matplotlib 如何使极坐标图中的Angular 顺时针方向且顶部为0°

2fjabf4q  于 2023-05-29  发布在  Angular
关注(0)|答案(5)|浏览(187)

我使用matplotlib和numpy来绘制极坐标图。下面是一些示例代码:

import numpy as N
import matplotlib.pyplot as P

angle = N.arange(0, 360, 10, dtype=float) * N.pi / 180.0
arbitrary_data = N.abs(N.sin(angle)) + 0.1 * (N.random.random_sample(size=angle.shape) - 0.5)

P.clf()
P.polar(angle, arbitrary_data)
P.show()

你会注意到0°在图上的3点钟位置,Angular 是逆时针方向的。对于我的数据可视化目的来说,在12点钟位置设置0°并使Angular 顺时针旋转会更有用。除了旋转数据和手动更改轴标签之外,还有其他方法吗?

unhi4e5o

unhi4e5o1#

更新这个问题,在Matplotlib 1.1中,现在PolarAxes中有两种方法可以设置theta方向(CW/CCW)和theta=0的位置。
查看http://matplotlib.sourceforge.net/devel/add_new_projection.html#matplotlib.projections.polar.PolarAxes
具体参见set_theta_direction()set_theta_offset()
很多人试图做指南针一样的情节,似乎。

i1icjdpr

i1icjdpr2#

要展开klimaat'sanswer,请使用以下示例:

from math import radians
import matplotlib.pyplot as plt

angle=[0.,5.,10.,15.,20.,25.,30.,35.,40.,45.,50.,55.,60.,65.,70.,75.,\
       80.,85.,90.,95.,100.,105.,110.,115.,120.,125.]

angle = [radians(a) for a in angle]

lux=[12.67,12.97,12.49,14.58,12.46,12.59,11.26,10.71,17.74,25.95,\
     15.07,7.43,6.30,6.39,7.70,9.19,11.30,13.30,14.07,15.92,14.70,\
     10.70,6.27,2.69,1.29,0.81]

plt.clf()
sp = plt.subplot(1, 1, 1, projection='polar')
sp.set_theta_zero_location('N')
sp.set_theta_direction(-1)
plt.plot(angle, lux)
plt.show()

  • matplotlib:创建缩放和转换的开发人员指南
d8tt03nd

d8tt03nd3#

我发现matplotlib允许你创建自定义投影。我创建了一个从PolarAxes继承的。

import numpy as np
import matplotlib.pyplot as plt

from matplotlib.projections import PolarAxes, register_projection
from matplotlib.transforms import Affine2D, Bbox, IdentityTransform

class NorthPolarAxes(PolarAxes):
    '''
    A variant of PolarAxes where theta starts pointing north and goes
    clockwise.
    '''
    name = 'northpolar'

    class NorthPolarTransform(PolarAxes.PolarTransform):
        def transform(self, tr):
            xy   = np.zeros(tr.shape, np.float_)
            t    = tr[:, 0:1]
            r    = tr[:, 1:2]
            x    = xy[:, 0:1]
            y    = xy[:, 1:2]
            x[:] = r * np.sin(t)
            y[:] = r * np.cos(t)
            return xy

        transform_non_affine = transform

        def inverted(self):
            return NorthPolarAxes.InvertedNorthPolarTransform()

    class InvertedNorthPolarTransform(PolarAxes.InvertedPolarTransform):
        def transform(self, xy):
            x = xy[:, 0:1]
            y = xy[:, 1:]
            r = np.sqrt(x*x + y*y)
            theta = np.arctan2(y, x)
            return np.concatenate((theta, r), 1)

        def inverted(self):
            return NorthPolarAxes.NorthPolarTransform()

        def _set_lim_and_transforms(self):
            PolarAxes._set_lim_and_transforms(self)
            self.transProjection = self.NorthPolarTransform()
            self.transData = (self.transScale + self.transProjection + (self.transProjectionAffine + self.transAxes))
            self._xaxis_transform = (self.transProjection + self.PolarAffine(IdentityTransform(), Bbox.unit()) + self.transAxes)
            self._xaxis_text1_transform = (self._theta_label1_position + self._xaxis_transform)
            self._yaxis_transform = (Affine2D().scale(np.pi * 2.0, 1.0) + self.transData)
            self._yaxis_text1_transform = (self._r_label1_position + Affine2D().scale(1.0 / 360.0, 1.0) + self._yaxis_transform)

register_projection(NorthPolarAxes)

angle = np.arange(0, 360, 10, dtype=float) * np.pi / 180.0
arbitrary_data = (np.abs(np.sin(angle)) + 0.1 * 
    (np.random.random_sample(size=angle.shape) - 0.5))

plt.clf()
plt.subplot(1, 1, 1, projection='northpolar')
plt.plot(angle, arbitrary_data)
plt.show()

7xllpg7q

7xllpg7q4#

你可以修改matplotlib/projects/polar.py。
上面写着:

def transform(self, tr):
        xy   = npy.zeros(tr.shape, npy.float_)
        t    = tr[:, 0:1]
        r    = tr[:, 1:2]
        x    = xy[:, 0:1]
        y    = xy[:, 1:2]
        x[:] = r * npy.cos(t)
        y[:] = r * npy.sin(t)
        return xy

写上:

def transform(self, tr):
        xy   = npy.zeros(tr.shape, npy.float_)
        t    = tr[:, 0:1]
        r    = tr[:, 1:2]
        x    = xy[:, 0:1]
        y    = xy[:, 1:2]
        x[:] = - r * npy.sin(t)
        y[:] = r * npy.cos(t)
        return xy

实际上我并没有尝试过,你可能需要根据自己的喜好调整x[:]和y[:]赋值。此更改将影响所有使用matplotlib polar plot的程序。

5lhxktic

5lhxktic5#

两个反转例程都应该使用转换的完整路径:

return NorthPolarAxes.InvertedNorthPolarTransform()

return NorthPolarAxes.NorthPolarTransform()

现在,自动创建的NorthPolarAxes子类(如NorthPolarAxesSubplot)可以访问变换函数。
希望这能帮上忙。

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