string manipulation using ifelse within a for loop

Question

I’m trying to solve a problem that at the beginning looked very simple but I've found myself stuck. So, I have an object that is character called “out”:

out<-c( " 59mn 59s", " 59mn 59s", " 1h 0mn",  " 1h 0mn",  " 1h 0mn",  " 1h 0mn",  " 1h 0mn", " 1h 0mn",   " 59mn 59s"," 59mn 59s", " 46mn 42s")

My final goal is to use the function “hms” from the package lubridate to convert the character into time object. The problem is that the data set is not consistent and in some cases I have hours and minutes others I have minutes and seconds. So I made an attempt to uniform the data set like this:

    out<-for(p in 1:length(out)) {                                         
 ifelse(!grepl("h",out[p]),paste("0h",out[p]),ifelse(!grepl("mn",out[p]),paste("0h 0mn",out[p]),ifelse(!grepl("s",out[p]),paste(out[p],"0s"))))                                                
                           }

The ifelse statement by itself works fine however in a for loop it just returns NULL. Does anyone have an idea of what is going wrong in this code? Another option would be a combination of the functions “hm” and “hms” from lubridate but my are skills won’t allow me to go on that path.

Many thanks

beginner has a good answer, but if you would not save the results of your for loop and instead print each iteration, you would see what was going on — rawr, Jul 30 '14 at 13:25
If you want to uniform your data you might want to remove the leading space or the last space present in some strings. ( http://stackoverflow.com/q/10502787/2886003 might be helpful) — llrs, Jul 30 '14 at 14:24

score 2 · Accepted Answer · answered Jul 30 '14 at 13:22

2

Since ifelse is vectorised, you can do this without using a loop:

ifelse(!grepl("h",out), paste("0h",out),
       ifelse(!grepl("mn",out), paste("0h 0mn",out),
              ifelse(!grepl("s",out), paste(out,"0s"), NA)))

#[1] "0h  59mn 59s" "0h  59mn 59s" " 1h 0mn 0s"   " 1h 0mn 0s"   " 1h 0mn 0s"   " 1h 0mn 0s"   " 1h 0mn 0s"  
#[8] " 1h 0mn 0s"   "0h  59mn 59s" "0h  59mn 59s" "0h  46mn 42s"

Is that what you expected?

(Note that I added an NA incase all others conditions are not TRUE and replaced out[p] inside the ifelse with only out.)

answered Jul 30 '14 at 13:22

talat

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I'm pretty sure the second line is actually redundant. – shadow Jul 30 '14 at 13:28
This as to be some kind of record the post was answered in less than 10 minutes and it works perfectly. Many thanks for you help – PatraoPedro Jul 30 '14 at 13:29
@shadow, you may be right, in fact, I didn't spend much time checking the logic of the operation. – talat Jul 30 '14 at 13:30
@PatraoPedro if it works perfectly and answers your question, feel free to accept the answer :) – talat Jul 30 '14 at 14:14

score 2 · Answer 2 · answered Jul 30 '14 at 13:32

2

Using some safer regular expression.

t<-c( " 59mn 59s", " 59mn 59s", " 1h 0mn",  " 1h 0mn",  " 1h 0mn",  " 1h 0mn",  " 1h 0mn", " 1h 0mn",   " 59mn 59s"," 59mn 59s", " 46mn 42s")
t[grep(" [0-9]{2}mn", t)] = paste( "0h", t[grep(" [0-9]{2}mn", t)], sep="")
t[grep("mn$", t)] = paste(t[grep("mn$", t)], " 0s", sep ="")

hms(t)
> [1] "59M 59S"  "59M 59S"  "1H 0M 0S" "1H 0M 0S" "1H 0M 0S" "1H 0M 0S" "1H 0M 0S"
> [8] "1H 0M 0S" "59M 59S"  "59M 59S"  "46M 42S"

answered Jul 30 '14 at 13:32

Vlo

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It does add the hours. You are looking at `hms(t)` instead of `t` – Vlo Jul 30 '14 at 15:10
True, ha. I've been making a point to learn perl's ridiculous regex, so I like answers that use any regular expressions. +1 – rawr Jul 30 '14 at 21:07

string manipulation using ifelse within a for loop

2 Answers2