CODEWITHSUNDEEP
rdata-conversion
user8687147
user8687147

Reputation: 13

R: How to convert min (string) to int

How do you convert a column with both minutes and hours to an int describing minutes like a df with

df$duration = ["1 h 10 min", "120 min",...]

duration
----------
1 h 10 min
120 min

to

df $duration = [70, 120, ...]

result
------
70
120

Upvotes: 1

Views: 215

Answers (3)

kdopen
kdopen

Reputation: 8215

Use the lubridate package, but you need to clean the data a little by getting all the values into a consistent format.

> df <- data.frame(duration=c("1 h 10 min","120 min"), stringsAsFactors = F)
> no_h<-!grepl("h", df$duration)
> df$duration[no_h] <- paste("0 h", df$duration[no_h])
> df$period <-hm(df$duration)
> df$minute <- hour(df$period)*60 + minute(df$period)
> df
     duration    period minute
1  1 h 10 min 1H 10M 0S     70
2 0 h 120 min   120M 0S    120
>

Upvotes: 2

d.b
d.b

Reputation: 32548

duration = c("1 h 10 min", "120 min")
sapply(strsplit(duration, " "), function(x){
    temp = as.numeric(x)
    if (length(temp) == 4){
        sum(as.numeric(temp[c(1, 3)]) * c(60, 1))
    }else{
        as.numeric(temp[1])
    }
    })
#[1]  70 120
#Warning messages:
#1: In FUN(X[[i]], ...) : NAs introduced by coercion
#2: In FUN(X[[i]], ...) : NAs introduced by coercion

Upvotes: 1

eipi10
eipi10

Reputation: 93851

Here's one option:

library(stringr)

d = c("1 h 10 min", "120 min", "2 h", "12 h 53 min")

na_to_0 = function(x) {x[is.na(x)] = 0; x}

to_minutes = function(s) {

  hr = na_to_0(60 *  as.numeric(str_replace(str_extract(s, "[0-9]{1,2} h"), " h", "")))
  min = na_to_0(as.numeric(str_replace(str_extract(s, "[0-9]{1,3} min"), " min", "")))

  hr + min
}

to_minutes(d)

[1] 70 120 120 773

Upvotes: 1

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