python - Plotly：如何创建奇数个子图？

Question

我希望第 5 个子图位于第三行两列的中心。（我已经尝试通过添加domain参数来做到这一点）。这是重现它的代码-

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

continent_df = pd.read_csv('https://raw.githubusercontent.com/vyaduvanshi/helper-files/master/continent.csv')
temp_cont_df = pd.pivot_table(continent_df, index='continent', aggfunc='last').reset_index()

fig = make_subplots(rows=3, cols=2, specs=[[{'type':'pie'},{'type':'pie'}],[{'type':'pie'},{'type':'pie'}],
                                           [{'type':'pie'},{'type':'pie'}]])

fig.add_pie(labels=continent_df.continent, values=continent_df.new_cases, row=1,col=1)
fig.add_pie(labels=continent_df.continent, values=continent_df.new_deaths, row=1,col=2)
fig.add_pie(labels=continent_df.continent, values=continent_df.new_recovered, row=2,col=1)
fig.add_pie(labels=continent_df.continent, values=continent_df.new_tests, row=2,col=2)
fig.add_pie(labels=temp_cont_df.continent, values=temp_cont_df.active_cases, row=3,col=1,domain={'x':[0.25,0.75],'y':[0,0.33]})

如果我没有在specs参数中包含第 6 个图，则会引发错误。

标签： pythonplotlydata-visualizationplotly-python

Answer 1

您可以通过正确设置domain. 这是一个示例，四个角各有一个图形，中间有一个图形。

阴谋

完整代码：

import plotly
import plotly.offline as py
import plotly.graph_objs as go

labels = ['Oxygen','Hydrogen','Carbon_Dioxide','Nitrogen']
values = [4500,2500,1053,500]
domains = [
    {'x': [0.0, 0.33], 'y': [0.0, 0.33]},
    {'x': [0.33, 0.66], 'y': [0.33, 0.66]},
    {'x': [0.0, 0.33], 'y': [0.66, 1.0]},
    {'x': [0.66, 1.00], 'y': [0.0, 0.33]},
    {'x': [0.66, 1.0], 'y': [0.66, 1.00]},
]
traces = []

for domain in domains:
    trace = go.Pie(labels = labels,
                   values = values,
                   domain = domain,
                   hoverinfo = 'label+percent+name')
    traces.append(trace)

layout = go.Layout(height = 600,
                   width = 600,
                   autosize = False,
                   title = 'Main title')
fig = go.Figure(data = traces, layout = layout)
#py.iplot(fig, show_link = False)
fig.show()