python - Plotly:如何为每个不同的子图应用不同的标题？

Question

鉴于我有来自 this link 的以下代码:

from plotly.subplots import make_subplots
import plotly.graph_objects as go

fig = make_subplots(rows=1, cols=2)

fig.add_trace(
    go.Scatter(x=[1, 2, 3], y=[4, 5, 6]),
    row=1, col=1
)

fig.add_trace(
    go.Scatter(x=[20, 30, 40], y=[50, 60, 70]),
    row=1, col=2
)

fig.update_layout(height=600, width=800, title_text="Subplots")
fig.show()

此代码中的问题是，xaxis 和 yaxis 没有任何标签。除此之外，当前代码仅对所有图应用一个标题，但我想对每个散点图应用不同的标题。

我怎样才能做到这一点？

Answer 1

The problem in this code is, the xaxis and yaxis does not have any label.

您可以通过对图形结构进行子集化来编辑任何轴:

fig['layout']['xaxis']['title']='Label x-axis 1'

Beside this, the current code applies only one title to all the plots

根据用户 shaik moeed 提到的绘图版本，您可以在图形定义中包含 subplot_titles:

fig = make_subplots(rows=1, cols=2, subplot_titles=('Subplot title1',  'Subplot title2'))

plotly :

代码:

from plotly.subplots import make_subplots
import plotly.graph_objects as go


# plotly fig setup
fig = make_subplots(rows=1,
                    cols=2,
                    subplot_titles=('Subplot title1',  'Subplot title2')) 

# traces
fig.add_trace(
    go.Scatter(x=[1, 2, 3], y=[4, 5, 6]),
    row=1, col=1
)

fig.add_trace(
    go.Scatter(x=[20, 30, 40], y=[50, 60, 70]),
    row=1, col=2
)

# edit axis labels
fig['layout']['xaxis']['title']='Label x-axis 1'
fig['layout']['xaxis2']['title']='Label x-axis 2'
fig['layout']['yaxis']['title']='Label y-axis 1'
fig['layout']['yaxis2']['title']='Label y-axis 2'

# plot it
fig.show()