如何在ImageGrid中删除刻度线？

Question

如何在将行标题保留在ImageGrid中的同时删除刻度线和数字？

grid[j].axis("off")

删除标签输入，地面真理，预测，而

grid[j].set_xticklabels([])
grid[j].set_yticklabels([])

仅删除数字，但不删除轴上的破折号。我查看了https://matplotlib.org/3.1.1/api/_as_gen/mpl_toolkits.axes_grid1.axes_grid.ImageGrid.html，但似乎没有刻度线。

for idx in range(0,1):
    fig  = plt.figure(idx, (15, 10))
    grid = ImageGrid(fig, 111, nrows_ncols=(3, 10), axes_pad=0.1)  
    grid[0].set_ylabel("Input",rotation=0)
    grid[10].set_ylabel("Ground Truth",rotation=0)
    grid[20].set_ylabel("Variational ConvLSTM",rotation=0)
    for j in range(10):
        #grid[j].set_xticklabels([])
        #grid[j].set_yticklabels([])
        #grid[j].axis("off") This removes the title
        grid[j].imshow(x_true[idx,j], cmap="gray")
        grid[j+10].imshow(x_true[idx,j+10], cmap="gray")
        grid[j+20].imshow(x_temp[idx,j], cmap="gray")

这里是我目前的输出。理想情况下，它应该没有刻度（破折号）并且没有数字。

Answer 1

简单的方法是先设置share_all = True然后>>

grid[0].get_yaxis().set_ticks([])
grid[0].get_xaxis().set_ticks([])
所以所有刻度都被禁用

for idx in range(0,1):
    fig  = plt.figure(idx, (15, 10))
    grid = ImageGrid(fig, 111, nrows_ncols=(3, 10), axes_pad=0.1, share_all=True)  

    grid[0].get_yaxis().set_ticks([])
    grid[0].get_xaxis().set_ticks([])

    # labels
    grid[0].set_ylabel("Input",rotation=0)
    grid[10].set_ylabel("Ground Truth",rotation=0)
    grid[20].set_ylabel("Prediction",rotation=0)
    for j in range(10):
        #grid[j].set_xticklabels([])
        #grid[j].set_yticklabels([])
        #grid[j].get_yaxis().set_ticks([])
        #grid[j].get_xaxis().set_ticks([])
        #grid[j].axis("off")
        grid[j].imshow(x_true[idx,j], cmap="gray")
        grid[j+10].imshow(x_true[idx,j+10], cmap="gray")
        grid[j+20].imshow(x_temp[idx,j], cmap="gray")