"ArgumentError: timestamps must be strictly monotonic" in Julia TimeArray constructor - julia

I want to figure out where is the duplicate data which cause this error, but how?
using DataFrames, TimeSeries, CSV
s = "2019-12-25,3
2020-01-01,6
2019-12-25,9
2020-01-02,10
2020-01-03,11
2020-01-04,12
2020-01-02,13
2020-01-02,14"
df=CSV.read(IOBuffer(s), types=[Date,Int], header=["timestamp","V")
ta = TimeArray(df, timestamp=:timestamp)
error message
ERROR: ArgumentError: timestamps must be strictly monotonic
Stacktrace:
[1] (::TimeSeries.var"#_#1#2")(::Bool, ::Type{TimeArray{Int64,1,Date,Array{Int64,1}}}, ::Array{Date,1}, ::Array{Int64,1}, ::Array{Symbol,1}, ::DataFrame) at /home/dlin/.julia/packages/TimeSeries/8Z5Is/src/timearray.jl:81
[2] TimeArray at /home/dlin/.julia/packages/TimeSeries/8Z5Is/src/timearray.jl:65 [inlined]
[3] #TimeArray#3 at /home/dlin/.julia/packages/TimeSeries/8Z5Is/src/timearray.jl:89 [inlined]
[4] TimeArray(::Array{Date,1}, ::Array{Int64,1}, ::Array{Symbol,1}, ::DataFrame) at /home/dlin/.julia/packages/TimeSeries/8Z5Is/src/timearray.jl:89
[5] #TimeArray#3(::Symbol, ::Type{TimeArray}, ::DataFrame) at /home/dlin/.julia/packages/TimeSeries/8Z5Is/src/tables.jl:70
[6] (::Core.var"#kw#Type")(::NamedTuple{(:timestamp,),Tuple{Symbol}}, ::Type{TimeArray}, ::DataFrame) at ./none:0
[7] top-level scope at REPL[239]:1
I want to find out which index caused the error, may similar to
│ Row │ timestamp │ V │
│ │ Date │ Int64 │
├─────┼────────────┼───────┤
│ 1 │ 2019-12-25 │ 3 │
│ 3 │ 2019-12-25 │ 9 │
Or even better find out all non unique value rows
│ Row │ timestamp │ V │
│ │ Date │ Int64 │
├─────┼────────────┼───────┤
│ 1 │ 2019-12-25 │ 3 │
│ 3 │ 2019-12-25 │ 9 │
│ 4 │ 2020-01-02 │ 10 │
│ 7 │ 2020-01-02 │ 13 │
│ 8 │ 2020-01-02 │ 14 │

Remove duplicates and than pass DataFrame to TimeArray:
julia> TimeArray(aggregate(df, :timestamp, minimum, sort=true), timestamp=:timestamp)
2×1 TimeArray{Int64,1,Date,Array{Int64,1}} 2019-12-25 to 2020-01-01
│ │ V_minimum │
├────────────┼───────────┤
│ 2019-12-25 │ 3 │
│ 2020-01-01 │ 6 │
If you have a DataFrame and just want to identify duplicate date values use the nonunique function.
julia> nonunique(df,:timestamp)
3-element Array{Bool,1}:
0
0
1
If you want just the rows unique to the date:
julia> unique(df,:timestamp)
2×2 DataFrame
│ Row │ timestamp │ V │
│ │ Date │ Int64 │
├─────┼────────────┼───────┤
│ 1 │ 2019-12-25 │ 3 │
│ 2 │ 2020-01-01 │ 6 │

By #Przemyslaw Szufel's answer, I figure out the way to find the content, but it is still not perfect, it can't show the original row index and only show the first non unique content.
julia> v=nonunique(df,1)
8-element Array{Bool,1}:
0
0
1
0
0
0
1
1
julia> f=findfirst(v)
3
julia> df[df.Column1 .== df.Column1[f],:]
2×2 DataFrame
│ Row │ Column1 │ Column2 │
│ │ Date │ Int64 │
├─────┼────────────┼─────────┤
│ 1 │ 2019-12-25 │ 3 │
│ 2 │ 2019-12-25 │ 9 │
BTW, I found the "ArgumentError: timestamps must be strictly monotonic" message is not only monotonic, but also "sorted" after check the source code of timearray.jl.

Related

How to read record format json in Julia?

