Convert age entered as 'X Weeks, Y Days, Z hours' in R

Question

I have an age variable containing observations that follow this (inconsistent) format:

3 weeks, 2 days, 4 hours
4 weeks, 6 days, 12 hours
3 days, 18 hours
4 days, 3 hours
7 hours
8 hours

I need to convert each observation to hours using R.

I have used strsplit(vector, ',') to split the variable at each comma.

I am running trouble because splitting each observation at the ',' yields anywhere from 1 to 3 entries for each observation. I do not know how to properly index these entries so that I end up with one row for each observation.

I am guessing that once I am able to store these values in sensible rows, I can extract the numeric data from each column in a row and convert accordingly, then sum the entire row.

I am also open to any different methods of approaching this problem.

Answer 1

After you split your data you can parse the resulting list for the keywords defining the times like 'hours', 'weeks', 'days' and create a dataframe containing the relevant value (or 0 if there is no value for a certain keyword). You can achieve that with something like this:

library(dplyr)
vector = c("3 weeks, 2 days, 4 hours", "4 weeks, 6 days, 12 hours", "3 days, 18 hours", "4 days, 3 hours", "7 hours", "8 hours")
split_vector = strsplit(vector, ",", fixed = TRUE)


parse_string = function(i){
  x = split_vector[[i]]
  data_frame(ID = i) %>% 
    mutate(hours = ifelse(any(grepl("hours", x)), as.numeric(gsub("\\D", "", x[grepl("hours", x)])), 0),
           days = ifelse(any(grepl("days", x)), as.numeric(gsub("\\D", "", x[grepl("days", x)])), 0),
           weeks = ifelse(any(grepl("weeks", x)), as.numeric(gsub("\\D", "", x[grepl("weeks", x)])), 0))
}

all_parsed = lapply(1:length(split_vector),  parse_string)
all_parsed = rbind_all(all_parsed) %>% 
  mutate(final_hours = hours + days * 24 + weeks * 7 * 24)

Answer 2

Hadleyverse comes to the rescue again:

library(lubridate)
library(stringr)

dat <- readLines(textConnection(" 3   weeks,   2  days,  4 hours
 4 week,  6 days,  12 hours 
3 days, 18 hours
4 day, 3 hours
 7 hours
8  hour"))

sapply(str_split(str_trim(dat), ",[ ]*"), function(x) {
  sum(sapply(x, function(y) {
    bits <- str_split(str_trim(y), "[ ]+")[[1]]
    duration(as.numeric(bits[1]), bits[2])
  })) / 3600
})

## [1] 556 828  90  99   7   8

I whacked the data a bit to show it's also somewhat flexible in how it parses things. I rly don't think the second str_trim is absolutely necessary but didn't have cycles to verify.

The exposition is that it trims the original vector then splits it into components (which makes a list of vectors). That list is then iterated over and the individual vector elements are further trimmed and split into # and unit duration. That's passed to lubridate and the value is returned and automatically converted to numeric seconds by the call to sum and we then make it into hours.