在 Matplotlib 中选择标记大小 [英] Choosing marker size in Matplotlib

问题描述

我正在matplotlib中使用正方形标记绘制散点图，如下所示:

I am doing a scatter plot with square marker in matplotlib like this one:

.

我想实现这样的目标:

这意味着我必须调整标记大小和图形大小/比例，以便标记之间没有空白.每个索引单元也应该有一个标记(x 和 y 都是整数)所以如果 y 从 60 到 100，应该有y 方向有 40 个标记.目前我正在手动调整它.知道实现这一目标的最佳方法是什么吗?

Which means I have to adjust the marker size and the figure size/ratio in such a way that there are no white space between markers. Also there should be a marker per index unit (x and y are both integers) so if y goes from 60 to 100, there should be 40 markers in y direction. At the moment I am tuning it manually. Any idea on what is the best way to achieve this?

推荐答案

我发现了两种解决方法:

I found two ways to go about this:

第一个基于 这个答案.基本上，您可以确定相邻数据点之间的像素数，并使用它来设置标记大小. scatter 中的标记大小以面积给出.

The first is based on this answer. Basically, you determine the number of pixels between the adjacent data-points and use it to set the marker size. The marker size in scatter is given as area.

fig = plt.figure()
ax = fig.add_subplot(111, aspect='equal')

# initialize a plot to determine the distance between the data points in pixel:    
x = [1, 2, 3, 4, 2, 3, 3]
y = [0, 0, 0, 0, 1, 1, 2]
s = 0.0
points = ax.scatter(x,y,s=s,marker='s')
ax.axis([min(x)-1., max(x)+1., min(y)-1., max(y)+1.])

# retrieve the pixel information:
xy_pixels = ax.transData.transform(np.vstack([x,y]).T)
xpix, ypix = xy_pixels.T

# In matplotlib, 0,0 is the lower left corner, whereas it's usually the upper 
# right for most image software, so we'll flip the y-coords
width, height = fig.canvas.get_width_height()
ypix = height - ypix

# this assumes that your data-points are equally spaced
s1 = xpix[1]-xpix[0]

points = ax.scatter(x,y,s=s1**2.,marker='s',edgecolors='none')
ax.axis([min(x)-1., max(x)+1., min(y)-1., max(y)+1.])

fig.savefig('test.png', dpi=fig.dpi)

第一种方法的缺点是符号重叠.我无法找到方法中的缺陷.我可以手动调整 s1 到

The downside of this first approach is, that the symbols overlap. I wasn't able to find the flaw in the approach. I could manually tweak s1 to

s1 = xpix[1]-xpix[0] - 13. 

为了提供更好的结果，但我无法确定 13. 背后的逻辑.

to give better results, but I couldn't determine a logic behind the 13..

因此，第二种方法基于这个答案.在这里，在绘图上绘制了各个正方形并相应地调整了大小.在某种程度上，它是一个手动散点图(使用循环来构建图形)，因此根据数据集可能需要一段时间.

Hence, a second approach based on this answer. Here, individual squares are drawn on the plot and sized accordingly. In a way it's a manual scatter plot (a loop is used to construct the figure), so depending on the data-set it could take a while.

这种方法使用patches而不是scatter，所以一定要包含

This approach uses patchesinstead of scatter, so be sure to include

from matplotlib.patches import Rectangle  

同样，使用相同的数据点:

Again, with the same data-points:

x = [1, 2, 3, 4, 2, 3, 3]
y = [0, 0, 0, 0, 1, 1, 2]
z = ['b', 'g', 'r', 'c', 'm', 'y', 'k'] # in your case, this is data
dx = [x[1]-x[0]]*len(x)   # assuming equally spaced data-points

# you can use the colormap like this in your case:
# cmap = plt.cm.hot 

fig = plt.figure()
ax = fig.add_subplot(111, aspect='equal')
ax.axis([min(x)-1., max(x)+1., min(y)-1., max(y)+1.])

for x, y, c, h in zip(x, y, z, dx):
    ax.add_artist(Rectangle(xy=(x-h/2., y-h/2.), 
                  color=c,             # or, in your case: color=cmap(c)                  
                  width=h, height=h))  # Gives a square of area h*h

fig.savefig('test.png')

对 Rectangle 的注释:坐标是左下角，因此是 x-h/2.
这种方法给出了连接的矩形.如果我仔细观察这里的输出，它们似乎仍然重叠了一个像素-同样，我不确定这是否有帮助.

One comment on the Rectangle: The coordinates are the lower left corner, hence x-h/2.
This approach gives connected rectangles. If I looked closely at the output here, they still seemed to overlap by one pixel - again, I'm not sure this can be helped.

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