I am able to read a json file and convert into dataframe using below code.
df = open(jsontable, "normal.json") |> DataFrame
normal.json looks like below,
{"col1":["thasin", "hello", "world"],"col2":[1,2,3],"col3":["abc", "def", "ghi"]}
So final df has,
3×3 DataFrame
│ Row │ col1 │ col2 │ col3 │
│ │ String │ Int64 │ String │
├─────┼────────┼───────┼────────┤
│ 1 │ thasin │ 1 │ abc │
│ 2 │ hello │ 2 │ def │
│ 3 │ world │ 3 │ ghi │
But, the same code is not working for record formatted json file.
the format is list like {column -> value}, … , {column -> value}
My sample json
{"billing_account_id":"0139A","credits":[],"invoice":{"month":"202003"},"cost_type":"regular"}
{"billing_account_id":"0139A","credits":[1.45],"invoice":{"month":"202003"},"cost_type":"regular"}
{"billing_account_id":"0139A","credits":[2.00, 3.56],"invoice":{"month":"202003"},"cost_type":"regular"}
Expected output:
billing_account_id cost_type credits invoice
0 0139A regular [] {'month': '202003'}
1 0139A regular [1.45] {'month': '202003'}
2 0139A regular [2.0, 3.56] {'month': '202003'}
This can be done in python like below,
data = []
for line in open("sample.json", 'r'):
data.append(json.loads(line))
print(data)
df=pd.DataFrame(data)
How to do this in Julia?
Note that your file is not a valid JSON (its lines are valid JSON, not the whole file).
You can do this like this:
julia> using DataFrames, JSON3
julia> df = JSON3.read.(eachline("sample.json")) |> DataFrame;
julia> df.credits = Vector{Float64}.(df.credits);
julia> df.invoice = Dict{Symbol,String}.(df.invoice);
julia> df
3×4 DataFrame
│ Row │ billing_account_id │ credits │ invoice │ cost_type │
│ │ String │ Array{Float64,1} │ Dict{Symbol,String} │ String │
├─────┼────────────────────┼────────────────────────────┼────────────────────────┼───────────┤
│ 1 │ 0139A │ 0-element Array{Float64,1} │ Dict(:month=>"202003") │ regular │
│ 2 │ 0139A │ [1.45] │ Dict(:month=>"202003") │ regular │
│ 3 │ 0139A │ [2.0, 3.56] │ Dict(:month=>"202003") │ regular │
The transformations on :credits and :invoice columns are to make them of type that is easy to work with (otherwise they use types that are defined internally by JSON3.jl).
A more advanced option is to do it in one shot by specifying the row schema using a NamedTuple type e.g.:
julia> df = JSON3.read.(eachline("sample.json"),
NamedTuple{(:billing_account_id, :credits, :invoice, :cost_type),Tuple{String,Vector{Float64},Dict{String,String},String}}) |>
DataFrame
3×4 DataFrame
│ Row │ billing_account_id │ credits │ invoice │ cost_type │
│ │ String │ Array{Float64,1} │ Dict{String,String} │ String │
├─────┼────────────────────┼────────────────────────────┼─────────────────────────┼───────────┤
│ 1 │ 0139A │ 0-element Array{Float64,1} │ Dict("month"=>"202003") │ regular │
│ 2 │ 0139A │ [1.45] │ Dict("month"=>"202003") │ regular │
│ 3 │ 0139A │ [2.0, 3.56] │ Dict("month"=>"202003") │ regular │
Unrelated to the julia answer, but in python you can do
pd.read_json("sample.json", lines=True)

