Matplotlib中极值轮廓图上的插值差异 [英] Interpolation differences on polar contour plots in Matplotlib

问题描述

我试图在极坐标图上生成等高线图，并在matlab中进行了一些快速脚本编写以获得一些结果.出于好奇，我还想使用matplotlib在python中尝试相同的操作，但是不知何故，我看到了相同输入数据的不同轮廓图集.我试图弄清楚发生了什么，是否有什么可以在我的python代码中进行调整的方式，以在两种情况下均获得相似的结果.

I am trying to generate contour plots on a polar plot and did some quick scripting in matlab to get some results. Out of curiosity I also wanted to try out the same thing in python using the matplotlib but somehow I am seeing different sets of contour plots for the same input data. I am trying to figure out whats going on and if there is anything I could tweak in my python code to get similar results in both cases.

此处是matlab结果的屏幕截图:

A screenshot of the matlab results is here:

在Matlab代码中，我使用了scatteredinterpolant函数来获取插值数据，我假设由于所使用的插值函数而出现了差异吗?

In the matlab code I used the scatteredinterpolant function to get the interpolated data, I am assuming the differences are occurring due to the interpolation function used?

输入数据为-

Angles = [-180, -90, 0 , 90, 180, -135, -45,45, 135, 180,-90, 0, 90, 180 ]

Radii = [0,0.33,0.33,0.33,0.33,0.5,0.5,0.5,0.5,0.5,0.6,0.6,0.6,0.6]

Values = [30.42,24.75, 32.23, 34.26, 26.31, 20.58, 23.38, 34.15,27.21, 22.609, 16.013, 22.75, 27.062, 18.27]

这是在spyder上使用python 2.7完成的.我已经尝试了scipy.interpolate.griddata和matplotlib.mlab.griddata，结果是相似的.我无法在mlab.griddata中使用nn方法，因为它一直为我提供被屏蔽的数据.

This was done using python 2.7, on spyder. I have tried both scipy.interpolate.griddata as well as matplotlib.mlab.griddata and the results are similar. I was unable to get the nn method working in mlab.griddata because it kept giving me masked data.

很抱歉，如果我错过任何相关内容-如果需要其他信息，请告诉我，我将更新我的帖子.

Apologies if I am missing anything relevant - please let me know if anyother info is required I will update my post.

线性Scipt网格数据图像如下所示:

The linear scipt griddata image looks like:

立方立体图像看起来像

And the cubic scipy image looks like

关于代码，这是代码-我将插值类型字符串传递到存在该代码的函数中.因此，线性"和三次"是2个输入.

As for the code, here is the code - I pass the interpolation type string into the function where this code is present. So 'linear' and 'cubic' are the 2 inputs.

val = np.array(list(values[i]))
radius = np.array(list(gamma[i]))    
ang = [math.radians(np.array(list(theta[i]))[x]) for x in xrange(0,len(theta[i]))]
radiiGrid = np.linspace(min(radius),max(radius),100)
anglesGrid = np.linspace(min(ang),max(ang),100)
radiiGrid, anglesGrid = np.meshgrid(radiiGrid, anglesGrid)
zgrid = griddata((ang,radius),val,(anglesGrid,radiiGrid), method=interpType)

角度输入是np.array(list(theta[i]))[x]的结果-这是因为角度信息存储在元组列表中(这是因为我正在读取数据并对其进行排序).我看了一下代码，以确保数据正确并且似乎对齐. γ对应于半径，值是我提供的样本数据中的值. 希望这可以帮助！

The angle input is what comes out of np.array(list(theta[i]))[x] - this is because the angle information is stored in a list of tuples (this is because I am reading in and sorting data). I took a look at the code to make sure the data is correct and it seems to line up. gamma corresponds to radii and values are the values in the sample data I provided. Hope this helps!

推荐答案

matplotlib中的极坐标图可能会比较棘手.发生这种情况时，一种快速的解决方案是将半径和角度转换为x，y，并以法线投影进行绘制.然后使一个空的极轴叠加在其上:

Polar plots in matplotlib can get tricky. When that happens, a quick solution is to convert radii and angle to x,y, plot in a normal projection. Then make a empty polar axis to superimpose on it:

from scipy.interpolate import griddata


Angles = [-180, -90, 0 , 90, 180, -135, 
          -45,45, 135, 180,-90, 0, 90, 180 ]

Radii = [0,0.33,0.33,0.33,0.33,0.5,0.5,
         0.5,0.5,0.5,0.6,0.6,0.6,0.6]

Angles = np.array(Angles)/180.*np.pi
x = np.array(Radii)*np.sin(Angles)
y = np.array(Radii)*np.cos(Angles)

Values = [30.42,24.75, 32.23, 34.26, 26.31, 20.58, 
          23.38, 34.15,27.21, 22.609, 16.013, 22.75, 27.062, 18.27]

Xi = np.linspace(-1,1,100)
Yi = np.linspace(-1,1,100)

#make the axes
f = plt.figure()
left, bottom, width, height= [0,0, 1, 0.7]
ax  = plt.axes([left, bottom, width, height])
pax = plt.axes([left, bottom, width, height], 
                projection='polar',
                axisbg='none')
cax = plt.axes([0.8, 0, 0.05, 1])
ax.set_aspect(1)
ax.axis('Off')


# grid the data.
Vi = griddata((x, y), Values, (Xi[None,:], Yi[:,None]), method='cubic')
cf = ax.contour(Xi,Yi,Vi, 15, cmap=plt.cm.jet)

#make a custom colorbar, because the default is ugly
gradient = np.linspace(1, 0, 256)
gradient = np.vstack((gradient, gradient))
cax.xaxis.set_major_locator(plt.NullLocator())
cax.yaxis.tick_right()
cax.imshow(gradient.T, aspect='auto', cmap=plt.cm.jet)
cax.set_yticks(np.linspace(0,256,len(cf1.get_array())))
cax.set_yticklabels(map(str, cf.get_array())[::-1])

这篇关于Matplotlib中极值轮廓图上的插值差异的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！