CODEWITHSUNDEEP
dataframejulia
Morpheu5
Morpheu5

Reputation: 2801

Group DataFrame by binning a column::Float64, in Julia

Say I have a DataFrame with a column of Float64s, I'd like to group the dataframe by binning that column. I hear the cut function might help, but it's not defined over dataframes. Some work has been done (https://gist.github.com/tautologico/3925372), but I'd rather use a library function rather than copy-pasting code from the Internet. Pointers?

EDIT Bonus karma for finding a way of doing this by month over UNIX timestamps :)

Upvotes: 2

Views: 1124

Answers (1)

Ian Marshall
Ian Marshall

Reputation: 739

You could bin dataframes based on a column of Float64s like this. Here my bins are increments of 0.1 from 0.0 to 1.0, binning the dataframe based on a column of 100 random numbers between 0.0 and 1.0.

using DataFrames #load DataFrames
df = DataFrame(index = rand(Float64,100)) #Make a DataFrame with some random Float64 numbers
df_array = map(x->df[(df[:index] .>= x[1]) .& (df[:index] .<x[2]),:],zip(0.0:0.1:0.9,0.1:0.1:1.0)) #Map an anonymous function that gets every row between two numbers specified by a tuple called x, and map that anonymous function to an array of tuples generated using the zip function.

This will produce an array of 10 dataframes, each one with a different 0.1-sized bin.

As for the UNIX timestamp question, I'm not as familiar with that side of things, but after playing around a bit maybe something like this could work:

using Dates

df = DataFrame(unixtime = rand(1E9:1:1.1E9,100)) #Make a dataframe with floats containing pretend unix time stamps
df[:date] = Dates.unix2datetime.(df[:unixtime]) #convert those timestamps to DateTime types
df[:year_month] = map(date->string(Dates.Year.(date))*" "*string(Dates.Month.(date)),df[:date]) #Make a string for every month in your time range
df_array = map(ym->df[df[:year_month] .== ym,:],unique(df[:year_month])) #Bin based on each unique year_month string

Upvotes: 2

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