How to convert String Series to Datetime Series in Julia

I have a csv file which looks like below,
20×2 DataFrame
│ Row │ Id │ Date │
│ │ Int64 │ String │
├─────┼───────┼────────────┤
│ 1 │ 1 │ 01-01-2010 │
│ 2 │ 2 │ 02-01-2010 │
│ 3 │ 3 │ 03-01-2010 │
│ 4 │ 4 │ 04-01-2010 │
│ 5 │ 5 │ 05-01-2010 │
│ 6 │ 6 │ 06-01-2010 │
│ 7 │ 7 │ 07-01-2010 │
│ 8 │ 8 │ 08-01-2010 │
│ 9 │ 9 │ 09-01-2010 │
│ 10 │ 10 │ 10-01-2010 │
│ 11 │ 11 │ 11-01-2010 │
│ 12 │ 12 │ 12-01-2010 │
│ 13 │ 13 │ 13-01-2010 │
│ 14 │ 14 │ 14-01-2010 │
│ 15 │ 15 │ 15-01-2010 │
│ 16 │ 16 │ 16-01-2010 │
│ 17 │ 17 │ 17-01-2010 │
│ 18 │ 18 │ 18-01-2010 │
│ 19 │ 19 │ 19-01-2010 │
│ 20 │ 20 │ 20-01-2010 │
after reading the csv file date columns is in String type. How to externally convert a string series into Datetime series. In Julia Data Frame docs doesn't talk Anything about TimeSeries.
How to externally convert a series or vector into Datetime format?
Is there anyway I can mention timeseries columns while reading a CSV File?
When reading-in a CSV file you can specify dateformat kwarg in CSV.jl:
CSV.File("your_file_name.csv", dateformat="dd-mm-yyyy") |> DataFrame
On the other hand if your data frame is called df then to convert String to Date in your case use:
using Dates
df.Date = Date.(df.Date, "dd-mm-yyyy")
Here is how I have done it:
First a helper function to convert different string formats.
parse_date(d::AbstractString) = DateTime(d, dateformat"yyyy-mm-dd HH:MM:SS")
parse_date(v::Vector{AbstractString}) = parse_date.(v)
parse_date(v::Vector{String}) = parse_date.(v)
parse_date(v::Vector{String31}) = parse_date(String.(v))
using Pipe, TimeSeries
prices = #pipe CSV.File(filename; header = 1, delim = ",") |>
TimeArray(_; timestamp = :Date, timeparser = parse_date)

Aggregate a data frame by values of column that meet a condition

Suppose I have the following DataFrame in Julia, named A:
│ Row │ x1 │ x2 │
├──────┼─────┼─────────┤
│ 1 │ 1.0 │ 5.78341 │
│ 2 │ 2.0 │ 5.05401 │
│ 3 │ 3.0 │ 4.79754 │
│ 4 │ 4.0 │ 4.4126 │
│ 5 │ 5.0 │ 4.29433 │
│ 6 │ 6.0 │ 4.14306 │
│ 7 │ 1.0 │ 5.94811 │
│ 8 │ 2.0 │ 5.0432 │
│ 9 │ 3.0 │ 4.78697 │
│ 10 │ 4.0 │ 4.40384 │
│ 11 │ 5.0 │ 4.29901 │
?
│ 3933 │ 2.0 │ 4.90528 │
│ 3934 │ 3.0 │ 4.57429 │
│ 3935 │ 4.0 │ 4.3988 │
│ 3936 │ 5.0 │ 4.19076 │
│ 3937 │ 6.0 │ 4.09517 │
│ 3938 │ 7.0 │ 3.96192 │
│ 3939 │ 1.0 │ 5.88878 │
│ 3940 │ 2.0 │ 5.87492 │
│ 3941 │ 3.0 │ 4.9453 │
│ 3942 │ 4.0 │ 4.39047 │
│ 3943 │ 5.0 │ 4.28096 │
│ 3944 │ 6.0 │ 4.13686 │
I want to calculate the mean of x2 values by x1 values only if the number of repetitions of x1 values in less or equal than 500, for example. I tried the following code, but it didn't work:
aggregate(A,length(:x1).<=500,mean)
If for example, only the values 1,2 and 3 meet the condition, the result should be:
│ Row │ x1 │ x2 │
├──────┼─────┼─────────┤
│ 1 │ 1.0 │ 5.85264 │
│ 2 │ 2.0 │ 5.15852 │
│ 3 │ 3.0 │ 4.92586 │
where the x2 values are the corresponding mean values.
Any suggestions?
I would use DataFramesMeta.jl here as it will be cleaner than using DataFrames.jl only functionality (I give two ways to obtain the desired result as examples):
using DataFramesMeta
# I generate a smaller DataFrame with cutoff of 15 for the example
df = DataFrame(x1=repeat([1,1,2,2,3], inner=10), x2=rand(50))
# first way to do it
#linq df |>
groupby(:x1) |>
where(length(:x1)>15) |>
#based_on(x2=mean(:x2))
# other way to do the same
#linq df |>
by(:x1, x2=mean(:x2), n=length(:x2)) |>
where(:n.>15) |>
select(:x1, :x2)

