Question

我正在开展一个项目，我想在图像上选择一个区域，以便在这个区域进行一些处理。所以我发现了matplotlib这个奇妙的工具RectangleSelector。但我想要的是将矩形设置在图像的初始位置，而不是等待用户点击第一个区域（例如图像顶部/左侧的10x10像素的矩形）。

起初我以为我必须使用state_modifier_keys，但它似乎不是这样。

那么，有可能吗？

一个例子（带有情节，但它是同一个问题）可以是：

from __future__ import print_function
from matplotlib.widgets import RectangleSelector
import numpy as np
import matplotlib.pyplot as plt


def line_select_callback(eclick, erelease):
    'eclick and erelease are the press and release events'
    x1, y1 = eclick.xdata, eclick.ydata
    x2, y2 = erelease.xdata, erelease.ydata
    print("(%3.2f, %3.2f) --> (%3.2f, %3.2f)" % (x1, y1, x2, y2))
    print(" The button you used were: %s %s" % (eclick.button, erelease.button))


def toggle_selector(event):
    print(' Key pressed.')
    if event.key in ['Q', 'q'] and toggle_selector.RS.active:
        print(' RectangleSelector deactivated.')
        toggle_selector.RS.set_active(False)
    if event.key in ['A', 'a'] and not toggle_selector.RS.active:
        print(' RectangleSelector activated.')
        toggle_selector.RS.set_active(True)


fig, current_ax = plt.subplots()                 # make a new plotting range
N = 100000                                       # If N is large one can see
x = np.linspace(0.0, 10.0, N)                    # improvement by use blitting!

plt.plot(x, +np.sin(.2*np.pi*x), lw=3.5, c='b', alpha=.7)  # plot something
plt.plot(x, +np.cos(.2*np.pi*x), lw=3.5, c='r', alpha=.5)
plt.plot(x, -np.sin(.2*np.pi*x), lw=3.5, c='g', alpha=.3)

print("\n      click  -->  release")

# drawtype is 'box' or 'line' or 'none'
toggle_selector.RS = RectangleSelector(current_ax, line_select_callback,
                                       drawtype='box', useblit=True,
                                       button=[1, 3],  # don't use middle button
                                       minspanx=5, minspany=5,
                                       spancoords='pixels',
                                       interactive=True)
plt.connect('key_press_event', toggle_selector)
plt.show()

Answer 1

没有blitting

如果您不想使用blitting，解决方案是将RectangleSelector的补丁设置为可见（RS.to_draw.set_visible(True)）并将其范围设置为您希望的数据（{ {1}}）。

RS.extents = (0,4,0,1)

随着blitting

不幸的是，当使用blitting时，上述情况将不起作用使用blitting的解决方案需要复制背景，然后将矩形设置为可见，但将from __future__ import print_function from matplotlib.widgets import RectangleSelector import numpy as np import matplotlib.pyplot as plt def line_select_callback(eclick, erelease): x1, y1 = eclick.xdata, eclick.ydata x2, y2 = erelease.xdata, erelease.ydata print("(%3.2f, %3.2f) --> (%3.2f, %3.2f)" % (x1, y1, x2, y2)) fig, current_ax = plt.subplots() N = 100000 x = np.linspace(0.0, 10.0, N) plt.plot(x, +np.sin(.2*np.pi*x), lw=3.5, c='b', alpha=.7) # plot something plt.plot(x, +np.cos(.2*np.pi*x), lw=3.5, c='r', alpha=.5) plt.plot(x, -np.sin(.2*np.pi*x), lw=3.5, c='g', alpha=.3) # drawtype is 'box' or 'line' or 'none' RS = RectangleSelector(current_ax, line_select_callback, drawtype='box', useblit=False, button=[1, 3], # don't use middle button minspanx=5, minspany=5, spancoords='pixels', interactive=True) RS.to_draw.set_visible(True) fig.canvas.draw() RS.extents = (0,4,0,1) plt.show()属性设置为false。然后第一次绘制画布，显示矩形。但是为了以后的使用，我们需要再次将animated属性设置为True。因此，第一次在图中发生单击，我们需要将animated属性设置为True，并使用在显示矩形之前保存的背景blit画布。

animated

Matplotlib rectangleSelector - 设置初始位置

1 个答案:

没有blitting

随着blitting