最有效的径向剖面计算方法

Question

我需要优化图像处理应用程序的这一部分.
它基本上是由它们与中心点的距离合并的像素的总和.

def radial_profile(data, center):
    y,x = np.indices((data.shape)) # first determine radii of all pixels
    r = np.sqrt((x-center[0])**2+(y-center[1])**2)
    ind = np.argsort(r.flat) # get sorted indices
    sr = r.flat[ind] # sorted radii
    sim = data.flat[ind] # image values sorted by radii
    ri = sr.astype(np.int32) # integer part of radii (bin size = 1)
    # determining distance between changes
    deltar = ri[1:] - ri[:-1] # assume all radii represented
    rind = np.where(deltar)[0] # location of changed radius
    nr = rind[1:] - rind[:-1] # number in radius bin
    csim = np.cumsum(sim, dtype=np.float64) # cumulative sum to figure out sums for each radii bin
    tbin = csim[rind[1:]] - csim[rind[:-1]] # sum for image values in radius bins
    radialprofile = tbin/nr # the answer
    return radialprofile

img = plt.imread('crop.tif', 0)
# center, radi = find_centroid(img)
center, radi = (509, 546), 55
rad = radial_profile(img, center)

plt.plot(rad[radi:])
plt.show()

输入图片:

径向轮廓:

通过提取结果图的峰值,我可以准确地找到外环的半径,这是这里的最终目标.

编辑:为了进一步参考,我将发布我的最终解决方案.cython与接受的答案相比,我使用了大约15-20倍的速度.

import numpy as np
cimport numpy as np
cimport cython
from cython.parallel import prange
from libc.math cimport sqrt, ceil


DTYPE_IMG = np.uint8
ctypedef np.uint8_t DTYPE_IMG_t
DTYPE = np.int
ctypedef np.int_t DTYPE_t


@cython.boundscheck(False)
@cython.wraparound(False)
@cython.nonecheck(False)
cdef void cython_radial_profile(DTYPE_IMG_t [:, :] img_view, DTYPE_t [:] r_profile_view, int xs, int ys, int x0, int y0) nogil:

    cdef int x, y, r, tmp

    for x in prange(xs):
        for y in range(ys):
            r =<int>(sqrt((x - x0)**2 + (y - y0)**2))
            tmp = img_view[x, y]
            r_profile_view[r] +=  tmp 


@cython.boundscheck(False)
@cython.wraparound(False)
@cython.nonecheck(False)
def radial_profile(np.ndarray img, int centerX, int centerY):


    cdef int xs, ys, r_max
    xs, ys = img.shape[0], img.shape[1]

    cdef int topLeft, topRight, botLeft, botRight

    topLeft = <int> ceil(sqrt(centerX**2 + centerY**2))
    topRight = <int> ceil(sqrt((xs - centerX)**2 + (centerY)**2))
    botLeft = <int> ceil(sqrt(centerX**2 + (ys-centerY)**2))
    botRight = <int> ceil(sqrt((xs-centerX)**2 + (ys-centerY)**2))

    r_max = max(topLeft, topRight, botLeft, botRight)

    cdef np.ndarray[DTYPE_t, ndim=1] r_profile = np.zeros([r_max], dtype=DTYPE)
    cdef DTYPE_t [:] r_profile_view = r_profile
    cdef DTYPE_IMG_t [:, :] img_view = img

    with nogil:
        cython_radial_profile(img_view, r_profile_view, xs, ys, centerX, centerY)
    return r_profile

Answer 1

看起来你可以在这里使用numpy.bincount:

import numpy as np

def radial_profile(data, center):
    y, x = np.indices((data.shape))
    r = np.sqrt((x - center[0])**2 + (y - center[1])**2)
    r = r.astype(np.int)

    tbin = np.bincount(r.ravel(), data.ravel())
    nr = np.bincount(r.ravel())
    radialprofile = tbin / nr
    return radialprofile