Question

我想有一个分为4个扇区的窗口：在（0,0）imshow图像（ax1）; （1,0）一个子图图像，它使用twinx（）图像来划分窗口（ax2＆amp; ax3）; （1,1）常规情节图像（ax4）;和图的迭代部分（0,1）应该给出“number_of_subplots”图一个在另一个之上（ax5）。希望没有xticklabels但是最后一个。

This is how the frame should look like before the iterative subplot creation.

我的问题：在迭代创建窗口右上方空间的子图时，子图跨越该空间并消除ax4

This is how the window looks after the "for" cyle for the subplot creation

下面你会发现我正在使用的代码的简化，这样你就可以更好地看到它。我已经用随机数替换了我的实验数据，因此您可以轻松地复制它。

你可以给我一个关于我做错的提示吗？我仍然不支持python中的所有处理程序。几年前我曾在matlab做类似的事情。

    import matplotlib.pyplot as plt
    from matplotlib.gridspec import GridSpec
    import numpy as np
    import pdb

    pos = [1,2,3,4,5]
    N = 50
    x = np.random.rand(N)
    y = np.random.rand(N)

    xx = np.linspace(0, 20, 1000)

    fig1 = plt.figure()
    number_of_subplots = len(pos) #number between 1-7
    ax1 = plt.subplot2grid((number_of_subplots+1,2),(0,0),rowspan = number_of_subplots-1) # Here the idea is to "dinamically" create the division of the grid, making space at the bottom of it for the image in the bottom left.
    ax1.scatter(x,y)

    ax2 = plt.subplot2grid((number_of_subplots+1,2),(number_of_subplots-1,0), rowspan = 2)
    ax2.plot(xx,np.sin(xx),label = 'sin(x)',color = 'b')
    ax3 = ax2.twinx()
    ax3.plot(xx,np.cos(xx), label = 'cos(x)', color = 'r')

    ax4 = plt.subplot2grid((number_of_subplots+1,2),(number_of_subplots-1,1), rowspan = 2)
    ax4.plot(xx,np.tan(xx), label = 'tan(x)', color = 'g')

    for i,v in enumerate(xrange(number_of_subplots)):
            v = v+1
            ax5 = plt.subplot2grid((number_of_subplots+1,2),(v-1,1))
            ax5.plot(np.sin(xx+3.1416*v/2)) # Grafica los perfiles, asociandoles el mismo color que para los cortes en la imagen 2D
            if (i % 2 == 0): #Even
                    ax5.yaxis.tick_left()
            else:
                    ax5.yaxis.tick_right()
    plt.draw()
    plt.show()

Answer 1

使用GridSpec解决了这个问题，因为它应该被使用。下面是代码的实现，它提供了以下解决方案。

This is the correct way the image should look like and the implementation is below on the code.

    import matplotlib.pyplot as plt
    import matplotlib.gridspec as gridspec
    import numpy as np
    import pdb

    pos = [1,2,3,4,5]

    N = 50
    x = np.random.rand(N)
    y = np.random.rand(N)

    xx = np.linspace(0, 20, 1000)

    number_of_subplots = len(pos) #number between 1-7

    fig1 = plt.figure()

    gs0 = gridspec.GridSpec(2,2,height_ratios=[3,1],hspace=0.1)
    ax1 = plt.subplot(gs0[0,0])
    ax2 = plt.subplot(gs0[-1,0])
    ax4 = plt.subplot(gs0[-1,-1])

    gs2 = gridspec.GridSpecFromSubplotSpec(number_of_subplots, 1, subplot_spec=gs0[1],wspace=0.0, hspace=0.0)

    ax1.scatter(x,y)

    ax2.plot(xx,np.sin(xx),label = 'sin(x)',color = 'b')
    ax3 = ax2.twinx()
    ax3.plot(xx,np.cos(xx), label = 'cos(x)', color = 'r')

    ax4.plot(xx,np.tan(xx), label = 'tan(x)', color = 'g')

    for i in enumerate(xrange(number_of_subplots)):
            ax5 = plt.subplot(gs2[i,:])
            ax5.plot(np.sin(xx+3.1416*i/2)) 
            if (i % 2 == 0): #Even
                    ax5.yaxis.tick_left()
            else:
                    ax5.yaxis.tick_right()
    plt.draw()
    plt.show()

在一个subplot2grid中创建迭代子图

1 个答案: