在Matplotlib中的内联标签 [英] Inline labels in Matplotlib

查看:145
本文介绍了在Matplotlib中的内联标签的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在Matplotlib中,做一个传奇并不难( example_legend(),下面),但我认为更好的风格, (如 example_inline(),如下)。这可以非常fiddly,因为我必须手动指定坐标,如果我重新格式化的情节,我可能要重新定位标签。有没有办法在Matplotlib的曲线上自动生成标签?

In Matplotlib, it's not too tough to make a legend (example_legend(), below), but I think it's better style to put labels right on the curves being plotted (as in example_inline(), below). This can be very fiddly, because I have to specify coordinates by hand, and, if I re-format the plot, I probably have to reposition the labels. Is there a way to automatically generate labels on curves in Matplotlib? Bonus points for being able to orient the text at an angle corresponding to the angle of the curve.

import numpy as np
import matplotlib.pyplot as plt

def example_legend():
    plt.clf()
    x = np.linspace(0, 1, 101)
    y1 = np.sin(x * np.pi / 2)
    y2 = np.cos(x * np.pi / 2)
    plt.plot(x, y1, label='sin')
    plt.plot(x, y2, label='cos')
    plt.legend()

def example_inline():
    plt.clf()
    x = np.linspace(0, 1, 101)
    y1 = np.sin(x * np.pi / 2)
    y2 = np.cos(x * np.pi / 2)
    plt.plot(x, y1, label='sin')
    plt.plot(x, y2, label='cos')
    plt.text(0.08, 0.2, 'sin')
    plt.text(0.9, 0.2, 'cos')

推荐答案

很好的问题,前一段时间我已经尝试了一下,但没有使用它很多,因为它仍然不防弹。我将绘图区域划分为32x32网格,并根据以下规则为每一行计算标签的最佳位置的潜在字段:

Nice question, a while ago I've experimented a bit with this, but haven't used it a lot because it's still not bulletproof. I divided the plot area into a 32x32 grid and calculated a 'potential field' for the best position of a label for each line according the following rules:


  • 空白是标签的好地方

  • 标签应该靠近相应的行

  • 标签应该远离其他行

  • white space is a good place for a label
  • Label should be near corresponding line
  • Label should be away from the other lines

代码如下:

import matplotlib.pyplot as plt
import numpy as np
from scipy import ndimage


def my_legend(axis = None):

    if axis == None:
        axis = plt.gca()

    N = 32
    Nlines = len(axis.lines)
    print Nlines

    xmin, xmax = axis.get_xlim()
    ymin, ymax = axis.get_ylim()

    # the 'point of presence' matrix
    pop = np.zeros((Nlines, N, N), dtype=np.float)    

    for l in range(Nlines):
        # get xy data and scale it to the NxN squares
        xy = axis.lines[l].get_xydata()
        xy = (xy - [xmin,ymin]) / ([xmax-xmin, ymax-ymin]) * N
        xy = xy.astype(np.int32)
        # mask stuff outside plot        
        mask = (xy[:,0] >= 0) & (xy[:,0] < N) & (xy[:,1] >= 0) & (xy[:,1] < N)
        xy = xy[mask]
        # add to pop
        for p in xy:
            pop[l][tuple(p)] = 1.0

    # find whitespace, nice place for labels
    ws = 1.0 - (np.sum(pop, axis=0) > 0) * 1.0 
    # don't use the borders
    ws[:,0]   = 0
    ws[:,N-1] = 0
    ws[0,:]   = 0  
    ws[N-1,:] = 0  

    # blur the pop's
    for l in range(Nlines):
        pop[l] = ndimage.gaussian_filter(pop[l], sigma=N/5)

    for l in range(Nlines):
        # positive weights for current line, negative weight for others....
        w = -0.3 * np.ones(Nlines, dtype=np.float)
        w[l] = 0.5

        # calculate a field         
        p = ws + np.sum(w[:, np.newaxis, np.newaxis] * pop, axis=0)
        plt.figure()
        plt.imshow(p, interpolation='nearest')
        plt.title(axis.lines[l].get_label())

        pos = np.argmax(p)  # note, argmax flattens the array first 
        best_x, best_y =  (pos / N, pos % N) 
        x = xmin + (xmax-xmin) * best_x / N       
        y = ymin + (ymax-ymin) * best_y / N       


        axis.text(x, y, axis.lines[l].get_label(), 
                  horizontalalignment='center',
                  verticalalignment='center')


plt.close('all')

x = np.linspace(0, 1, 101)
y1 = np.sin(x * np.pi / 2)
y2 = np.cos(x * np.pi / 2)
y3 = x * x
plt.plot(x, y1, 'b', label='blue')
plt.plot(x, y2, 'r', label='red')
plt.plot(x, y3, 'g', label='green')
my_legend()
plt.show()

结果图:

And the resulting plot:

这篇关于在Matplotlib中的内联标签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!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