我有一个极(r,theta)网格(这意味着每个单元格是一个环形区域)包含一些物理量(例如温度)的值,我想重新网格化(或重新投影,或者将这些值重新采样到笛卡尔网格上。有没有可以做到这一点的Python软件包?
我对将细胞中心的坐标从极地转换为笛卡尔并不感兴趣 - 这很容易。相反,我正在寻找一个可以实际重新网格化数据的软件包。
感谢您的任何建议!
答案 0 :(得分:8)
感谢您的回答 - 在仔细考虑了这一点之后,我想出了以下代码:
import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as mpl
from scipy.interpolate import interp1d
from scipy.ndimage import map_coordinates
def polar2cartesian(r, t, grid, x, y, order=3):
X, Y = np.meshgrid(x, y)
new_r = np.sqrt(X*X+Y*Y)
new_t = np.arctan2(X, Y)
ir = interp1d(r, np.arange(len(r)), bounds_error=False)
it = interp1d(t, np.arange(len(t)))
new_ir = ir(new_r.ravel())
new_it = it(new_t.ravel())
new_ir[new_r.ravel() > r.max()] = len(r)-1
new_ir[new_r.ravel() < r.min()] = 0
return map_coordinates(grid, np.array([new_ir, new_it]),
order=order).reshape(new_r.shape)
# Define original polar grid
nr = 10
nt = 10
r = np.linspace(1, 100, nr)
t = np.linspace(0., np.pi, nt)
z = np.random.random((nr, nt))
# Define new cartesian grid
nx = 100
ny = 200
x = np.linspace(0., 100., nx)
y = np.linspace(-100., 100., ny)
# Interpolate polar grid to cartesian grid (nearest neighbor)
fig = mpl.figure()
ax = fig.add_subplot(111)
ax.imshow(polar2cartesian(r, t, z, x, y, order=0), interpolation='nearest')
fig.savefig('test1.png')
# Interpolate polar grid to cartesian grid (cubic spline)
fig = mpl.figure()
ax = fig.add_subplot(111)
ax.imshow(polar2cartesian(r, t, z, x, y, order=3), interpolation='nearest')
fig.savefig('test2.png')
这不是严格的重新网格化,但可以满足我的需求。只需发布代码,以防其他人有用。随意提出改进建议!
答案 1 :(得分:3)
您可以使用scipy.ndimage.geometric_transform
更紧凑地执行此操作。以下是一些示例代码:
import numpy as N
import scipy as S
import scipy.ndimage
temperature = <whatever>
# This is the data in your polar grid.
# The 0th and 1st axes correspond to r and θ, respectively.
# For the sake of simplicity, θ goes from 0 to 2π,
# and r's units are just its indices.
def polar2cartesian(outcoords, inputshape, origin):
"""Coordinate transform for converting a polar array to Cartesian coordinates.
inputshape is a tuple containing the shape of the polar array. origin is a
tuple containing the x and y indices of where the origin should be in the
output array."""
xindex, yindex = outcoords
x0, y0 = origin
x = xindex - x0
y = yindex - y0
r = N.sqrt(x**2 + y**2)
theta = N.arctan2(y, x)
theta_index = N.round((theta + N.pi) * inputshape[1] / (2 * N.pi))
return (r,theta_index)
temperature_cartesian = S.ndimage.geometric_transform(temperature, polar2cartesian,
order=0,
output_shape = (temperature.shape[0] * 2, temperature.shape[0] * 2),
extra_keywords = {'inputshape':temperature.shape,
'center':(temperature.shape[0], temperature.shape[0])})
您可以根据需要更改order=0
以获得更好的插值效果。输出数组temperature_cartesian
在这里是2r乘2r,但你可以指定你喜欢的任何大小和原点。
答案 2 :(得分:2)
我在前一段时间试图做类似的事情,这就是将极地数据重新投入笛卡尔网格,反之亦然。这里提出的解决方案很好。但是,执行坐标转换需要一些时间。我只是想分享另一种方法,可以将处理时间缩短50倍或更多。
该算法使用scipy.ndimage.interpolation.map_coordinates
函数。
让我们看一个小例子:
import numpy as np
# Auxiliary function to map polar data to a cartesian plane
def polar_to_cart(polar_data, theta_step, range_step, x, y, order=3):
from scipy.ndimage.interpolation import map_coordinates as mp
# "x" and "y" are numpy arrays with the desired cartesian coordinates
# we make a meshgrid with them
X, Y = np.meshgrid(x, y)
