numpy二进制光栅图像到多边形的转换 [英] numpy binary raster image to polygon transformation

问题描述

我想将2d numpy数组转换为多边形.性能对我来说非常重要，但我想避免进行C扩展. 可以通过腐蚀制作二进制轮廓图.然后我发现

I want to transform a 2d numpy array into a polygon. Performance is very important for me, but I want to avoid making a C extension. A binary outline image can be made with erosion. Then I found this. It was too slow and didn't cope with the spikes created by erosion sometimes. A spike:

000100
000100
000100
111011

我的第一次尝试:

mat = mat.copy() != 0
mat = mat - scipy.ndimage.binary_erosion(mat)

vertices = np.argwhere(mat)
minx = vertices.min(axis=0)[0]
maxx = vertices.max(axis=0)[0]

vertices_sorted = {}
for x in xrange(minx - 1, maxx + 2):
    vertices_sorted[x] = []

for vertex in vertices:
    vertices_sorted[vertex[0]].append(vertex[1])

vertex_loop = [(minx, vertices_sorted[minx][0])]
while True:
    x, y = vertex_loop[-1]
    for column, row in ((x, y + 1), (x, y - 1), 
    (x + 1, y), (x + 1, y + 1), (x + 1, y - 1),
    (x - 1, y), (x - 1, y + 1), (x - 1, y - 1)):
        if row in vertices_sorted[column]:
            vertices_sorted[column].remove(row)
            vertex_loop.append((column, row))
            break
    else:
        vertex_loop.pop()

    if vertex_loop[-1] == vertex_loop[0]:
        break
return vertex_loop[:-1]

大多数情况下都可以运行，但是速度不够快. 我的第二个代码很少起作用，但是我还没有修复它，因为它比第一个慢许多倍:

It works most of the time, but it isn't fast enough. My second code works only rarely, but I havent fixed it, because it is multiple times slower than the first one:

mat = mat.copy() != 0
mat = mat - scipy.ndimage.binary_erosion(mat)

xs, ys = np.nonzero(mat)
ys = np.ma.array(ys)

vertex_loop = [(xs[0], ys[0])]
ys[0] = np.ma.masked
while True:
    x, y = vertex_loop[-1]
    start = np.searchsorted(xs, x-1, side="left")
    end = np.searchsorted(xs, x+1, side="right")

    for i in xrange(start, end):
        if ys[i] == y or ys[i] == y + 1 or ys[i] == y - 1:
            vertex_loop.append((xs[i], ys[i]))
            ys[i] = np.ma.masked
            break
    else:
        if np.all(ys.mask):
            break
        else:
            vertex_loop.pop()
return vertex_loop

如何进一步提高速度?

似乎numpy掩码数组非常慢.此实现几乎与第一个实现一样快:

It seems that numpy masked arrays are extremely slow. This implementation is almost exactly as fast as the first one:

#import time
#t1 = time.time()
mat = mat.copy() != 0
mat = mat - scipy.ndimage.binary_erosion(mat)

xs, ys = np.nonzero(mat)
#t2 = time.time()
minx = xs[0]
maxx = xs[-1]

# Ketju pakosti käy läpi kaikki rivit minx:n ja maxx:n välissä, sillä se ON KETJU
xlist = range(minx - 1, maxx + 2)
# starts ja ends ovat dictit jotka kertovat missä slicessä x == key
tmp = np.searchsorted(xs, xlist, side="left")
starts = dict(zip(xlist, tmp))
tmp = np.searchsorted(xs, xlist, side="right")
ends = dict(zip(xlist, tmp))

unused = np.ones(len(xs), dtype=np.bool)
#t3 = time.time()
vertex_loop = [(xs[0], ys[0])]
unused[0] = 0
count = 0
while True:
    count += 1
    x, y = vertex_loop[-1]
    for i in xrange(starts[x - 1], ends[x + 1]):
        row = ys[i]
        if unused[i] and (row == y or row == y + 1 or row == y - 1):
            vertex_loop.append((xs[i], row))
            unused[i] = 0
            break
    else:
        if abs(x - xs[0]) <= 1 and abs(y - ys[0]) <= 1:
            break
        else:
            vertex_loop.pop()
#t4 = time.time()
#print abs(t1-t2)*1000, abs(t2-t3)*1000, abs(t3-t4)*1000
return vertex_loop

我想知道是否有一种简便的方法来完成我没偶然发现的scipy.

