python策略堆叠+分组条形图

Question

我试图在python中使用plotly创建一个条形图，它既是堆叠的，又是分组的。
玩具示例（不同年份的花费和收入）：

import pandas as pd
import plotly.graph_objs as go

data = pd.DataFrame(
    dict(
        year=[2000,2010,2020],
        var1=[10,20,15],
        var2=[12,8,18],
        var3=[10,17,13],
        var4=[12,11,20],
    )
)

fig = go.Figure(
    data = [
        go.Bar(x=data['year'], y=data['var1'], offsetgroup=0, name='spent on fruit'),
        go.Bar(x=data['year'], y=data['var2'], offsetgroup=0, base=data['var1'], name='spent on toys'),
        go.Bar(x=data['year'], y=data['var3'], offsetgroup=1, name='earned from stocks'),
        go.Bar(x=data['year'], y=data['var4'], offsetgroup=1, base=data['var3'], name='earned from gambling'),
    ]
)
fig.show()   

更新：根据这个Github的问题，堆叠的，分组的酒吧情节正在为未来的情节版本开发，所以这可能不会再是一个问题了。

Answer 1

似乎没有一种方法来创建堆叠和分组的条形图，但有一个变通办法，可以解决您的问题。您将需要创建子组，然后使用Plotly中的堆叠条形图一次绘制一个条形图，用子组1绘制var1和var2，用子组2绘制var3和var4。

此解决方案为您提供了所需的功能，但更改了条形图的格式和美观。每个条之间将有相等的间距，因为从Plotly的角度来看，这些条是堆叠的（而不是分组的），我无法想出一种方法来消除subgroup1和subgroup2文本，而不同时消除x轴刻度中的年份。任何阴谋专家，请随时插话和完善我的答案！

import pandas as pd
import plotly.graph_objs as go

df = pd.DataFrame(
    dict(
        year=[2000,2010,2020],
        var1=[10,20,15],
        var2=[12,8,18],
        var3=[10,17,13],
        var4=[12,11,20],
    )
)
        
fig = go.Figure()

fig.update_layout(
    template="simple_white",
    xaxis=dict(title_text="Year"),
    yaxis=dict(title_text="Count"),
    barmode="stack",
)

groups = ['var1','var2','var3','var4']
colors = ["blue","red","green","purple"]
names = ['spent on fruit','spent on toys','earned from stocks','earned from gambling']

i = 0
for r, n, c in zip(groups, names, colors):
    ## put var1 and var2 together on the first subgrouped bar
    if i <= 1:
        fig.add_trace(
            go.Bar(x=[df.year, ['subgroup1']*len(df.year)], y=df[r], name=n, marker_color=c),
        )
    ## put var3 and var4 together on the first subgrouped bar
    else:
        fig.add_trace(
            go.Bar(x=[df.year, ['subgroup2']*len(df.year)], y=df[r], name=n, marker_color=c),
        )
    i+=1

fig.show()