用Bokeh-Python绘制直线方程

Question

我在博基创建一个XY图表，有一条1:1的线，理想情况下还有两条线，误差为+/-10%和+/-20%。目前我的图表工作，但似乎不太和谐，显示太多的传奇条目。 当前代码：

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)

所以我想把定义q到u的那一部分替换掉，用一些方程来绘制每一条误差线的两对点，r/s（+/-10%误差）和t/u（+/-20%误差）来代替q。那样的话，我会得到一个单一的图例条目。你知道吗

但这会引发一个错误：

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

对于每种错误类型，我仍然会得到重复的条目

Answer 1

你不能用一个浮点数乘以一个列表。如果我理解正确，像这样的事情应该会得到你想要的结果：

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

并将*1.1替换为0.9、1.2和0.8，用于您希望引用q的其他变体