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Python Plotly:不使用Plotly express时,如何设置散布轨迹的标题颜色?_Python_Plotly - Fatal编程技术网

Python Plotly:不使用Plotly express时,如何设置散布轨迹的标题颜色?

Python Plotly:不使用Plotly express时,如何设置散布轨迹的标题颜色?,python,plotly,Python,Plotly,当我做这样的散点图时: fig2 = go.Figure() for num in range(len(dy)): data = dz[0:,num] fig2.add_trace(go.Scatter(x=dx, y=data, name=str(dy[num]), mode='lines')) 如何设置颜色编码标题,如在plotly.express.line中执行color=时 您可以通过图UpdateGend\u title=dictfont=dictcolor='Bl

当我做这样的散点图时:

fig2 = go.Figure()

for num in range(len(dy)):
    data = dz[0:,num]
    fig2.add_trace(go.Scatter(x=dx, y=data, name=str(dy[num]), mode='lines'))
如何设置颜色编码标题,如在plotly.express.line中执行color=时


您可以通过图UpdateGend\u title=dictfont=dictcolor='Blue'编辑字体颜色、大小和系列:

完整代码:


你只是想知道这里痕迹的颜色,对吗?而不是任何东西的标题?或者你是在同时谈论遗迹的颜色和传说中名字的颜色吗?@Vesland我谈论遗迹传说上方的标题。例如,在下面的回答中,您有px.linedf,x=year,y=lifeExp,color='country',它将把单词'country'放在痕迹的图例上方。这个词正是我想要的。那么请在你的问题中加入一个截图。@Vesland完成。你的阴谋版本是什么?
import plotly.express as px
import plotly.graph_objects as go

df = px.data.gapminder().query("continent=='Oceania'")

countries = df['country'].unique()
countries

fig2 = go.Figure()
for country in countries:
    df2 = df[df['country']==country]
    fig2.add_traces(go.Scatter(x=df2['year'], y=df2['lifeExp'], name = country,
                              mode='lines',
                              ))

fig2.update_xaxes(title='Year')
fig2.update_yaxes(title='LiefExp')

fig2.update_layout(legend_title=dict(text='Country',
                                     font=dict(family="sans-serif",
                                     size = 18,
                                     color='blue')))


fig2.show()