Categorize numeric variable into group/ bins/ breaks

Question

I am trying to categorize a numeric variable (age) into groups defined by intervals so it will not be continuous. I have this code:

data$agegrp(data$age >= 40 & data$age <= 49) <- 3
data$agegrp(data$age >= 30 & data$age <= 39) <- 2
data$agegrp(data$age >= 20 & data$age <= 29) <- 1

the above code is not working under survival package. It's giving me:

invalid function in complex assignment

Can you point me where the error is? data is the dataframe I am using.

Orchidectomy

Use [ for subsetting, not (. – Apomorphine

The function you'll want to use is cut. – Romanist

@joan can you show me how it is done using cut? – Orchidectomy

The answer depends on What result do you want? a) just an integer (or NA) b) factor labels or indeed c) an array of dichotomized/ dummy variables? findInterval() can only do the first, whereas cut() does both. findInterval() is faster (O(log(no. of bins)) although that's rarely an issue. – Letti

Answer 1

I would use findInterval() here:

First, make up some sample data

set.seed(1)
ages <- floor(runif(20, min = 20, max = 50))
ages
# [1] 27 31 37 47 26 46 48 39 38 21 26 25 40 31 43 34 41 49 31 43

Use findInterval() to categorize your "ages" vector.

findInterval(ages, c(20, 30, 40))
# [1] 1 2 2 3 1 3 3 2 2 1 1 1 3 2 3 2 3 3 2 3

Alternatively, as recommended in the comments, cut() is also useful here:

cut(ages, breaks=c(20, 30, 40, 50), right = FALSE)
cut(ages, breaks=c(20, 30, 40, 50), right = FALSE, labels = FALSE)

Allanadale

@leian, have you tried the code? It should. However, when asking questions here in the R tag at SO, it is best to include a minimal reproducible example if you want more targeted help. – Allanadale

but what will be the variable name of the result of this findInterval()? – Orchidectomy

Whatever you want it to be! From your example, I would assume you would do something like data$agegrp <- findInterval(data$age, c(20, 30, 40)). – Allanadale

Answer 2

We can use dplyr:

library(dplyr)

data <- data %>% mutate(agegroup = case_when(age >= 40  & age <= 49 ~ '3',
                                             age >= 30  & age <= 39 ~ '2',
                                             age >= 20  & age <= 29 ~ '1')) # end function

Compared to other approaches, dplyr is easier to write and interpret.

Abstention

You can also use cut in mutate instead of case_when. Eg. data %>% mutate(agegroup = cut(ages, breaks = c(20, 30, 40, 50), right = T, labels = F)) – Filigree

@Filigree This is /such/ a good answer, many thanks. labels=TRUE will even give reasonable labels. – Immix

Answer 3

This answer provides two ways to solve the problem using the data.table package, which would greatly improve the speed of the process. This is crucial if one is working with large data sets.

1s Approach: an adaptation of the previous answer but now using data.table + including labels:

library(data.table)

agebreaks <- c(0,1,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,500)
agelabels <- c("0-1","1-4","5-9","10-14","15-19","20-24","25-29","30-34",
               "35-39","40-44","45-49","50-54","55-59","60-64","65-69",
               "70-74","75-79","80-84","85+")

setDT(data)[ , agegroups := cut(age, 
                                breaks = agebreaks, 
                                right = FALSE, 
                                labels = agelabels)]

2nd Approach: This is a more wordy method, but it also makes it more clear what exactly falls within each age group:

setDT(data)[age <1, agegroup := "0-1"]
data[age >0 & age <5, agegroup := "1-4"]
data[age >4 & age <10, agegroup := "5-9"]
data[age >9 & age <15, agegroup := "10-14"]
data[age >14 & age <20, agegroup := "15-19"]
data[age >19 & age <25, agegroup := "20-24"]
data[age >24 & age <30, agegroup := "25-29"]
data[age >29 & age <35, agegroup := "30-34"]
data[age >34 & age <40, agegroup := "35-39"]
data[age >39 & age <45, agegroup := "40-44"]
data[age >44 & age <50, agegroup := "45-49"]
data[age >49 & age <55, agegroup := "50-54"]
data[age >54 & age <60, agegroup := "55-59"]
data[age >59 & age <65, agegroup := "60-64"]
data[age >64 & age <70, agegroup := "65-69"]
data[age >69 & age <75, agegroup := "70-74"]
data[age >74 & age <80, agegroup := "75-79"]
data[age >79 & age <85, agegroup := "80-84"]
data[age >84, agegroup := "85+"]

Although the two approaches should give the same result, I prefer the 1st one for two reasons. (a) It is shorter to write and (2) the age groups are ordered in the correct way, which is crucial when it comes to visualizing the data.

Year

The second approach doesn't seem to work with R 3.2. It gives an error could not find function ":=" – Looker

It works for me. Make sure you load the data.table library library(data.table) ; and that you are working with a data.table (not a data frame) setDT(your_dataframe) # convert your DF into a data.table – Year

Answer 4

Let's say that your ages were stored in the dataframe column labeled age. Your dataframe is df, and you want a new column age_grouping containing the "bucket" that your ages fall in.

In this example, suppose that your ages ranged from 0 -> 100, and you wanted to group them every 10 years. The following code would accomplish this by storing these intervals in a new age grouping column:

df$age_grouping <- cut(df$age, seq(0, 100, 10))

Malvoisie