How to plot histogram subplots of a DataFrame?

folleyz · March 23, 2022, 9:08pm

Hello Everybody!

I have a question, i’m refatoring my old python projects to Julia and i have a doubt in a situation.

Looks how i plot using python matplotlib:

df.hist(figsize=(12,10));

And now using Julia StatsPlots

begin 	

	p1 = Plots.histogram(df.Age; 
		plot_title = "Age",plot_titlefontsize = 11)
	p2 = Plots.histogram(df.Balance; 
		plot_title = "Balance",plot_titlefontsize = 11)
	p3 = Plots.histogram(df.CreditScore; 
		plot_title = "CreditScore",plot_titlefontsize = 11)
	p4 = Plots.histogram(df.CustomerId; 
		plot_title = "CustomerId",plot_titlefontsize = 11)
	p5 = Plots.histogram(df.EstimatedSalary; 
		plot_title = "EstimatedSalary",plot_titlefontsize = 11)
	p6 = Plots.histogram(df.Exited; 
		plot_title = "Exited",plot_titlefontsize = 11)
	p7 = Plots.histogram(df.HasCrCard; 
		plot_title = "HasCrCard",plot_titlefontsize = 11)
	p8 = Plots.histogram(df.IsActiveMember; 
		plot_title = "IsActiveMember",plot_titlefontsize = 11)
	p9 = Plots.histogram(df.NumOfProducts; 
		plot_title = "NumOfProducts",plot_titlefontsize = 11)
	p10 = Plots.histogram(df.RowNumber; 
		plot_title = "RowNumber",plot_titlefontsize = 11)
	p11 = Plots.histogram(df.Tenure; 
		plot_title = "Tenure",plot_titlefontsize = 11)
	


	@df df Plots.plot(p1,p2,p3,p4,p5,p6,p7,p8,p9,p10,p11;
		layout = 11,
		size=(1000,600)
	)
end

How can i improve the Julia code to get that same result without writing a lot of code?

Thx for Help!

and sorry, I’m Julia newbie

icweaver · March 23, 2022, 9:50pm

Hey! Would using the legend as the column label work instead? That way we could just do something like this:

using DataFrames, StatsPlots

df = DataFrame(randn(100, 3), [:Tenure, :Balance, :CreditScore])

N = ncol(df)

@df df histogram(cols(1:N); layout=N)

index

folleyz · March 23, 2022, 10:03pm

Thx @icweaver !

I tried that but returned an error.

begin
	N = ncol(df)

	@df df Plots.histogram(cols(1:N); layout=N)
end

Cannot convert Matrix{Any} to series data for plotting

    error(::String)@error.jl:33
    _prepare_series_data(::Matrix{Any})@series.jl:8
    _series_data_vector(::Matrix{Any}, ::Dict{Symbol, Any})@series.jl:27
    macro expansion@series.jl:127[inlined]
    apply_recipe(::AbstractDict{Symbol, Any}, ::Type{RecipesPipeline.SliceIt}, ::Any, ::Any, ::Any)@RecipesBase.jl:289
    _process_userrecipes!(::Any, ::Any, ::Any)@user_recipe.jl:36
    recipe_pipeline!(::Any, ::Any, ::Any)@RecipesPipeline.jl:70
    _plot!(::Plots.Plot, ::Any, ::Any)@plot.jl:208
    #plot#135@plot.jl:91[inlined]
    var"#histogram#419"(::Base.Pairs{Symbol, V, Tuple{Vararg{Symbol, N}}, NamedTuple{names, T}} where {V, N, names, T<:Tuple{Vararg{Any, N}}}, ::typeof(Plots.histogram), ::Any)@RecipesBase.jl:410
    var"#add_label#19"(::Base.Pairs{Symbol, Int64, Tuple{Symbol}, NamedTuple{(:layout,), Tuple{Int64}}}, ::typeof(StatsPlots.add_label), ::Vector{Matrix{Symbol}}, ::Function, ::Matrix{Any})@df.jl:155
    (::var"#405#406"{Module, Colon, Int64})(::DataFrames.DataFrame)@none:0
    top-level scope@Local: 4[inlined]

What’s my mystake?

icweaver · March 23, 2022, 10:08pm

That is very odd. Could you share a MWE, including the df you are using?

