Question

这个问题是this的后续问题。我如何布置注释，以便在标记点完全或几乎重合时它们仍然可读？我需要一个程序化的解决方案，手动调整偏移量不是一个选项。带丑陋标签的样本：

import numpy as np
import matplotlib.pyplot as plt

np.random.seed(0)
N = 10
data = np.random.random((N, 4))
data[1, :2] = data[0, :2]
data[-1, :2] = data[-2, :2] + .01
labels = ['point{0}'.format(i) for i in range(N)]
plt.subplots_adjust(bottom = 0.1)
plt.scatter(
    data[:, 0], data[:, 1], marker = 'o', c = data[:, 2], s = data[:, 3]*1500,
    cmap = plt.get_cmap('Spectral'))
for label, x, y in zip(labels, data[:, 0], data[:, 1]):
    plt.annotate(
        label,
        xy = (x, y), xytext = (-20, 20),
        textcoords = 'offset points', ha = 'right', va = 'bottom',
        bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
        arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))

plt.show()

Answer 1

取最后一个点之间的距离，设置阈值保持，然后相应地翻转x，y文本。见下文。

import numpy as np 
import matplotlib.pyplot as plt

np.random.seed(0)
N = 10
data = np.random.random((N, 4))
data[1, :2] = data[0, :2]
data[-1, :2] = data[-2, :2] + .01
labels = ['point{0}'.format(i) for i in range(N)]
plt.subplots_adjust(bottom = 0.1)
plt.scatter(
    data[:, 0], data[:, 1], marker = 'o', c = data[:, 2], s = data[:, 3]*1500,
    cmap = plt.get_cmap('Spectral'))

old_x = old_y = 1e9 # make an impossibly large initial offset
thresh = .1 #make a distance threshold

for label, x, y in zip(labels, data[:, 0], data[:, 1]):
    #calculate distance
    d = ((x-old_x)**2+(y-old_y)**2)**(.5)

    #if distance less than thresh then flip the arrow
    flip = 1
    if d < .1: flip=-2

    plt.annotate(
        label,
        xy = (x, y), xytext = (-20*flip, 20*flip),
        textcoords = 'offset points', ha = 'right', va = 'bottom',
        bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
        arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))
    old_x = x
    old_y = y

plt.show()

导致：

Answer 2

这是我最终的结果。对于所有情况都不是完美的，对于这个示例问题，它甚至不能顺利运行，但我认为它足以满足我的需求。感谢Dan的回答，指出了我正确的方向。

import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial import cKDTree


def get_label_xy(tree, thresh, data, i):
    neighbors = tree.query_ball_point([data[i, 0], data[i, 1]], thresh)
    if len(neighbors) == 1:
        xy = (-30, 30)
    else:
        mean = np.mean(data[:, :2][neighbors], axis=0)

        if mean[0] == data[i, 0] and mean[1] == data[i, 1]:
            if i < np.max(neighbors):
                xy = (-30, 30)
            else:
                xy = (30, -30)
        else:
            angle = np.arctan2(data[i, 1] - mean[1], data[i, 0] - mean[0])

            if angle > np.pi / 2:
                xy = (-30, 30)
            elif angle > 0:
                xy = (30, 30)
            elif angle > -np.pi / 2:
                xy = (30, -30)
            else:
                xy = (-30, -30)
    return xy


def labeled_scatter_plot(data, labels):
    plt.subplots_adjust(bottom = 0.1)
    plt.scatter(
        data[:, 0], data[:, 1], marker = 'o', c = data[:, 2], s = data[:, 3]*1500,
        cmap = plt.get_cmap('Spectral'))

    tree = cKDTree(data[:, :2])
    thresh = .1

    for i in range(data.shape[0]):
        xy = get_label_xy(tree, thresh, data, i)

        plt.annotate(
            labels[i],
            xy = data[i, :2], xytext = xy,
            textcoords = 'offset points', ha = 'center', va = 'center',
            bbox = dict(boxstyle = 'round,pad=0.5', fc = 'yellow', alpha = 0.5),
            arrowprops = dict(arrowstyle = '->', connectionstyle = 'arc3,rad=0'))


np.random.seed(0)
N = 10
data = np.random.random((N, 4))
data[1, :2] = data[0, :2]
data[-1, :2] = data[-2, :2] + .01
data[5, :2] = data[4, :2] + [.05, 0]
data[6, :2] = data[4, :2] + [.05, .05]
data[7, :2] = data[4, :2] + [0, .05]
labels = ['point{0}'.format(i) for i in range(N)]

labeled_scatter_plot(data, labels)

plt.show()

具有重合点/重叠注释的Pyplot标签散点图

2 个答案: