I'm trying to have a plot title which contains variable values and also characters with subscripts, however when I try:
title = "ηₛ = $η̂[Pa S] , μₛ = $μ̂[Pa], μₚ = $μ̂ₚ[Pa] , ηₚ = $η̂ₚ[Pa S] \n α = $α̂ , ζ = $ζ̂"
Inside the plot function, the title appears with X marks where the subscripts are. I tried to use LaTeX ```title = L" .." but then the variable values don't appear.
Is there any way to have both in the title I need?
If you want a fully working solution this is what I think you need to do, note that %$ is used for interpolation:
title = L"\eta_1 = %$(η̂[Pa, S])"
The reason is that, while some of the characters will be rendered correctly as Bill noted, not all of them will unless you use LaTeXStrings.jl.
See:
help?> LaTeXStrings.#L_str
L"..."
Creates a LaTeXString and is equivalent to latexstring(raw"..."), except that %$ can be used for interpolation.
julia> L"x = \sqrt{2}"
L"$x = \sqrt{2}$"
julia> L"x = %$(sqrt(2))"
L"$x = 1.4142135623730951$"
I am learning how to create plots with slider bars. Here is my code based off the first example of this tutorial
using Plots
gr()
using GLMakie
function plotLaneEmden(log_delta_xi=-4, n=3)
fig = Figure()
ax = Axis(fig[1, 1])
sl_x = Slider(fig[2, 1], range = 0:0.01:4.99, startvalue = 3)
sl_y = Slider(fig[1, 2], range = -6:0.01:0.1, horizontal = false, startvalue = -2)
point = lift(sl_x.value, sl_y.value) do n, log_delta_xi
Point2f(n, log_delta_xi)
end
plot(n, 1 .- log_delta_xi.^2/6, linecolor = :green, label="n = $n")
xlabel!("ξ")
ylabel!("θ")
end
plotLaneEmden()
When I run this, it gives UndefVarError: plot not defined. What am I missing here?
It looks like you are trying to mix and match Plots.jl and Makie.jl. Specifically, the example from your link is entirely for Makie (specifically, with the GLMakie backend), while the the plot function you are trying to add uses syntax specific to the Plots.jl version of plot (specifically including linecolor and label keyword arguments).
Plots.jl and Makie.jl are two separate and unrelated plotting libraries, so you have to pick one and stick with it. Since both libraries export some of the same function names, using both at once will lead to ambiguity and UndefVarErrors if not disambiguated.
The other potential problem is that it looks like you are trying to make a line plot with only a single x and y value (n and log_delta_xi are both single numbers in your code as written). If that's what you want, you'll need a scatter plot instead of a line plot; and if that's not what you want you'll need to make those variables vectors instead somehow.
Depending on what exactly you want, you might try something more along the lines of (in a new session, using only Makie and not Plots):
using GLMakie
function plotLaneEmden(log_delta_xi=-4, n=3)
fig = Figure()
ax = Axis(fig[1, 1], xlabel="ξ", ylabel="θ")
sl_x = Slider(fig[2, 1], range = 0:0.01:4.99, startvalue = n)
sl_y = Slider(fig[1, 2], range = -6:0.01:0.1, horizontal = false, startvalue = log_delta_xi)
point = lift(sl_x.value, sl_y.value) do n, log_delta_xi
Point2f(n, 1 - log_delta_xi^2/6)
end
sca = scatter!(point, color = :green, markersize = 20)
axislegend(ax, [sca], ["n = $n"])
fig
end
plotLaneEmden()
Or, below, a simple example for interactively plotting a line rather than a point:
using GLMakie
function quadraticsliders(x=-5:0.01:5)
fig = Figure()
ax = Axis(fig[1, 1], xlabel="X", ylabel="Y")
sl_a = Slider(fig[2, 1], range = -3:0.01:3, startvalue = 0.)
sl_b = Slider(fig[1, 2], range = -3:0.01:3, horizontal = false, startvalue = 0.)
points = lift(sl_a.value, sl_b.value) do a, b
Point2f.(x, a.*x.^2 .+ b.*x)
end
l = lines!(points, color = :blue)
onany((a,b)->axislegend(ax, [l], ["$(a)x² + $(b)x"]), sl_a.value, sl_b.value)
limits!(ax, minimum(x), maximum(x), -10, 10)
fig
end
quadraticsliders()
ETA: A couple examples closer to what you might be looking for
I would like to plot a two variable function(s) (e_pos and e_neg in the code). Here, t and a are constants which I have given the value of 1.
