How use Dataframes columns as a z axis for 3d plot

New to Julia
1

Hello,
I have a TimeArray in which the first column is the timestamp and all other columns are related to energy cost for a load over time. I have about 100 other columns. I’m trying to plot the other columns as a z axis, so I have something like: x_axis as TimeStamp, y_axis as values, z_axis as columns names/labels. I have named all data columns as [“c1”, “c2” etc].

Just don’t have idea if its possible or how to do this.

Anyone could help on this?

Thanks in advance!

2

DataFrame columns are just vectors. So nothing about DataFrames is needed here. Maybe this example in Plots.jl can help. Instead of x, y, and z, use the columns you want.

3

Comparativo gráficos
Comparativo gráficos1343×438 49.2 KB

Hi, thanks for the answer. But I’m afraid its not what I’m looking for. Heatmap gave me the black graph on the right. I want something more like the left graph but in 3D and for all 121 columns.
In the left graph i showed only 5 of them.

4

You mean like this? https://www.r-graph-gallery.com/294-basic-ridgeline-plot.html

5

visualy this 3D graph would do the job! But not using the density function. Data I have isn’t a distribution or related. I tried for the last 2 days to draw this graph but its not working.

I’m trying to plot now with CairoMakie.jl Any other suggestion?

7

I don’t think I’ve ever seen one of those plots that aren’t a density plots, in R or Julia. The best option might be to abandon the ridge plot and just do a traditional line plot with different colors.

8

so I have something like: x_axis as TimeStamp, y_axis as values, z_axis as columns names/labels

Can you provide an example of your data? I’m not sure I understand what you are hoping to portray on the Z-Axis.

9

Would successive 3D line plots be adequate?
Plot3d_DataFrame

using Plots, Dates, DataFrames

dts = today():Year(1):today()+Year(20)
m = length(dts)
n = 100
vs = rand(m,n) .+ vec(1:n)' .- n÷2
df = DataFrame([dts vs], :auto)
p = plot(legend=false, xlabel="Date",ylabel="curve#",zlabel="values")
[plot!(df[:,1], fill(i,m), df[:,i+1]) for i in 1:n]
display(p)
1 Like
10

This code worked fine.

function plot_curvas_custo(custo::TimeArray)
	m = length(timestamp(custo))
	n = length(colnames(custo))
	df = values(custo)
	p = plot(legend=false, xlabel="Horário",ylabel="#Curva",zlabel="R\$")
	[plot!(timestamp(custo), fill(i,m), df[:,i]) for i in 1:div(n,10):n]
	plot(p)
end

Gráficos 2
Gráficos 2709×411 47.8 KB

Thanks to all that reply!!!

1 Like