Question

我制作了一个简单的matplotlib代码，我正在生成散点图。现在我想打开对应于点击该点时的点的图像。例如，当我点击第一点时它应该打开图像一，对于第二点它应该打开图像二。这是我的代码，

import matplotlib.image as mpimg
import numpy as np
import matplotlib.pyplot as plt

x=[1,2,3,4]
y=[1,4,9,16]

fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(x,y,'o')
coords = []

def onclick(event):
  global ix, iy
  ix, iy = event.xdata, event.ydata
  print 'x = %f, y = %f'%(ix, iy)

  global coords
  coords.append((ix, iy))
  print len(coords)
  z=len(coords)-1
  print coords[z][1]
  per = (10*coords[z][1])/100
  errp = abs(coords[z][1]+per)
  errn = abs(coords[z][1]-per)
  print "errn=%f, errp=%f"%(errn, errp)

  for i in range(len(x)):
    if abs(float(y[i])) >= errn and abs(float(y[i])) <= errp :
      print y[i]
      fig2 = plt.figure()
      img=mpimg.imread('white.png')
      line2 = plt.imshow(img)
      fig2.show()
  return coords

cid = fig.canvas.mpl_connect('button_press_event', onclick)
plt.show()

现在的问题是当我放大y值非常高的情节时（即使我点击远离该点，也会打开10 ^ 3的数量。

点击点本身而不是附近区域的某个地方时，如何获得所需图像？

[如果这是重复的问题，请给我一个原始问题的链接]

编辑：忘记添加图片white.png

Answer 1

如果你想在缩放时使用它，我会使点击必须发生的距离是轴限制的函数：

import matplotlib.image as mpimg
import numpy as np
import matplotlib.pyplot as plt

plt.close('all')

x=[1,2,3,4]
y=[1,4,9,16]

fig = plt.figure()
ax  = fig.add_subplot(111)
ax.plot(x, y, 'o')

def onclick(event):
    ix, iy = event.xdata, event.ydata
    print("I clicked at x={0:5.2f}, y={1:5.2f}".format(ix,iy))

    # Calculate, based on the axis extent, a reasonable distance 
    # from the actual point in which the click has to occur (in this case 5%)
    ax = plt.gca()
    dx = 0.05 * (ax.get_xlim()[1] - ax.get_xlim()[0])
    dy = 0.05 * (ax.get_ylim()[1] - ax.get_ylim()[0])

    # Check for every point if the click was close enough:
    for i in range(len(x)):
        if(x[i] > ix-dx and x[i] < ix+dx and y[i] > iy-dy and y[i] < iy+dy):
            print("You clicked close enough!")

cid = fig.canvas.mpl_connect('button_press_event', onclick)
plt.show()

Matplotlib散点图：打开与我点击的点对应的图像

1 个答案: