如何使matplotlib极坐标图中的角度在顶部顺时针旋转0°？

Question

我正在使用matplotlib和numpy制作一个极地情节.以下是一些示例代码:

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()

您会注意到0°位于绘图上的3点处,并且角度逆时针方向.对于我的数据可视化目的而言,在12点位置具有0°并且角度顺时针方向将更有用.除了旋转数据和手动更改轴标签之外,有没有办法做到这一点？

Answer 1

更新这个问题,在Matplotlib 1.1中,现在有两种方法PolarAxes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

具体来说,请参见set_theta_direction()和set_theta_offset().

似乎有很多人试图做类似罗盘的情节.

Answer 2

我发现了 - matplotlib允许您创建自定义投影.我创建了一个继承自的PolarAxes.

import numpy as N
import matplotlib.pyplot as P

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   = N.zeros(tr.shape, N.float_)
            t    = tr[:, 0:1]
            r    = tr[:, 1:2]
            x    = xy[:, 0:1]
            y    = xy[:, 1:2]
            x[:] = r * N.sin(t)
            y[:] = r * N.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 = N.sqrt(x*x + y*y)
            theta = N.arctan2(y, x)
            return N.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(N.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 = 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.subplot(1, 1, 1, projection='northpolar')
P.plot(angle, arbitrary_data)
P.show()

Answer 3

通过一个例子扩展klimaat的答案:

import math
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 = [math.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]

import matplotlib.pyplot as P
import matplotlib
P.clf()
sp = P.subplot(1, 1, 1, projection='polar')
sp.set_theta_zero_location('N')
sp.set_theta_direction(-1)
P.plot(angle, lux)
P.show()