问题描述
我有一个.txt文件,其中包含2D地图中规则间隔的点的x,y值,第三个坐标是该点的密度.
I have a .txt file containing the x,y values of regularly spaced points in a 2D map, the 3rd coordinate being the density at that point.
4.882812500000000E-004 4.882812500000000E-004 0.9072267
1.464843750000000E-003 4.882812500000000E-004 1.405174
2.441406250000000E-003 4.882812500000000E-004 24.32851
3.417968750000000E-003 4.882812500000000E-004 101.4136
4.394531250000000E-003 4.882812500000000E-004 199.1388
5.371093750000000E-003 4.882812500000000E-004 1278.898
6.347656250000000E-003 4.882812500000000E-004 1636.955
7.324218750000000E-003 4.882812500000000E-004 1504.590
8.300781250000000E-003 4.882812500000000E-004 814.6337
9.277343750000000E-003 4.882812500000000E-004 273.8610
当我使用以下命令在gnuplot中绘制此密度图时:
When I plot this density map in gnuplot, with the following commands:
set palette rgbformulae 34,35,0
set size square
set pm3d map
splot "dens_map.map" u 1:2:(log10($3+10.)) title "Density map"`
哪个给我这张美丽的照片:
Which gives me this beautiful image:
现在,我希望使用matplotlib获得相同的结果.
Now I would like to have the same result with matplotlib.
推荐答案
我的目标是提供更完整的答案,包括选择颜色图和颜色轴的对数归一化.
Here is my aim at a more complete answer including choosing the color map and a logarithmic normalization of the color axis.
import matplotlib.pyplot as plt
import matplotlib.cm as cm
from matplotlib.colors import LogNorm
import numpy as np
x, y, z = np.loadtxt('data.txt', unpack=True)
N = int(len(z)**.5)
z = z.reshape(N, N)
plt.imshow(z+10, extent=(np.amin(x), np.amax(x), np.amin(y), np.amax(y)),
cmap=cm.hot, norm=LogNorm())
plt.colorbar()
plt.show()
我在这里假设您的数据可以通过简单的整形转换为2d数组.如果不是这种情况,那么您就需要更加努力地以这种形式获取数据.如果您的数据位于网格上,那么使用imshow而不是pcolormesh会更有效(看起来确实如此).上面的代码片段生成了以下图像,非常接近您想要的图像:
I assume here that your data can be transformed into a 2d array by a simple reshape. If this is not the case than you need to work a bit harder on getting the data in this form. Using imshow and not pcolormesh is more efficient here if you data lies on a grid (as it seems to do). The above code snippet results in the following image, that comes pretty close to what you wanted:
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