java - 在java opencv中使用dilate时更改大小

Question

我在java中使用opencv。我面临这个奇怪的问题，在我执行 dilate 功能后，我的图像改变了它的大小。在 documentation明确表示:

dst - output image of the same size and type as src.

我的代码是:

Log.d(TAG, "dilate size1  " + dilate.size().height + " " + dilate.size().width);
Imgproc.GaussianBlur(warpg, smooth, new Size(3, 3), 3);
Log.d(TAG, "smooth size1  " + smooth.size().height + " " + smooth.size().width);
Imgproc.adaptiveThreshold(smooth, thresh, 255, 0, 1, 5, 2);
Log.d(TAG, "thresh size1  " + thresh.size().height + " " + thresh.size().width);
kernel = Imgproc.getStructuringElement(Imgproc.MORPH_CROSS, new Size(3, 3));
Imgproc.erode(thresh, erode, kernel, new Point(-1, -1), 1);
Log.d(TAG, "erode size1  " + erode.size().height + " " + erode.size().width);
Imgproc.dilate(erode, dilate, kernel, new Point(-1, -1), 1);
Imgproc.findContours(thresh, contours, hierarchy,
        Imgproc.RETR_TREE, Imgproc.CHAIN_APPROX_SIMPLE);
Log.d(TAG, "dilate size2  " + dilate.size().height + " " + dilate.size().width);

输出为:

dilate size1  450.0 450.0
smooth size1  450.0 450.0
thresh size1  450.0 450.0
erode size1  450.0 450.0
dilate size2  1.0 313.0

有什么想法吗？

Answer 1

我解决了做:

dilate = new Mat(erode.rows(), erode.cols(), erode.type());
erode.copyTo(dilate);
Imgproc.dilate(dilate, dilate, kernel, new Point(-1, -1), 1);

干杯。