Passing a vector or dataframe column string date convert to Date format

I have loaded data into a .csv within Julia.
I wish to convert my string Date to Date format:
julia> head(df)
6×7 DataFrames.DataFrame
│ Row │ Date │ Open │ High │ Low │ Close │ Adj_Close │ Volume │
├─────┼────────────┼─────────┼─────────┼─────────┼─────────┼───────────┼─────────┤
│ 1 │ 1993-01-29 │ 43.9687 │ 43.9687 │ 43.75 │ 43.9375 │ 27.6073 │ 1003200 │
│ 2 │ 1993-02-01 │ 43.9687 │ 44.25 │ 43.9687 │ 44.25 │ 27.8036 │ 480500 │
│ 3 │ 1993-02-02 │ 44.2187 │ 44.375 │ 44.125 │ 44.3437 │ 27.8625 │ 201300 │
│ 4 │ 1993-02-03 │ 44.4062 │ 44.8437 │ 44.375 │ 44.8125 │ 28.1571 │ 529400 │
│ 5 │ 1993-02-04 │ 44.9687 │ 45.0937 │ 44.4687 │ 45.0 │ 28.2749 │ 531500 │
│ 6 │ 1993-02-05 │ 44.9687 │ 45.0625 │ 44.7187 │ 44.9687 │ 28.2552 │ 492100 │
The type is:
julia> showcols(df)
6258×7 DataFrames.DataFrame
│ Col # │ Name │ Eltype │ Missing │ Values │
├───────┼───────────┼──────────────────────────────────┼─────────┼───────────────────────────┤
│ 1 │ Date │ Union{Missings.Missing, String} │ 0 │ 1993-01-29 … 2017-12-01 │
│ 2 │ Open │ Union{Float64, Missings.Missing} │ 0 │ 43.9687 … 264.76 │
│ 3 │ High │ Union{Float64, Missings.Missing} │ 0 │ 43.9687 … 265.31 │
│ 4 │ Low │ Union{Float64, Missings.Missing} │ 0 │ 43.75 … 260.76 │
│ 5 │ Close │ Union{Float64, Missings.Missing} │ 0 │ 43.9375 … 264.46 │
│ 6 │ Adj_Close │ Union{Float64, Missings.Missing} │ 0 │ 27.6073 … 264.46 │
│ 7 │ Volume │ Union{Int64, Missings.Missing} │ 0 │ 1003200 … 159947700 │
Right now the Date is a string.
So wish to convert the column a Date format.
trying:
df[:Date, DateFormat("yyyy-mm-dd")]
and
df[df[:Date] = DateFormat("yyyy-mm-dd")]
with error:
MethodError: Cannot convert an object of type DateFormat{Symbol("yyyy-mm-dd"),Tuple{Base.Dates.DatePart{'y'},Base.Dates.Delim{Char,1},Base.Dates.DatePart{'m'},Base.Dates.Delim{Char,1},Base.Dates.DatePart{'d'}}} to an object of type String
This may have arisen from a call to the constructor String(...),
since type constructors fall back to convert methods.
in setindex! at DataFrames\src\dataframe\dataframe.jl:376
in fill! at base\multidimensional.jl:841
REPL
encase my syntax is wrong I make a vector x from the date column:
x = df[:Date]
Date(x, "yyyy-mm-dd")
MethodError: Cannot convert an object of type Array{Union{Missings.Missing, String},1} to an object of type Int64
REPL
This is easy with R but Julia can not find that great of information, any assistance appreciated.
I am also following this link:
https://docs.julialang.org/en/release-0.4/manual/dates/
Here is an example:
julia> df = Dates.DateFormat("y-m-d");
julia> dt = Date("2015-01-01",df)
2015-01-01
julia> dt2 = Date("2015-01-02",df)
2015-01-02
Why cant I pass a vector or data frame column through this??
Update:
This works when I pass one element from the vector:
julia> Date(x[1], Dates.DateFormat("yyyy-mm-dd"))
1993-01-29
I just want to convert every element to this format and store in the data frame
Simply write Date.(x, Dates.DateFormat("yyyy-mm-dd")) to get what you want.
Notice the . after Date - it tells Julia to apply Date function to all elements of x and Dates.DateFormat("yyyy-mm-dd") will be reused in every call as it is a scalar.