I wonder if there is an easy way to do this with scipy that I've failed to stumble upon.

在pygame中，有一个遮罩对象可以在0.025毫秒内满足我的需要，而我的解决方案则需要35毫秒，而我在互联网上某个地方发现的find_contours可以在4-5毫秒内完成它.我将修改pygame.mask.outline的源代码以使用numpy数组并将其发布到此处.

In pygame there is a mask object that does just what I need in 0.025 ms while my solution requires 35 ms and find_contours that I found on the internet somewhere does it in 4-5 ms. I'm going to modify the source code for pygame.mask.outline to use a numpy array and post it here.

推荐答案

在这里是:一种获取二进制numpy数组轮廓的极其快速的方法.

Here it is: an extremely fast way of getting the outline of a binary numpy array.

outline.py:

outline.py:

from scipy.weave import inline, converters

_code = open("outline.c", "r").read()

def outline(data, every):
    width, height = data.shape
    return inline(_code, ['data', 'width', 'height', 'every'], type_converters=converters.blitz)

outline.c:

outline.c:

/*
Modifioitu pygame.mask.Mask.outline
Input: data, width, height, every
*/

PyObject *plist, *value;
int x, y, e, firstx, firsty, secx, secy, currx, curry, nextx, nexty, n;
int a[14], b[14];
a[0] = a[1] = a[7] = a[8] = a[9] = b[1] = b[2] = b[3] = b[9] = b[10] = b[11]= 1;
a[2] = a[6] = a[10] = b[4] = b[0] = b[12] = b[8] = 0;
a[3] = a[4] = a[5] = a[11] = a[12] = a[13] = b[5] = b[6] = b[7] = b[13] = -1;

plist = NULL;
plist = PyList_New (0);
/*if (!plist) En ymmärrä mihin tätä tarvii
    return NULL;*/

every = 1;
n = firstx = firsty = secx = x = 0;

/*if(!PyArg_ParseTuple(args, "|i", &every)) {
    return NULL;
}

 by copying to a new, larger mask, we avoid having to check if we are at
   a border pixel every time.  
bitmask_draw(m, c, 1, 1); */

e = every;

/* find the first set pixel in the mask */
for (y = 1; y < height-1; y++) {
    for (x = 1; x < width-1; x++) {
        if (data(x, y)) {
             firstx = x;
             firsty = y;
             value = Py_BuildValue("(ii)", x-1, y-1);
             PyList_Append(plist, value);
             Py_DECREF(value);
             break;
        }
    }
    if (data(x, y))
        break;
}



/* covers the mask having zero pixels or only the final pixel
Pikseleitä on ainakin kymmenen
if ((x == width-1) && (y == height-1)) {
    return plist;
}        */

/* check just the first pixel for neighbors */
for (n = 0;n < 8;n++) {
    if (data(x+a[n], y+b[n])) {
        currx = secx = x+a[n];
        curry = secy = y+b[n];
        e--;
        if (!e) {
            e = every;
            value = Py_BuildValue("(ii)", secx-1, secy-1);
            PyList_Append(plist, value);
            Py_DECREF(value);
        }
        break;
    }
}       

/* if there are no neighbors, return
Pikseleitä on ainakin kymmenen
if (!secx) {
    return plist;
}*/

/* the outline tracing loop */
for (;;) {
    /* look around the pixel, it has to have a neighbor */
    for (n = (n + 6) & 7;;n++) {
        if (data(currx+a[n], curry+b[n])) {
            nextx = currx+a[n];
            nexty = curry+b[n];
            e--;
            if (!e) {
                e = every;
                if ((curry == firsty && currx == firstx) && (secx == nextx && secy == nexty)) {
                    break;
                }
                value = Py_BuildValue("(ii)", nextx-1, nexty-1);
                PyList_Append(plist, value);
                Py_DECREF(value);
            }
            break;
        }
    }
    /* if we are back at the first pixel, and the next one will be the
       second one we visited, we are done */
    if ((curry == firsty && currx == firstx) && (secx == nextx && secy == nexty)) {
        break;
    }

    curry = nexty;
    currx = nextx;
}

return_val = plist;

这篇关于numpy二进制光栅图像到多边形的转换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！