My code to plot this function is the following:
t = 1
a = 1
kx = ky = range(3.14/a, step=0.1, 3.14/a)
# Doing a meshgrid for values of k
KX, KY = kx'.*ones(size(kx)[1]), ky'.*ones(size(ky)[1])
e_pos = +t.*sqrt.((3 .+ (4).*cos.((3)*KX*a/2).*cos.(sqrt(3).*KY.*a/2) .+ (2).*cos.(sqrt(3).*KY.*a)));
e_neg = -t.*sqrt.((3 .+ (4).*cos.((3)*KX*a/2).*cos.(sqrt(3).*KY.*a/2) .+ (2).*cos.(sqrt(3).*KY.*a)));
using Plots
plot(KX,KY,e_pos, st=:surface,cmap="inferno")
If I use Plots this way, sometimes I get an empty 3D plane without the surface. What am I doing wrong? I think it may have to do with the meshgrids I did for kx and ky, but I am unsure.
Edit: I also get the following error:
I changed some few things in my code.
First, I left the variables as ranges. Second, I simply computed the functions I needed without mapping the variables onto them. Here's the code:
t = 2.8
a = 1
kx = range(-pi/a,stop = pi/a, length=100)
ky = range(-pi/a,stop = pi/a, length=100)
#e_pos = +t*np.sqrt(3 + 4*np.cos(3*KX*a/2)*np.cos(np.sqrt(3)*KY*a/2) + 2*np.cos(np.sqrt(3)*KY*a))
e_pos(kx,ky) = t*sqrt(3+4cos(3*kx*a/2)*cos(sqrt(3)*ky*a/2) + 2*cos(sqrt(3)*ky*a))
e_neg(kx,ky) = -t*sqrt(3+4cos(3*kx*a/2)*cos(sqrt(3)*ky*a/2) + 2*cos(sqrt(3)*ky*a))
# Sort of broadcasting?
e_posfunc = e_pos.(kx,ky);
e_negfunc = e_neg.(kx,ky);
For the plotting I simply used the GR backend:
using Plots
gr()
plot(kx,ky,e_pos,st=:surface)
plot!(kx,ky,e_neg,st=:surface, xlabel="kx", ylabel="ky",zlabel="E(k)")
I got what I wanted!
I'm trying to do something that I'd normally consider trivial but seems to be very difficult in bokeh: Adding a vertical colorbar to a plot and then having the title of the colorbar (a.k.a. the variable behind the colormapping) appear to one side of the colorbar but rotated 90 degrees clockwise from horizontal.
From what I can tell of the bokeh ColorBar() interface (looking at both documentation and using the python interpreter's help() function for this element), this is not possible. In desperation I have added my own Label()-based annotation. This works but is klunky and displays odd behavior when deployed in a bokeh serve situation--that the width of the data window on the plot varies inversely with the length of the title colorbar's title string.
Below I've included a modified version of the bokeh server mpg example. Apologies for its complexity, but I felt this was the best way to illustrate the problem using infrastructure/data that ships with bokeh. For those unfamiliar with bokeh serve, this code snippet needs to saved to a file named main.py that resides in a directory--for the sake of argument let's say CrossFilter2--and in the parent directory of CrossFilter2 one needs to invoke the command
bokeh serve --show CrossFilter2
this will then display in a browser window (localhost:5006/CrossFilter2) and if you play with the color selection widget you will see what I mean, namely that short variable names such as 'hp' or 'mpg' result in a wider data display windows than longer variable names such as 'accel' or 'weight'. I suspect that there may be a bug in how label elements are sized--that their x and y dimensions are swapped--and that bokeh has not understood that the label element has been rotated.
My questions are:
Must I really have to go to this kind of trouble to get a simple colorbar label feature that I can get with little-to-no trouble in matplotlib/plotly?
