簡體   English   中英

在Bokeh中將方程式繪制為線-Python

[英]Plotting equations as lines in Bokeh - Python

我正在Bokeh中創建一條1:1線的XY圖表,理想情況下再創建兩條線以實現+/- 10%的誤差和+/- 20%的誤差。 目前,我的圖表可以運行,但看起來有些不合常理,並且顯示了太多圖例條目。 目前的代碼:

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)

因此,我想替換一下我通過u定義q的部分,以便使用一些繪制r / s(+/- 10%誤差)和t / u(誤差為+/- 20%)。 這樣,我將為每個結局輸入一個圖例條目。

但這會引發錯誤:

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

而且我仍然會為每種錯誤類型輸入重復的條目

您不能將列表乘以浮點數。 如果我正確理解,類似這樣的東西應該會得到想要的結果:

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

並將* 1.1與0.9、1.2和0.8替換為您希望引用q的其他變體。

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM