Python Plotly - 多个下拉图，每个都有子图

Question

问题

我正在尝试结合两个 Python Plotly 功能。其中之一是下拉菜单，用户可以在其中切换绘图 (link to examples)。另一个特征是子图。

我的尝试

我有使用下拉菜单的工作代码，但它没有子图。所以我查找了如何创建子图 here我认为我可以像对待 go.Box 图一样对待 tools.make_subplots 图。如果这看起来很幼稚，我很抱歉，我实际上是 Plotly 的新手。

但是，这种尝试根本不起作用，我看到两个异常上升，我已将它们放在最底部。

我的工作代码(无子图)

# NOTE: This code goes in between 'START' and 'END' below
traces = []
for data in datas:
    traces.append(go.Box(
                            x=data.index,
                            y=data.values, 
                            showlegend=False
                        ))

我的子图代码

import plotly.offline as py
import plotly.graph_objs as go
from plotly.offline import init_notebook_mode, iplot, plot
from plotly import tools
init_notebook_mode(connected=True)

### Create individual figures
# START
traces = []
for data in datas:
    fig = tools.make_subplots(rows=1, cols=2)

    trace1 = go.Box(
                    x=data.head(10).index,
                    y=data.head(10).values, 
                    showlegend=False
                )
    trace2 = go.Box(
                    x=data.tail(10).index,
                    y=data.tail(10).values, 
                    showlegend=False
                )   
    fig.append_trace(trace1, 1, 1)
    fig.append_trace(trace2, 1, 2)
    traces.append(fig)
# END

### Create buttons for drop down menu
buttons = []
for i, label in enumerate(labels):
    visibility = [i==j for j in range(len(labels))]
    button = dict(
                 label =  label,
                 method = 'update',
                 args = [{'visible': visibility},
                     {'title': label}])
    buttons.append(button)

updatemenus = list([
    dict(active=-1,
         x=-0.15,
         buttons=buttons
    )
])

layout = dict(title='Title', 
              showlegend=False,
              updatemenus=updatemenus)

fig = dict(data=traces, layout=layout)

iplot(fig, filename='dropdown')

错误 1

'layout' is not allowed in 'scatter'

Path To Error: ['data'][0]['layout']

错误2

PlotlyError: Invalid 'figure_or_data' argument. Plotly will not be
able to properly parse the resulting JSON. If you want to send this 'figure_or_data' to Plotly anyway (not recommended), you can set 'validate=False' as a plot option.

Here's why you're seeing this error:

'layout' is not allowed in 'scatter'

...

注意:当我将 validate=False 作为绘图选项传递时，我看到的只是一个带有下拉菜单功能的空绘图

Answer 1

Naren 是对的，只需要制作一个子图。而要创建下拉菜单效果，您只需要像这样在同一位置添加两条走线..

fig.append_trace(trace1, 1, 1)
fig.append_trace(trace2, 1, 1)

工作代码

import plotly.offline as py
import plotly.graph_objs as go
from plotly.offline import init_notebook_mode, iplot, plot
from plotly import tools
init_notebook_mode(connected=True)

x = [i for i in range(100)]
df_1 = pd.DataFrame([(i, 1+i) for i in range(100)], columns=["X", "Y"])
df_2 = pd.DataFrame([(i, i*i) for i in range(100)], columns=["X", "Y"])
labels = ["Plus one", "Square"]

### Create individual figures
# START
fig = tools.make_subplots(rows=1, cols=2)

trace1 = go.Bar(
                x=df_1.head(10).X,
                y=df_1.head(10).Y, 
                showlegend=False
            )
trace2 = go.Bar(
                x=df_2.head(10).X,
                y=df_2.head(10).Y, 
                showlegend=False
            )

fig.append_trace(trace1, 1, 1)
fig.append_trace(trace2, 1, 1)

trace1 = go.Bar(
                x=df_1.tail(10).X,
                y=df_1.tail(10).Y, 
                showlegend=False
            )
trace2 = go.Bar(
                x=df_2.tail(10).X,
                y=df_2.tail(10).Y, 
                showlegend=False
            )   
fig.append_trace(trace1, 1, 2)
fig.append_trace(trace2, 1, 2)
# END

### Create buttons for drop down menu
buttons = []
for i, label in enumerate(labels):
    visibility = [i==j for j in range(len(labels))]
    button = dict(
                 label =  label,
                 method = 'update',
                 args = [{'visible': visibility},
                     {'title': label}])
    buttons.append(button)

updatemenus = list([
    dict(active=-1,
         x=-0.15,
         buttons=buttons
    )
])

fig['layout']['title'] = 'Title'
fig['layout']['showlegend'] = False
fig['layout']['updatemenus'] = updatemenus

iplot(fig, filename='dropdown')

谢谢

非常感谢 Naren Murali 的帮助!稍微编辑您的代码给了我答案。

我正在寻找创造这个..

但是你的代码给了我这些 plotly ......