在Bokeh中将方程式绘制为线-Python

Question

I'm creating an XY chart in Bokeh with a 1:1 line and ideally two more lines for +/- 10% error and +/- 20% errors. At the moment my chart works but seems unpythonic and shows too many legend entries. The code at present:

import pandas as pd
from bokeh.plotting import figure, output_file, save
from bokeh.io import show, output_notebook
from bokeh.models import Span, HoverTool, ColumnDataSource
import numpy as np
# Call up duplicate plot

TOOLTIPS=[
    ("Sample", "@Sample"),
    ("Batch", "@Batch_No"),
    ("Source", "@Hole_ID"),
    ("Type", "@QC_Category")]

pdup = figure(title='Duplicate QC Review', x_axis_label='Duplicate', y_axis_label='Original', tools=tools_to_show, 
           tooltips=TOOLTIPS, outline_line_width=olwidth)

q = [0, 10000]
r = [0, 11000]
s = [0, 9000]
t = [0, 12000]
u = [0, 8000]

# 1:1 line, 10% and 20% error lines both above and below 1:1 line
pdup.line(q, q, color='green', legend='1:1')
pdup.line(q, r, color='orange', legend = '10%')
pdup.line(q, s, color='orange', legend = '-10%')
pdup.line(q, t, color='red', legend = '20%')
pdup.line(q, u, color='red', legend = '-20%')

pdup.circle(x='Copper_ppm', y='Cu_Duplicate', source=srcdup, size=10, color='green', legend='Copper (ppm)')
pdup.triangle(x='Gold_ppm', y='Au_Duplicate', source=srcdup, color='orange', size=10, legend='Gold (ppm)')
pdup.square(x='Molybdenum_ppm', y='Mo_Duplicate', source=srcdup, color='purple', size=10, legend='Molybdenum (ppm)')
pdup.diamond(x='Sulphur_ppm', y='S_Duplicate', source=srcdup, color='gray', size=10, legend='Sulphur (%)')

# Legend settings
# Make a series or connecting lines hidden by clicking on the legend entry
pdup.legend.click_policy='hide'
pdup.legend.border_line_color = "black"
pdup.legend.background_fill_color = "white"
pdup.legend.location = 'top_left'

show(pdup)

So I'd like to replace the section where I'm defining q through u in order to plot two pairs of points for each error line with some equations that plot r/s (+/-10% error) and t/u (+/-20% error) for a qiven q instead. That way I'll end up with a single legend entry for each.

This throws an error though:

q = [0, 10000]
r = [q + (0.1 * q)]

And I'd still end up with duplicate entries for each error type

Answer 1

You can't multiply a list by a float. If I understand correctly, something like this should get the result you want:

q = [0, 10000]
r = [q[0],q[1]*1.1]

And replace the * 1.1 with 0.9, 1.2 and 0.8 for the other variations you wish to reference to q.