Matplotlib rectangleSelector - 设置初始位置

Question

我正在做一个项目，我想在图像上选择一个区域，以便对该区域进行一些处理。所以我找到了matplotlib的这个很棒的工具RectangleSelector。但我想要的是将矩形设置在图像的初始位置，而不是等待用户单击第一个区域（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，但似乎不是这个。

那么，有可能吗？

一个例子（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想要的数据 (RS.extents = (0,4,0,1))。

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()

带位传送

不幸的是，当使用 blitting 时，上述内容将不起作用。
使用 blitting 的解决方案需要涉及复制背景，然后将矩形设置为可见，但将 animated 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，第一次在图中发生点击时，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对画布进行blit。

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=True,
                                       button=[1, 3],  # don't use middle button
                                       minspanx=5, minspany=5,
                                       spancoords='pixels',
                                       interactive=True)
fig.canvas.draw()
bg = fig.canvas.copy_from_bbox(RS.ax.bbox)
RS.set_visible(True)

ext = (0,4,0,1)
RS.draw_shape(ext)
# Update displayed handles
RS._corner_handles.set_data(*RS.corners)
RS._edge_handles.set_data(*RS.edge_centers)
RS._center_handle.set_data(*RS.center)
for artist in RS.artists:
    RS.ax.draw_artist(artist)
    artist.set_animated(False)
fig.canvas.draw()

def initclick(evt):
    RS.background = bg
    RS.update()
    for artist in RS.artists:
        artist.set_animated(True)
    fig.canvas.mpl_disconnect(cid)

cid = fig.canvas.mpl_connect("button_press_event",initclick)

plt.show()

Answer 2

对于 matplotlib v1.21.1 和 py3.7 这有效（RS useblit=True 和 False）：

RS.to_draw.set_visible(True)
RS.extents = (x1,x2,y1,y2)

注意矩形的坐标顺序：x1、x2、y1、y2（不是x1、y1、x2、y2！）