3

问题

我有一个带有 2 个 y 轴的图，每个轴对应一组线。实线对应于左侧 y 轴，虚线对应于右侧 y 轴。我也有一个图例，我希望它只使用实线作为键，因为虚线具有相同的标签，具体取决于它们的颜色。

问题是当我为实线绘制图例，然后为虚线绘制代码时，网格线通过图例显示。我需要为两个轴指定网格线，否则它们不会显示，如果我将图例移动到虚线，它会使用虚线作为键。我也不想改变我的绘图顺序。

代码和情节

#Plot
x= np.arange(0,3)
fig,ax = plt.subplots(figsize=(6,6))
#DOD
dod1 = ax.plot(x, ctrl_dod,  color='r',       label='CTRL'  )
dod2 = ax.plot(x, mfkc_dod,  color='#e68a00', label='MFKC'  )
dod3 = ax.plot(x, gses_dod,  color='green',   label='GSES'  )
dod4 = ax.plot(x, gses3_dod, color='blue',    label='GSES-3')
dod5 = ax.plot(x, gses4_dod, color='purple',  label='GSES-4')
dod6 = ax.plot(x, mera_dod,  color='brown',   label='MERRA2')
ax.xaxis.grid(True)
ax.set_ylim([0.02,0.044])
ax.set_yticks(np.arange(0.02,0.045,0.004))
ax.set_xlabel('Month')
ax.set_ylabel('Dust Optical Depth (550 nm)')
ax.set_title('Global Mean DOD and DCM')
legend = ax.legend()
legend.get_frame().set_facecolor('white')

#DCM
ax2  = ax.twinx()
dcm1 = ax2.plot(x, ctrl_dcm*1e6,  color='r',       linestyle='--', label='CTRL'  )
dcm2 = ax2.plot(x, mfkc_dcm*1e6,  color='#e68a00', linestyle='--', label='MFKC'  )
dcm3 = ax2.plot(x, gses_dcm*1e6,  color='green',   linestyle='--', label='GSES'  )
dcm4 = ax2.plot(x, gses3_dcm*1e6, color='blue',    linestyle='--', label='GSES-3')
dcm5 = ax2.plot(x, gses4_dcm*1e6, color='purple',  linestyle='--', label='GSES-4')
dcm6 = ax2.plot(x, mera_dcm*1e6,  color='brown',   linestyle='--', label='MERRA2')
ax2.xaxis.grid(True)
ax2.yaxis.grid(True)
ax2.set_xlabel('Month')
ax2.set_ylabel('Dust Column Mass (mg m-2)')

#Limits
axes = plt.gca()
axes.set_xlim([-0.25,2.25])

#Labels
axes.set_xticks(x)
axes.set_xticklabels(['June','July','August'])

#Save
pylab.savefig('dod+dcm.png')

在此处输入图像描述

问题

我怎样才能

a) 让图例键使用实线和

b) 有传说不透明白色的背景吗？

4

2 回答 2

5

您可以为第二个轴创建图例，但使用第一个轴的句柄。

h, l = ax.get_legend_handles_labels()
legend = ax2.legend(h,l, facecolor="white")
于 2017-07-26T20:05:00.720 回答
0

我来到这里，因为我有同样的问题，但想要两个独立的图例，一个用于双轴。

@ImportanceOfBeingErnest 提供的已接受答案不再有效，因为它只允许将一个图例附加到ax2.

在这种情况下，@ImportanceOfBeingErnest 在https://github.com/matplotlib/matplotlib/issues/3706#issuecomment-378407795上也提供了另一种解决方案。

我将它们都转换为我通常在这种情况下使用的一个函数，并认为在这里提供它会很有用：

def legend_to_ax( ax, ax_placein=None, method=2, **kwargs ):
    """ Wrapper around ax.legend(**kwargs) which permits to have the legend
    placed in the axis ax_placein.
    This is useful when drawing legends for multiple axes, e.g.produced with
    ax2 = ax1.twinx(), in order to have all legends on top of all axes.
    In this case provide the uppermost axis, e.g. ax2 here, as ax_placein.
    Args:
     - ax         : axis for which the legend is created
     - ax_placein : axis which the legend should be placed in (optional)
     - method     : method by which to handle the placement of the legend in
                    ax_placein.
                    method=1 ... based on 
                                 https://stackoverflow.com/a/45336414/7042795
                                 This method fails when multiple legends should
                                 be added to ax_placein.
                    method=2 ... based on
                                 https://github.com/matplotlib/matplotlib/issues/3706#issuecomment-378407795
                                 This adds the legend as an artist to ax_placein
                                 making it no longer appear via ax.get_legend().
                                 Instead parse ax_placein.artists()
     - **kwargs   : kwargs of ax.legend()
    Returns:
     - leg        : legend handle
    """

    if ax_placein is None:
        leg = ax.legend(**kwargs)
    
    elif method==1:
        # based on https://stackoverflow.com/a/45336414/7042795
        h, l = ax.get_legend_handles_labels()
        if ax_placein is not None:
            ax = ax_placein
        leg =  ax.legend( h, l, **kwargs)
    
    elif method==2:
        # based on https://github.com/matplotlib/matplotlib/issues/3706#issuecomment-378407795
        leg = ax.legend(**kwargs)
        if ax_placein is not None:
            leg.remove()
            ax_placein.add_artist(leg)
    
    return leg

所以，有了这个功能，你可以简单地做

legend = legend_to_ax( ax, ax_placein=ax2, method=1, facecolor="white")

或者

legend = legend_to_ax( ax, ax_placein=ax2, method=2, facecolor="white")

如果您确实想要两个图例，这应该可以：

legend1 = legend_to_ax( ax,  ax_placein=ax2, method=2, facecolor="white", loc="upper left")
legend2 = legend_to_ax( ax2, ax_placein=ax2, method=2, facecolor="white", loc="upper right")

我希望这是一个合适的地方来分享这些信息。积分归功于@ImportanceOfBeingErnest。我只将他们的解决方案合并到这个函数中。

于 2021-02-18T09:26:04.317 回答