The details are explained here https://docs.julialang.org/en/latest/base/arrays/#Broadcast-and-vectorization-1.
As a side note if you use latest version of CSV.jl package then it should detect Date type automatically:
julia> data="""Date,Open,High,Low,Close,Adj_Close,Volume
1993-01-29,43.9687,43.9687,43.75,43.9375,27.6073,1003200
1993-02-01,43.9687,44.25,43.9687,44.25,27.8036,480500
1993-02-02,44.2187,44.375,44.125 ,44.3437,27.8625,201300"""
"Date,Open,High,Low,Close,Adj_Close,Volume\n1993-01-29,43.9687,43.9687,43.75,43.9375,27.6073,1003200\n1993-02-01,43.9687,44.25,43.9687,44.25,27.8036,480500\n1993-02-02,44.2187,44.375,44.125 ,44.3437,27.8625,201300"
julia> showcols(CSV.read(IOBuffer(data)))
3×7 DataFrames.DataFrame
│ Col # │ Name │ Eltype │ Missing │ Values │
├───────┼───────────┼──────────────────────────────────┼─────────┼───────────────────────────┤
│ 1 │ Date │ Union{Date, Missings.Missing} │ 0 │ 1993-01-29 … 1993-02-02 │
│ 2 │ Open │ Union{Float64, Missings.Missing} │ 0 │ 43.9687 … 44.2187 │
│ 3 │ High │ Union{Float64, Missings.Missing} │ 0 │ 43.9687 … 44.375 │
│ 4 │ Low │ Union{Float64, Missings.Missing} │ 0 │ 43.75 … 44.125 │
│ 5 │ Close │ Union{Float64, Missings.Missing} │ 0 │ 43.9375 … 44.3437 │
│ 6 │ Adj_Close │ Union{Float64, Missings.Missing} │ 0 │ 27.6073 … 27.8625 │
│ 7 │ Volume │ Union{Int64, Missings.Missing} │ 0 │ 1003200 … 201300 │
And even if it would not you can pass types argument (in example below it avoids an union with Missing if you would not want this for some reason):
julia> showcols(CSV.read(IOBuffer(data), types=[String; fill(Float64, 5); Int]))
3×7 DataFrames.DataFrame
│ Col # │ Name │ Eltype │ Missing │ Values │
├───────┼───────────┼─────────┼─────────┼───────────────────────────┤
│ 1 │ Date │ String │ 0 │ 1993-01-29 … 1993-02-02 │
│ 2 │ Open │ Float64 │ 0 │ 43.9687 … 44.2187 │
│ 3 │ High │ Float64 │ 0 │ 43.9687 … 44.375 │
│ 4 │ Low │ Float64 │ 0 │ 43.75 … 44.125 │
│ 5 │ Close │ Float64 │ 0 │ 43.9375 … 44.3437 │
│ 6 │ Adj_Close │ Float64 │ 0 │ 27.6073 … 27.8625 │
│ 7 │ Volume │ Int64 │ 0 │ 1003200 … 201300 │
EDIT: Under DataFrames.jl version 0.14 or later use describe instead of showcols.
Here is what I came up with:
# Pull date column and store in vector
x = df[:Date]
# loop to iterate through each element in vector, converting to Date format
v = []
for i in 1:length(x)
z = Date(x[i], Dates.DateFormat("yyyy-mm-dd"))
push!(v,z)
end
# Check format
julia> v[1] - v[3]
-4 days
# cbind() R equivalent hcat() to existing data frame
df = hcat(df,v)
With the output:
julia> head(df)
6×8 DataFrames.DataFrame
│ Row │ Date │ Open │ High │ Low │ Close │ Adj_Close │ Volume │ x1 │
├─────┼────────────┼─────────┼─────────┼─────────┼─────────┼───────────┼─────────┼────────────┤
│ 1 │ 1993-01-29 │ 43.9687 │ 43.9687 │ 43.75 │ 43.9375 │ 27.6073 │ 1003200 │ 1993-01-29 │
│ 2 │ 1993-02-01 │ 43.9687 │ 44.25 │ 43.9687 │ 44.25 │ 27.8036 │ 480500 │ 1993-02-01 │
│ 3 │ 1993-02-02 │ 44.2187 │ 44.375 │ 44.125 │ 44.3437 │ 27.8625 │ 201300 │ 1993-02-02 │
│ 4 │ 1993-02-03 │ 44.4062 │ 44.8437 │ 44.375 │ 44.8125 │ 28.1571 │ 529400 │ 1993-02-03 │
│ 5 │ 1993-02-04 │ 44.9687 │ 45.0937 │ 44.4687 │ 45.0 │ 28.2749 │ 531500 │ 1993-02-04 │
│ 6 │ 1993-02-05 │ 44.9687 │ 45.0625 │ 44.7187 │ 44.9687 │ 28.2552 │ 492100 │ 1993-02-05 │

