python 将极坐标重新投影到笛卡尔网格

Reprojecting polar to cartesian grid

提问by astrofrog

I have a polar (r,theta) grid (which means that each cell is an annulus section) containing values of some physical quantity (e.g. temperature), and I would like to re-grid (or re-project, or resample) these values onto a cartesian grid. Are there any Python packages that can do this?

我有一个极坐标 (r,theta) 网格（这意味着每个单元格都是一个环面部分），其中包含一些物理量（例如温度）的值，我想重新网格（或重新投影或重新采样）这些值到笛卡尔网格。是否有任何 Python 包可以做到这一点？

I am not interested in converting the coordinates of the centers of the cells from polar to cartesian - this is very easy. Instead, I'm looking for a package that can actually re-grid the data properly.

我对将单元格中心的坐标从极坐标转换为笛卡尔坐标不感兴趣 - 这很容易。相反，我正在寻找一个可以真正正确地重新排列数据的包。

Thanks for any suggestions!

感谢您的任何建议！

回答by astrofrog

Thanks for your answers - after thinking a bit more about this I came up with the following code:

感谢您的回答 - 在对此进行了更多思考后，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')

Which is not strictly re-gridding, but works fine for what I need. Just posting the code in case it is useful to anyone else. Feel free to suggest improvements!

这不是严格的重新网格化，但可以很好地满足我的需要。只需发布代码，以防对其他人有用。随时提出改进建议！

回答by ptomato

You can do this more compactly with scipy.ndimage.geometric_transform. Here is some sample code:

您可以使用scipy.ndimage.geometric_transform. 下面是一些示例代码：

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])})

You can change order=0as desired for better interpolation. The output array temperature_cartesianis 2r by 2r here, but you can specify any size and origin you like.

您可以order=0根据需要进行更改以获得更好的插值。这里的输出数组temperature_cartesian是 2r x 2r，但您可以指定任何您喜欢的大小和原点。

回答by I?igo Hernáez Corres

I came to this post some time ago when trying to do something similar, this is, reprojecting polar data into a cartesian grid and vice-versa. The solution proposed here works fine. However, it takes some time to perform the coordinate transform. I just wanted to share another approach which can reduce the processing time up to 50 times or more.

前段时间我在尝试做类似的事情时来到这篇文章，即将极地数据重新投影到笛卡尔网格中，反之亦然。这里提出的解决方案工作正常。但是，执行坐标变换需要一些时间。我只是想分享另一种可以将处理时间减少多达 50 倍或更多的方法。

The algorithm uses the scipy.ndimage.interpolation.map_coordinatesfunction.

该算法使用该scipy.ndimage.interpolation.map_coordinates函数。

Let's see a little example:

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)

I hope this helps someone in the same situation as me.

我希望这可以帮助与我处于相同情况的人。

回答by DanHickstein

Are there any Python packages that can do this?

是否有任何 Python 包可以做到这一点？

Yes! There is now – at least one – Python package that has a function to re-map a matrix from cartesian to polar coordinates: abel.tools.polar.reproject_image_into_polar(), which is part of the PyAbel package.

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 包的一部分。

(I?igo Hernáez Corres is correct, scipy.ndimage.interpolation.map_coordinatesis the fastest way that we have found so far to reproject from cartesian to polar coordinates.)

（I?igo Hernáez Corres 是正确的，scipy.ndimage.interpolation.map_coordinates是迄今为止我们发现的从笛卡尔坐标重新投影到极坐标的最快方法。）

PyAbel can be installed from PyPiby entering the following on the command line:

PyAbel 可以通过在命令行输入以下命令从PyPi安装：

pip install pyabel

Then, in python, you can use the following code to re-project an image into polar coordinates:

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)

[Depending on the application, you might consider passing the jacobian=Trueargument, which re-scales the intensities of the matrix to take into the account the stretching of the grid (changing "bin sizes") that takes place when you transform from Cartesian to polar coodinates.]

[根据应用程序，您可能会考虑传递jacobian=True参数，它重新调整矩阵的强度以考虑从笛卡尔坐标转换为极坐标时发生的网格拉伸（改变“bin 大小”） .]

Here is a complete example:

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()

Note: there is another good discussion (about re-mapping color images to polar coordinates) on SO: image information along a polar coordinate system

注意：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）关于 SO：沿着极坐标系的图像信息

回答by HanClinto

OpenCV 3.4 can do this now pretty easily with warpPolar()

OpenCV 3.4 现在可以使用warpPolar()很容易地做到这一点

Very simple to call:

调用非常简单：

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()