Plotly scatterplot legends not displaying legend title, but with = sign for each color-CodePudding

Plotly refuses to display the legend like it should. In my case, the color is dictated by a datapoint's "Name". The Legend displays Name=[name of the datapoint] instead of giving the Legend the title "Name".

It looks like the image above, but it should look like this:

Like the "species" title, the title "Name" should be displayed in my case. Any help?

CodePudding user response：

It's hard to tell exactly what's happening here without a reproducible example. To my knowledge, a proper application of

Plot 2 - After corrections:

Complete code:

import plotly.graph_objects as go
import plotly.express as px

# data
df = px.data.iris()

# adjust data to reproduce the problem
df['species'] = ['species='   s for s in df['species']]

# build figure
fig = px.scatter(df, x='petal_length', y = 'petal_width', color = 'species')

# change legend entries
fig.for_each_trace(lambda t: t.update(name=t.name.split("=")[-1]))

# edit title and show figure
fig.update_layout(legend_title_text = '<b>Custom title</b>')
fig.show()

CodePudding user response：

you have not included data or code to generate plot
I've sourced from kaggle and used px.scatter(dfs["Pokemon.csv"], x="Attack", y="Defense", color="Name") which generates figure as you want
plotly version: '5.3.1'

import kaggle.cli
import sys, requests
import pandas as pd
from pathlib import Path
from zipfile import ZipFile
import urllib
import plotly.express as px

# fmt: off
# download data set
url = "https://www.kaggle.com/abcsds/pokemon"
sys.argv = [sys.argv[0]]   f"datasets download {urllib.parse.urlparse(url).path[1:]}".split(" ")
kaggle.cli.main()
zfile = ZipFile(f'{urllib.parse.urlparse(url).path.split("/")[-1]}.zip')
dfs = {f.filename: pd.read_csv(zfile.open(f)) for f in zfile.infolist()}
# fmt: on

px.scatter(dfs["Pokemon.csv"], x="Attack", y="Defense", color="Name")