多级饼图散景

Question

我目前正在尝试将散景中的饼图组合在一起，该饼图使用楔形字形而不是预设的饼图（需要更好的方法）。我想制作一个只有一个半饼图的多级饼图，我发现我只需要乘以一个 pi 而不是两个来得到半周长。我还减去了我正在为下一层工作的相反角度以限制其范围，但是第二个到处都是，如果有人可以帮助我，请。提前谢谢。

attack1 = {'shots': ['home', 'away'], 'shots':[144, 48, 36, 128, 27, 41]}

attack2 = {'index':['shots','shots on target', 'goals'], 'home':[144,48,36],'away':[128,27,41]}


attack1, attack2 = pd.DataFrame(attack1, index=['home','away']), pd.DataFrame(attack2, index=['shots','shots on target','goals'])



c1 = YlGnBu

c2 = PuBu



attack1['colors'] = [c1[6][1], RdYlGn[6][3]]
attack1['angle'] = attack1['shots'] / attack1['shots'].sum() * pi # + 0.5*pi
attack1['pct'] = pct(attack1['shots'], attack1['shots'].sum())
attack1['pct'] = attack1['pct'].astype(str) + '%'
attack1['pct'] = attack1['pct'].str.pad(28, side='left')




attack2['colors'] = c2[6][2:5]
attack2['angle'] = attack2['home'] / attack2['home'].cumsum()  * (pi - attack1['angle'][0] -.255)          
attack2['pct'] = pct(attack2['home'], attack2['home'].sum())
attack2['pct'] = attack2['pct'].astype(str) + '%'

xP = 'home', 'away'
sourceAttack1 = ColumnDataSource(data=dict(x=xP, y=attack1['shots'], c=attack1['colors'], a=attack1['angle'], l=attack1['pct']))
sourceAttack2 = ColumnDataSource(data=dict(x=list(attack2.index.values), y=attack2['home'], c=attack2['colors'], a=attack2['angle'], l=attack2['pct']))



pA1 = figure(plot_height=450, plot_width=550, title='Attack',
       tools = 'pan, wheel_zoom, box_zoom, reset',  x_range=(-0.4, 0.9))

pA1.wedge(x=0, y=1, radius=0.4, start_angle=cumsum('a', include_zero=True), end_angle=cumsum('a'),
     line_color='white', fill_color='c', legend_field='x', source=sourceAttack1)

pA1.wedge(x=0, y=1, radius=0.3, start_angle=cumsum('a', include_zero=True), end_angle=cumsum('a'),
     line_color='white', fill_color='c', legend_field='x', source=sourceAttack2)

hA1 = HoverTool()
hA1.tooltips=[('location', '@x'), ('shots','@y'), ('pct', '@l')]

lA1 = LabelSet(x=0, y=1, text='l', text_color='black', text_font_size='15px', text_align='left',
    angle=cumsum('a', include_zero=True), source=sourceAttack1, render_mode='canvas') # y_offset=50


pA1.add_tools(hA1)
pA1.axis.axis_label=None
pA1.axis.visible=False
pA1.grid.grid_line_color = None
pA1.legend.background_fill_alpha=None
pA1.legend.border_line_alpha=0
pA1.legend.location= (0,50) #'center_right'
pA1.legend.label_text_font_size = "15px"
pA1.legend.orientation ='vertical'
pA1.title.text_font_size = '25px'
pA1.toolbar.autohide = True
pA1.outline_line_color=None



show(pA1)

这是目前的结果

Answer 1

事实证明，在为我正在寻找的内容创建一个最小的示例时，我最终按照我的意愿制作了它。谢谢@mosc9575，也许我只需要离开一会儿。

#values
attack = [200,80,34,190,85,40, 314,315 ]
overall = {'attack': attack}
index = ['shots','shotsT','goals','shotsA','shotsTA','goalsA','attackHome','attackAway']

#dataframe
dfpie = pd.DataFrame(overall, index = index)
dfpie['color'] = viridis(len(dfpie.index))

#df without overal which would be the first pie chart
dfpie1 = dfpie.iloc[:-2]

#df overall
dfpie2 = pd.DataFrame(dfpie.iloc[-2:])


#angle for both df * pi instead *2*pi to keep it half circumference C=2πr
dfpie1['angles'] = dfpie1['attack'] / dfpie1['attack'].sum() * pi
dfpie2['angles'] = dfpie2['attack'] / dfpie2['attack'].sum() * pi

#sources
sa1 = ColumnDataSource(data=dict(x=list(dfpie2.index.values), y=dfpie2['attack'], c=dfpie2['color'], a=dfpie2['angles']))
sa2 = ColumnDataSource(data=dict(x=list(dfpie1.index.values), y=dfpie1['attack'], c=dfpie1['color'], a=dfpie1['angles']))

#figure instance
piA = figure(plot_height=450, plot_width=550, title='Attack',
           tools = 'pan, wheel_zoom, box_zoom, reset',  x_range=(-0.4, 0.9))

#glyph
piA.wedge(x=0, y=1, radius=0.4, start_angle=cumsum('a', include_zero=True), end_angle=cumsum('a'),
         line_color='white', fill_color='c', legend_field='x', source=sa1)

#overlay glyph
piA.wedge(x=0, y=1, radius=0.3, start_angle=cumsum('a', include_zero=True), end_angle=cumsum('a'),
         line_color='white', fill_color='c', legend_field='x', source=sa2)


#remove grid lines
piA.axis.axis_label=None
piA.axis.visible=False
piA.grid.grid_line_color = None
piA.outline_line_color=None


show(piA)