angle of rotation.
My question is, what is the best way to obtain the angle of rotation of the data matrix? I originally thought that I could crop the image with the bounding box to get a segmented image of just the data matrix. Then I could use image thresholding or contouring to get an angle of rotation. However, I'm not sure how to get the correct bounding box, and even if I did I don't know how to use thresholding. I would also prefer to not use thresholding because it isn't always accurate. The data matrix always has a solid border on the bottom and left sides, so I think it may be possible to use it as a fiducial to align the image, however I was unable to find any libraries that were able to return the angle of rotation of the data matrix.
I am open to any suggestions. Thanks in advance.
答案1
得分: 3
感谢@flakes提供的建议。从PR和问题中结合代码，我创建了以下解决方案：
from pylibdmtx.pylibdmtx import _region, _decoder, _image, _pixel_data, _decoded_matrix_region
from pylibdmtx.wrapper import c_ubyte_p, DmtxPackOrder, DmtxVector2, dmtxMatrix3VMultiplyBy, DmtxUndefined
from ctypes import cast, string_at
from collections import namedtuple
import numpy
_pack_order = {
8: DmtxPackOrder.DmtxPack8bppK,
16: DmtxPackOrder.DmtxPack16bppRGB,
24: DmtxPackOrder.DmtxPack24bppRGB,
32: DmtxPackOrder.DmtxPack32bppRGBX,
}
Decoded = namedtuple('Decoded', 'data rect')
def decode_with_region(image):
results = []
pixels, width, height, bpp = _pixel_data(image)
with _image(cast(pixels, c_ubyte_p), width, height, _pack_order[bpp]) as img:
with _decoder(img, 1) as decoder:
while True:
with _region(decoder, None) as region:
if not region:
break
else:
res = _decode_region(decoder, region)
if res:
open_cv_image = numpy.array(image)
# Convert RGB to BGR
open_cv_image = open_cv_image[:, :, ::-1].copy()
height, width, _ = open_cv_image.shape
topLeft = (res.rect['01']['x'], height - res.rect['01']['y'])
topRight = (res.rect['11']['x'], height - res.rect['11']['y'])
bottomRight = (res.rect['10']['x'], height - res.rect['10']['y'])
bottomLeft = (res.rect['00']['x'], height - res.rect['00']['y'])
results.append(Decoded(res.data, (topLeft, topRight, bottomRight, bottomLeft)))
return results
def _decode_region(decoder, region):
with _decoded_matrix_region(decoder, region, DmtxUndefined) as msg:
if msg:
vector00 = DmtxVector2()
vector11 = DmtxVector2(1.0, 1.0)
vector10 = DmtxVector2(1.0, 0.0)
vector01 = DmtxVector2(0.0, 1.0)
dmtxMatrix3VMultiplyBy(vector00, region.contents.fit2raw)
dmtxMatrix3VMultiplyBy(vector11, region.contents.fit2raw)
dmtxMatrix3VMultiplyBy(vector01, region.contents.fit2raw)
dmtxMatrix3VMultiplyBy(vector10, region.contents.fit2raw)
return Decoded(
string_at(msg.contents.output),
{
'00': {
'x': int((vector00.X) + 0.5),
'y': int((vector00.Y) + 0.5)
},
'01': {
'x': int((vector01.X) + 0.5),
'y': int((vector01.Y) + 0.5)
},
'10': {
'x': int((vector10.X) + 0.5),
'y': int((vector10.Y) + 0.5)
},
'11': {
'x': int((vector11.X) + 0.5),
'y': int((vector11.Y) + 0.5)
}
}
)
else:
return None
要解码图像，请使用decode_with_region()而不是pylibdmtx的decode()。它会输出一个包含坐标的字典，我可以在图像上绘制并获得以下输出：
然后，我可以使用这些坐标来获取旋转角度：
def get_data_from_matrix(image):
decoded = decode_with_region(image)[0]
topLeft, topRight = decoded.rect[2], decoded.rect[3]
rotation = -math.atan2(topLeft[1] - topRight[1], topLeft[0] - topRight[0]) * (180 / math.pi)
image = image.rotate(rotation, expand=True)
英文:
Thank you to @flakes for the suggestion. Combining code from the PR and issue, I created the following solution:
from pylibdmtx.pylibdmtx import _region, _decoder, _image, _pixel_data, _decoded_matrix_region
from pylibdmtx.wrapper import c_ubyte_p, DmtxPackOrder, DmtxVector2, dmtxMatrix3VMultiplyBy, DmtxUndefined
from ctypes import cast, string_at
from collections import namedtuple
import numpy
_pack_order = {
8: DmtxPackOrder.DmtxPack8bppK,
16: DmtxPackOrder.DmtxPack16bppRGB,
24: DmtxPackOrder.DmtxPack24bppRGB,
32: DmtxPackOrder.DmtxPack32bppRGBX,
}
Decoded = namedtuple('Decoded', 'data rect')
def decode_with_region(image):
results = []
pixels, width, height, bpp = _pixel_data(image)
with _image(cast(pixels, c_ubyte_p), width, height, _pack_order[bpp]) as img:
with _decoder(img, 1) as decoder:
while True:
with _region(decoder, None) as region:
if not region:
break
else:
res = _decode_region(decoder, region)
if res:
open_cv_image = numpy.array(image)
# Convert RGB to BGR
open_cv_image = open_cv_image[:, :, ::-1].copy()
height, width, _ = open_cv_image.shape
topLeft = (res.rect['01']['x'], height - res.rect['01']['y'])
topRight = (res.rect['11']['x'], height - res.rect['11']['y'])
bottomRight = (res.rect['10']['x'], height - res.rect['10']['y'])
bottomLeft = (res.rect['00']['x'], height - res.rect['00']['y'])
results.append(Decoded(res.data, (topLeft, topRight, bottomRight, bottomLeft)))
return results
def _decode_region(decoder, region):
with _decoded_matrix_region(decoder, region, DmtxUndefined) as msg:
if msg:
vector00 = DmtxVector2()
vector11 = DmtxVector2(1.0, 1.0)
vector10 = DmtxVector2(1.0, 0.0)
vector01 = DmtxVector2(0.0, 1.0)
dmtxMatrix3VMultiplyBy(vector00, region.contents.fit2raw)
dmtxMatrix3VMultiplyBy(vector11, region.contents.fit2raw)
dmtxMatrix3VMultiplyBy(vector01, region.contents.fit2raw)
dmtxMatrix3VMultiplyBy(vector10, region.contents.fit2raw)
return Decoded(
string_at(msg.contents.output),
{
'00': {
'x': int((vector00.X) + 0.5),
'y': int((vector00.Y) + 0.5)
},
'01': {
'x': int((vector01.X) + 0.5),
'y': int((vector01.Y) + 0.5)
},
'10': {
'x': int((vector10.X) + 0.5),
'y': int((vector10.Y) + 0.5)
},
'11': {
'x': int((vector11.X) + 0.5),
'y': int((vector11.Y) + 0.5)
}
}
)
else:
return None
To decode an image, use decode_with_region() instead of pylibdmtx's decode(). It outputs a dictionary of coordinates, which I can plot on an image and get the following output:
<img src="https://i.stack.imgur.com/xaytm.png" width="300">
I can then use these coordinates to obtain an angle of rotation:
def get_data_from_matrix(image):
decoded = decode_with_region(image)[0]
topLeft, topRight = decoded.rect[2], decoded.rect[3]
rotation = -math.atan2(topLeft[1] - topRight[1], topLeft[0] - topRight[0]) * (180 / math.pi)
image = image.rotate(rotation, expand=True)
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