Edit: whoops, linked to to the wrong example above. Should be updated now

folleyz · March 23, 2022, 10:26pm

Here’s a df sample:

first(df,5)

	RowNumber	CustomerId	Surname	CreditScore	Geography	Gender	Age	Tenure	Balance	NumOfProducts	HasCrCard	IsActiveMember	EstimatedSalary	Exited
	Int64	Int64	String31	Int64	String7	String7	Int64	Int64	Float64	Int64	Int64	Int64	Float64	Int64
1	1	15634602	Hargrave	619	France	Female	42	2	0.0	1	1	1	1,01E+10	1

2	2	15647311	Hill	608	Spain	Female	41	1	83807.9	1	0	1	1,13E+10	0

3	3	15619304	Onio	502	France	Female	42	8	1,60E+10	3	1	0	1,14E+10	1

4	4	15701354	Boni	699	France	Female	39	1	0.0	2	0	0	93826.6	0

5	5	15737888	Mitchell	850	Spain	Female	43	2	1,26E+10	1	1	1	79084.1	0

icweaver · March 23, 2022, 10:33pm

Thanks, but I just meant a copy-and-pasteable code snippet that I can run to try and reproduce your error please =]

The package versions that you are using would also be really helpful! To work from a clean environment you can do the following:

julia> ]
pkg> activate --temp
(jl_jTP8xX) pkg> add <list of packages used here>
julia> <MWE here>

Then you can view the list of packages and the versions that were used by just typing st (short for status) into the packge repl (shown by the prompt that started with pkg>):

julia> ]
(jl_jTP8xX) pkg> st
<list of packages used should display here>

folleyz · March 23, 2022, 10:47pm

Code used

begin
	N = ncol(df)

	@df df Plots.histogram(cols(1:N); layout=N)
end

Error:

Cannot convert Matrix{Any} to series data for plotting

    error(::String)@error.jl:33
    _prepare_series_data(::Matrix{Any})@series.jl:8
    _series_data_vector(::Matrix{Any}, ::Dict{Symbol, Any})@series.jl:27
    macro expansion@series.jl:127[inlined]
    apply_recipe(::AbstractDict{Symbol, Any}, ::Type{RecipesPipeline.SliceIt}, ::Any, ::Any, ::Any)@RecipesBase.jl:289
    _process_userrecipes!(::Any, ::Any, ::Any)@user_recipe.jl:36
    recipe_pipeline!(::Any, ::Any, ::Any)@RecipesPipeline.jl:70
    _plot!(::Plots.Plot, ::Any, ::Any)@plot.jl:208
    #plot#135@plot.jl:91[inlined]
    var"#histogram#419"(::Base.Pairs{Symbol, V, Tuple{Vararg{Symbol, N}}, NamedTuple{names, T}} where {V, N, names, T<:Tuple{Vararg{Any, N}}}, ::typeof(Plots.histogram), ::Any)@RecipesBase.jl:410
    var"#add_label#19"(::Base.Pairs{Symbol, Int64, Tuple{Symbol}, NamedTuple{(:layout,), Tuple{Int64}}}, ::typeof(StatsPlots.add_label), ::Vector{Matrix{Symbol}}, ::Function, ::Matrix{Any})@df.jl:155
    (::var"#405#406"{Module, Colon, Int64})(::DataFrames.DataFrame)@none:0
    top-level scope@Local: 4[inlined]

Packages

(jl_tJabzV) pkg> st
      Status `C:\Users\Pedro Henrique\AppData\Local\Temp\jl_tJabzV\Project.toml`
  [336ed68f] CSV v0.10.3
  [a93c6f00] DataFrames v1.3.2
  [91a5bcdd] Plots v1.27.2
  [2913bbd2] StatsBase v0.33.16
  [f3b207a7] StatsPlots v0.14.33

I think i found the problem, and is in the non numerical columns.

begin
	N = ncol(df)

	@df df Plots.histogram(cols(7:N); layout=N)
end

Above i used only columns with numerical values to test and works!

Sorry about all the trouble, I’m not in the area of computer science

icweaver · March 23, 2022, 10:50pm

No worries at all, great find!

rafael.guerra · March 23, 2022, 11:25pm

For the original question, you could also feed the titles to the @df macro to obtain:

CODE

using DataFrames, StatsPlots

colnames = [:Age, :Balance, :CreditScore, :CustomerId, :EstimatedSalary, :Exited,
           :HasCrCard, :IsActiveMember, :NumOfProducts, :RowNumber, :tenure]
N = length(colnames)
df = DataFrame(randn(100, N), colnames)

@df df histogram(cols(1:N); layout=N, legend=false, title=permutedims(colnames),
    size=(1000,600), frame=:box, titlefontsize=11, c=:blues)

rafael.guerra · March 23, 2022, 11:28pm

It is recommended to start a new topic when question is unrelated.

icweaver · March 23, 2022, 11:31pm

As @rafael.guerra mentioned, we would be more than happy to follow up with this in a new thread! It just helps others find the relevant bits of the discussion in the future =]

folleyz · March 23, 2022, 11:40pm

Nice @rafael.guerra and @icweaver !

With your help i learned a lot and created a very simple function who solves my question.

function plot_df(df::AbstractDataFrame,graph_type::Symbol)
	numerical_cols = Symbol.(names(df,Real))

	@df df Plots.plot(cols(numerical_cols); layout = length(numerical_cols), seriestype = graph_type,
	title=permutedims(numerical_cols),
	size=(1200,600),
	frame=:box)
end

plot_df(df,:histogram)

I really appreciate your collaboration, and i will make a new topic for the countplot!

Thx!