layout - Plotly：仅在特定子图中堆积条形图

Question

我有两个带有两列的条形图，创建如下：

fig = tools.make_subplots(rows=1, cols=2)
trace1 = go.Bar(x=x1 , y=y1)
trace2 = go.Bar(x=x2, y=y2)
fig.append_trace(trace1, 1, 1)
fig.append_trace(trace2, 1, 1)
trace3 = go.Bar(x=x3 , y=y3)
trace4 = go.Bar(x=x4, y=y4)
fig.append_trace(trace3, 1, 2)
fig.append_trace(trace4, 1, 2)

我只想要第二个带有堆叠条的子图。我尝试了以下

fig['layout']["xaxis2"].update(barmode='stack')

但这当然行不通。如何将“堆栈”属性仅应用于第二个子图而不是整个图形？

谢谢

标签： layoutplotlysubplot

Answer 1

您可以使用offsetgroup = same（堆叠的所有图形中的相同数字）。
在第二个酒吧base=(first bar y)

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

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

fig.add_trace(go.Bar(name='dogs', x=[1, 2, 3], y=[4, 5, 6],
                     marker_color='blue',
                     marker=dict(color=[4, 5, 6], coloraxis="coloraxis")),
              row=1, col=1)

fig.add_trace(go.Bar(name='cats', x=[1, 2, 3], y=[1, 2, 3],
                     marker_color='yellow',
                     marker=dict(color=[4, 5, 6], coloraxis="coloraxis")),
              row=1, col=1)
female_y = [4,5,6]

fig.add_trace(go.Bar(name='female', x=[1, 2, 3], y=female_y,
                     offsetgroup=1, marker_color='green',
                     marker=dict(color=[4, 5, 6])),
              row=1, col=2)

fig.add_trace(go.Bar(name='male', x=[1, 2, 3], y=[2, 3, 5],
                     offsetgroup=1, marker_color='red',
                     base = female_y ,
                     marker=dict(color=[2, 3, 5])),
              row=1, col=2)

fig.add_trace(go.Bar(name='others', x=[1, 2, 3], y=[1, 1.2, 2.5],
                     offsetgroup=2, marker_color='rgb(0,255,255)',
                     marker=dict(color=[2, 3, 5])),
              row=1, col=2)

fig.show()