If I must go through the hassle you can see in my sample code, is there some other way I can do this that avoids the data window width problem?
import numpy as np
import pandas as pd
from bokeh.layouts import row, widgetbox
from bokeh.models import Select
from bokeh.models import HoverTool, ColorBar, LinearColorMapper, Label
from bokeh.palettes import Spectral5
from bokeh.plotting import curdoc, figure, ColumnDataSource
from bokeh.sampledata.autompg import autompg_clean as df
df = df.copy()
SIZES = list(range(6, 22, 3))
COLORS = Spectral5
# data cleanup
df.cyl = df.cyl.astype(str)
df.yr = df.yr.astype(str)
columns = sorted(df.columns)
discrete = [x for x in columns if df[x].dtype == object]
continuous = [x for x in columns if x not in discrete]
quantileable = [x for x in continuous if len(df[x].unique()) > 20]
def create_figure():
xs = df[x.value].tolist()
ys = df[y.value].tolist()
x_title = x.value.title()
y_title = y.value.title()
name = df['name'].tolist()
kw = dict()
if x.value in discrete:
kw['x_range'] = sorted(set(xs))
if y.value in discrete:
kw['y_range'] = sorted(set(ys))
kw['title'] = "%s vs %s" % (y_title, x_title)
p = figure(plot_height=600, plot_width=800,
tools='pan,box_zoom,wheel_zoom,lasso_select,reset,save',
toolbar_location='above', **kw)
p.xaxis.axis_label = x_title
p.yaxis.axis_label = y_title
if x.value in discrete:
p.xaxis.major_label_orientation = pd.np.pi / 4
if size.value != 'None':
groups = pd.qcut(df[size.value].values, len(SIZES))
sz = [SIZES[xx] for xx in groups.codes]
else:
sz = [9] * len(xs)
if color.value != 'None':
coloring = df[color.value].tolist()
cv_95 = np.percentile(np.asarray(coloring), 95)
mapper = LinearColorMapper(palette=Spectral5,
low=cv_min, high=cv_95)
mapper.low_color = 'blue'
mapper.high_color = 'red'
add_color_bar = True
ninety_degrees = pd.np.pi / 2.
color_bar = ColorBar(color_mapper=mapper, title='',
#title=color.value.title(),
title_text_font_style='bold',
title_text_font_size='20px',
title_text_align='center',
orientation='vertical',
major_label_text_font_size='16px',
major_label_text_font_style='bold',
label_standoff=8,
major_tick_line_color='black',
major_tick_line_width=3,
major_tick_in=12,
location=(0,0))
else:
c = ['#31AADE'] * len(xs)
add_color_bar = False
if add_color_bar:
source = ColumnDataSource(data=dict(x=xs, y=ys,
c=coloring, size=sz, name=name))
else:
source = ColumnDataSource(data=dict(x=xs, y=ys, color=c,
size=sz, name=name))
if add_color_bar:
p.circle('x', 'y', fill_color={'field': 'c',
'transform': mapper},
line_color=None, size='size', source=source)
else:
p.circle('x', 'y', color='color', size='size', source=source)
p.add_tools(HoverTool(tooltips=[('x', '#x'), ('y', '#y'),
('desc', '#name')]))
if add_color_bar:
color_bar_label = Label(text=color.value.title(),
angle=ninety_degrees,
text_color='black',
text_font_style='bold',
text_font_size='20px',
x=25, y=300,
x_units='screen', y_units='screen')
p.add_layout(color_bar, 'right')
p.add_layout(color_bar_label, 'right')
return p
def update(attr, old, new):
layout.children[1] = create_figure()
x = Select(title='X-Axis', value='mpg', options=columns)
x.on_change('value', update)
y = Select(title='Y-Axis', value='hp', options=columns)
y.on_change('value', update)
size = Select(title='Size', value='None',
options=['None'] + quantileable)
size.on_change('value', update)
color = Select(title='Color', value='None',
options=['None'] + quantileable)
color.on_change('value', update)
controls = widgetbox([x, y, color, size], width=200)
layout = row(controls, create_figure())
curdoc().add_root(layout)
curdoc().title = "Crossfilter"
You can add a vertical label to the Colorbar by plotting it on a separate axis and adding a title to this axis. To illustrate this, here's a modified version of Bokeh's standard Colorbar example (found here):
