R strip在数据帧中分割一列

Question

I have a 'data' frame, with multiple columns, one of them being 'Runtime' which has data in two formats:

Runtime
1 h 10 min
67 min
1 h 0 min
86 min
97 min

I want to convert all of them into Minutes. Have tried 'strsplit' and 'strip_split_fixed'. Can anyone show me a way to achieve my goal, split or any other method?

Thank you in advance !

Answer 1

I think I saw this kind of solution somewhere. Don't hit me.

df = data.frame(Runtime = c('1 h 10 min', '67 min', '1 h 0 min', '86 min', '97 min'))

df$exp <- gsub("h", "* 60 +", df$Runtime)
df$exp <- gsub("min", "* 1", df$exp)

sapply(df$exp, FUN = function(x) eval(parse(text = x)))

1 * 60 + 10 * 1          67 * 1  1 * 60 + 0 * 1          86 * 1          97 * 1 
             70              67              60              86              97 

Answer 2

You can get it one call using gsubfn and regex:

library(gsubfn)
gsubfn("^(?:(\\d+)\\s*h)?\\s*(\\d+)\\s*min.*$",
 ~ sum(as.numeric(x) * 60, as.numeric(y), as.numeric(z), na.rm=TRUE), x)
#[1] "70" "67" "60" "86" "97"

Answer 3

Here's an example of how you can do it:

# setting up your data.frame of interest
df = data.frame(Runtime = c('1 h 10 min', '67 min', '1 h 0 min', '86 min', '97 min'))



df$Runtime = gsub(' min', '', df$Runtime) # remove the min labels
hrs = grepl('h', x = df$Runtime) # which values are in an "x h y min" format?
runtime_sub = sapply(strsplit(df[hrs, 'Runtime'], ' h '), function(i) sum(as.numeric(i) * c(60, 1))) # convert the "x h y min" entries into numeric values in minutes
df$Runtime = as.numeric(df$Runtime) # convert the vector to numeric (yes, it's supposed to return a warning. Ignore it.
df[hrs, 'Runtime'] = runtime_sub # add the converted values

This results in:

  Runtime
1      70
2      67
3      60
4      86
5      97

Answer 4

1) Read df[[1]] and if the third column is NA then the first column gives the minutes; otherwise, 60 times the first column plus the third column gives the minutes:

with(read.table(text = as.character(df[[1]]), fill = TRUE), 
        ifelse(is.na(V3), V1, 60*V1 + V3))
## [1] 70 67 60 86 97

2) A variation is to paste "0 h" at the beginning of each component that does not have an h giving hm and read that computing 60 times the first column plus the third column.

hm <- paste(ifelse(grepl("h", df[[1]]), "", "0 h"), df[[1]])
with(read.table(text = hm), 60 * V1 + V3)
## [1] 70 67 60 86 97