# Now that we have the X and Y coordinates of each point in the output plane
# we can calculate their corresponding theta and range
Tc = np.degrees(np.arctan2(Y, X)).ravel()
Rc = (np.sqrt(X**2 + Y**2)).ravel()
# Negative angles are corrected
Tc[Tc < 0] = 360 + Tc[Tc < 0]
# Using the known theta and range steps, the coordinates are mapped to
# those of the data grid
Tc = Tc / theta_step
Rc = Rc / range_step
# An array of polar coordinates is created stacking the previous arrays
coords = np.vstack((Ac, Sc))
# To avoid holes in the 360º - 0º boundary, the last column of the data
# copied in the begining
polar_data = np.vstack((polar_data, polar_data[-1,:]))
# The data is mapped to the new coordinates
# Values outside range are substituted with nans
cart_data = mp(polar_data, coords, order=order, mode='constant', cval=np.nan)
# The data is reshaped and returned
return(cart_data.reshape(len(y), len(x)).T)
polar_data = ... # Here a 2D array of data is assumed, with shape thetas x ranges
# We create the x and y axes of the output cartesian data
x = y = np.arange(-100000, 100000, 1000)
# We call the mapping function assuming 1 degree of theta step and 500 meters of
# range step. The default order of 3 is used.
cart_data = polar_to_cart(polar_data, 1, 500, x, y)
我希望这可以帮助处理与我相同情况的人。
答案 3 :(得分:1)
是否有可以执行此操作的Python程序包?
是的!现在 - 至少有一个 - Python包具有将矩阵从笛卡尔坐标重新映射到极坐标的功能:abel.tools.polar.reproject_image_into_polar()
,它是PyAbel package的一部分。
(IñigoHernáezCorres是正确的,scipy.ndimage.interpolation.map_coordinates
是我们迄今为止从笛卡尔坐标到极坐标重新投影的最快方式。)
可以通过在命令行输入以下内容从PyPi安装PyAbel:
pip install pyabel
然后,在python中,您可以使用以下代码将图像重新投影到极坐标中:
import abel
abel.tools.polar.reproject_image_into_polar(MyImage)
[根据应用程序的不同,您可以考虑传递jacobian=True
参数,该参数会重新调整矩阵的强度,以便考虑到网格的拉伸(更改“bin size”)当你从笛卡儿变为极坐标时。]
这是一个完整的例子:
import numpy as np
import matplotlib.pyplot as plt
import abel
CartImage = abel.tools.analytical.sample_image(501)[201:-200, 201:-200]
PolarImage, r_grid, theta_grid = abel.tools.polar.reproject_image_into_polar(CartImage)
fig, axs = plt.subplots(1,2, figsize=(7,3.5))
axs[0].imshow(CartImage , aspect='auto', origin='lower')
axs[1].imshow(PolarImage, aspect='auto', origin='lower',
extent=(np.min(theta_grid), np.max(theta_grid), np.min(r_grid), np.max(r_grid)))
axs[0].set_title('Cartesian')
axs[0].set_xlabel('x')
axs[0].set_ylabel('y')
axs[1].set_title('Polar')
axs[1].set_xlabel('Theta')
axs[1].set_ylabel('r')
plt.tight_layout()
plt.show()
注意:在SO上还有另一个很好的讨论(关于将彩色图像重新映射到极坐标):image information along a polar coordinate system
答案 4 :(得分:0)
OpenCV 3.4现在可以通过warpPolar()轻松地做到这一点
非常简单的呼叫方式:
import numpy as np
import cv2
from matplotlib import pyplot as plt
# Read in our image from disk
image = cv2.imread('washington_quarter.png',0)
plt.imshow(image),plt.show()
margin = 0.9 # Cut off the outer 10% of the image
# Do the polar rotation along 1024 angular steps with a radius of 256 pixels.
polar_img = cv2.warpPolar(image, (256, 1024), (image.shape[0]/2,image.shape[1]/2), image.shape[1]*margin*0.5, cv2.WARP_POLAR_LINEAR)
# Rotate it sideways to be more visually pleasing
polar_img = cv2.rotate(polar_img, cv2.ROTATE_90_COUNTERCLOCKWISE)
plt.imshow(polar_img),plt.show()