r - How can I bin/bucket a data.frame by birthyears

Question

I have a data.frame with three columns: a token, year of birth and number of contacts. The birthyears range from 1934 to 2020 and I don't want individual years but 5-year groups like 2000-2005, 2006-2010 and so on to later visualize the contact count per age group.

I already found the cut function like this:

# set up cut-off values 
breaks <- c(0,2,4,6,8,10,12,14,16,18,20)
# specify interval/bin labels
tags <- c("[0-2)","[2-4)", "[4-6)", "[6-8)", "[8-10)", "[10-12)","[12-14)", "[14-16)","[16-18)", "[18-20)")
# bucketing values into bins
group_tags <- cut(v$MeanEducation, 
                  breaks=breaks, 
                  include.lowest=TRUE, 
                  right=FALSE, 
                  labels=tags)

However in this example I'd have to set a vector of breaks and labels manually.

Is there a solution to automize this? Like beginning the first bucket at the next lower by 5 dividable year than the minimum in my dataframe. Analogue at the top end.

Thanks in advance!

Answer 1

The tags that you have defined are created by default by cut function you don't have to add them manually. Moreover, you can use seq to create sequence of breaks and paste to generate labels programatically.

#Generate data
set.seed(123)
x <- sample(10)
x
#[1]  3 10  2  8  6  9  1  7  5  4
#Create breaks
breaks <- seq(0, 10, 2)
#Create labels
labels <- paste(head(breaks, -1), tail(breaks, -1), sep = '-')

#Without labels
cut(x, breaks)

#[1] (2,4]  (8,10] (0,2]  (6,8]  (4,6]  (8,10] (0,2]  (6,8]  (4,6]  (2,4] 
#Levels: (0,2] (2,4] (4,6] (6,8] (8,10]

#With labels
cut(x, breaks, labels)
#[1] 2-4  8-10 0-2  6-8  4-6  8-10 0-2  6-8  4-6  2-4 
#Levels: 0-2 2-4 4-6 6-8 8-10