How to change the color key using Gadfly's `Scale.color_discrete_manual` in Julia?

I have imported a DataFrame as below:
julia> df
100×3 DataFrames.DataFrame
│ Row │ ex1 │ ex2 │ admit │
├─────┼─────────┼─────────┼───────┤
│ 1 │ 34.6237 │ 78.0247 │ 0 │
│ 2 │ 30.2867 │ 43.895 │ 0 │
│ 3 │ 35.8474 │ 72.9022 │ 0 │
│ 4 │ 60.1826 │ 86.3086 │ 1 │
│ 5 │ 79.0327 │ 75.3444 │ 1 │
│ 6 │ 45.0833 │ 56.3164 │ 0 │
│ 7 │ 61.1067 │ 96.5114 │ 1 │
│ 8 │ 75.0247 │ 46.554 │ 1 │
⋮
│ 92 │ 90.4486 │ 87.5088 │ 1 │
│ 93 │ 55.4822 │ 35.5707 │ 0 │
│ 94 │ 74.4927 │ 84.8451 │ 1 │
│ 95 │ 89.8458 │ 45.3583 │ 1 │
│ 96 │ 83.4892 │ 48.3803 │ 1 │
│ 97 │ 42.2617 │ 87.1039 │ 1 │
│ 98 │ 99.315 │ 68.7754 │ 1 │
│ 99 │ 55.34 │ 64.9319 │ 1 │
│ 100 │ 74.7759 │ 89.5298 │ 1 │
I want to plot this DataFrame using ex1 as x-axis, ex2 as y-axis. In addition, the data is categorized by the third column :admit, so I want to give dots different colors based on the :admit value.
I used Scale.color_discrete_manual to set up colors, and I tried to use Guide.manual_color_key to change the color key legend. However it turns out Gadfly made two color keys.
p = plot(df, x = :ex1, y = :ex2, color=:admit,
Scale.color_discrete_manual(colorant"deep sky blue",
colorant"light pink"),
Guide.manual_color_key("Legend", ["Failure", "Success"],
["deep sky blue", "light pink"]))
My question is how to change the color key legend when using Scale.color_discrete_manual?
One related question is Remove automatically generated color key in Gadfly plot, where the best answer suggests to use two layers plus Guide.manual_color_key. Is there a better solution for using DataFrame and Scale.color_discrete_manual?
Currently, it looks like users cannot customize the color legend generated by color or Scale.color_discrete_manual based on the discussion.
From the same source, Mattriks suggested to use an extra column as "label". Although it is not "natural" for changing color key, it works pretty well.
Therefore, for the same dataset in the problem. We add one more column:
df[:admission] = map(df[:admit])do x
if x == 1
return "Success"
else
return "Failure"
end
end
julia> df
100×4 DataFrames.DataFrame
│ Row │ exam1 │ exam2 │ admit │ admission │
├─────┼─────────┼─────────┼───────┼───────────┤
│ 1 │ 34.6237 │ 78.0247 │ 0 │ "Failure" │
│ 2 │ 30.2867 │ 43.895 │ 0 │ "Failure" │
│ 3 │ 35.8474 │ 72.9022 │ 0 │ "Failure" │
│ 4 │ 60.1826 │ 86.3086 │ 1 │ "Success" │
│ 5 │ 79.0327 │ 75.3444 │ 1 │ "Success" │
│ 6 │ 45.0833 │ 56.3164 │ 0 │ "Failure" │
│ 7 │ 61.1067 │ 96.5114 │ 1 │ "Success" │
│ 8 │ 75.0247 │ 46.554 │ 1 │ "Success" │
⋮
│ 92 │ 90.4486 │ 87.5088 │ 1 │ "Success" │
│ 93 │ 55.4822 │ 35.5707 │ 0 │ "Failure" │
│ 94 │ 74.4927 │ 84.8451 │ 1 │ "Success" │
│ 95 │ 89.8458 │ 45.3583 │ 1 │ "Success" │
│ 96 │ 83.4892 │ 48.3803 │ 1 │ "Success" │
│ 97 │ 42.2617 │ 87.1039 │ 1 │ "Success" │
│ 98 │ 99.315 │ 68.7754 │ 1 │ "Success" │
│ 99 │ 55.34 │ 64.9319 │ 1 │ "Success" │
│ 100 │ 74.7759 │ 89.5298 │ 1 │ "Success" │
Then color the data using this new column Scale.color_discrete_manual:
plot(df, x = :exam1, y = :exam2, color = :admission,
Scale.color_discrete_manual(colorant"deep sky blue",
colorant"light pink"))

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