CODEWITHSUNDEEP
r
Harjeet Singh Bariar
Harjeet Singh Bariar

Reputation: 130

How to concatenate a character to a column based on a condition in R

Hi i have this data frame 

       Table_Number.130. ID_HOUR Date   Time_.EST.
1                   137     480  365       600
2                   340       0  365       1612
3                   340       0  365       1619
4                   340       0  365       657
5                   340       0  365       700
6                   129      60  365       700
7                   340       0  365       1703
8                   340       0  365       1709
9                   340       0  365       1740
10                  340       0  365       1755
11                  129      60  365       800
12                  340       0  365       804
13                  340       0  365       811
14                  340       0  365       1817
15                  340       0  365       1825
16                  340       0  365       1833
17                  340       0  365       1839
18                  340       0  365       1848
19                  340       0  365       1857
20                  129      60  365       1900

I am trying to convert Time_.EST. field into time format but when i run 

x[,"Time_.EST."]<-format(as.Date(x[,"Time_.EST."], format="%H%M" ), format = "%H:%M")

This command shows N/A where there are 3 characters in Time_.EST. field but works fine for 4 characters. Is there a workaround for this.

Upvotes: 1

Views: 243

Answers (2)

mmarks
mmarks

Reputation: 1219

You could pad out to a set number of digits using the sprintf function as below:

newtime <- sprintf("%04d",times$time)
#Make a new list with at least 4 digits: adds leading zeroes as needed
times <- cbind(newtime,times)
#Combine the two together

#A single step would also work
times$newtimes <- sprintf("%04d",times$time)

Then you will always have a 4 digit number to work with

Upvotes: 1

Gregor Thomas
Gregor Thomas

Reputation: 145765

This will paste a "0" on if there are just 3 characters in the column.

x[["Time_.EST."]] = ifelse(nchar(x[["Time_.EST."]]) == 3,
                           paste0("0", x[["Time_.EST."]]),
                           x[["Time_.EST."]])

Upvotes: 2

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