import numpy as np
from bokeh.plotting import figure, output_file, show
from bokeh.models import LogColorMapper, LogTicker, ColorBar
from bokeh.layouts import row
plot_height = 500
plot_width = 500
color_bar_height = plot_height + 11
color_bar_width = 180
output_file('color_bar.html')
def normal2d(X, Y, sigx=1.0, sigy=1.0, mux=0.0, muy=0.0):
z = (X-mux)**2 / sigx**2 + (Y-muy)**2 / sigy**2
return np.exp(-z/2) / (2 * np.pi * sigx * sigy)
X, Y = np.mgrid[-3:3:100j, -2:2:100j]
Z = normal2d(X, Y, 0.1, 0.2, 1.0, 1.0) + 0.1*normal2d(X, Y, 1.0, 1.0)
image = Z * 1e6
color_mapper = LogColorMapper(palette="Viridis256", low=1, high=1e7)
plot = figure(x_range=(0,1), y_range=(0,1), toolbar_location=None,
width=plot_width, height=plot_height)
plot.image(image=[image], color_mapper=color_mapper,
dh=[1.0], dw=[1.0], x=[0], y=[0])
Now, to make the Colorbar, create a separate dummy plot, add the Colorbar to the dummy plot and place it next to the main plot. Add the Colorbar label as the title of the dummy plot and center it appropriately.
color_bar = ColorBar(color_mapper=color_mapper, ticker=LogTicker(),
label_standoff=12, border_line_color=None, location=(0,0))
color_bar_plot = figure(title="My color bar title", title_location="right",
height=color_bar_height, width=color_bar_width,
toolbar_location=None, min_border=0,
outline_line_color=None)
color_bar_plot.add_layout(color_bar, 'right')
color_bar_plot.title.align="center"
color_bar_plot.title.text_font_size = '12pt'
layout = row(plot, color_bar_plot)
show(layout)
This gives the following output image:
One thing to look out for is that color_bar_width is set wide enough to incorporate both the Colorbar, its axes labels and the Colorbar label. If the width is set too small, you will get an error and the plot won't render.
As of Bokeh 0.12.10 there is no built in label available for colorbars. In addition to your approach or something like it, another possibility would be a custom extension, though that is similarly not trivial.
Offhand, a colobar label certainly seems like a reasonable thing to consider. Regarding the notion that it ought to be trivially available, if you polled all users about what they consider should be trivially available, there will be thousands of different suggestions for what to prioritize. As is very often the case in the OSS world, there are far more possible things to do, than there are people to do them (less than 3 in this case). So, would first suggest a GitHub Issue to request the feature, and second, if you have the ability, volunteering to help implement it. Your contribution would be valuable and appreciated by many.
I'm trying to draw sigmoid function using this code on scilab, but the result I got is not from the equation. what's wrong with my code?
x = -6:1:6;
y = 1/(1+%e^-x)
y =
0.0021340
0.0007884
0.0002934
0.0001113
0.0000443
0.0000196
0.0000106
0.0000072
0.0000060
0.0000055
0.0000054
0.0000053
0.0000053
http://en.wikipedia.org/wiki/Sigmoid_function
thank you so much
Try:
-->function [y] = f(x)
--> y = 1/(1+%e^-x)
-->endfunction
-->x = -6:1:6;
-->fplot2d(x,f)
which yields:
Your approach calculates the pseudoinverse of the (1+%e.^x) vector. You can verify by executing: (1+%e^-x)*y
Here are two things you could do:
x = -6:1:6; y = ones(x)./(1+%e.^-x)
This gives the result you need. This performs element-wise division as expected.
Another approach is:
x = -6:1:6
deff("z = f(x)", "z = 1/(1+%e^-x)")
// The above line is the same as defining a function-
// just as a one liner on the interpreter.
y = feval(x, f)
Both approaches will yield the same result.
With Scilab ≥ 6.1.1, simply
x = (-6:1:6)';
plot(x, 1./(1+exp(-x)))