parse values based on groups in R - r

Question

I have a very large dataset and a sample of that looks something like the one below:
| Id | Name | Start_Date | End_Date |
|----|---------|------------|------------|
| 10 | Mark | 4/2/1999 | 7/5/2018 |
| 10 | | 1/1/2000 | 9/24/2018 |
| 25 | | 5/3/1968 | 6/3/2000 |
| 25 | | 6/6/2009 | 4/23/2010 |
| 25 | Anthony | 2/20/2010 | 7/21/2016 |
| 25 | | 9/12/2014 | 11/26/2019 |
I need to parse the names from Name column based on their Id such that the output table looks like:
| Id | Name | Start_Date | End_Date |
|----|---------|------------|------------|
| 10 | Mark | 4/2/1999 | 7/5/2018 |
| 10 | Mark | 1/1/2000 | 9/24/2018 |
| 25 | Anthony | 5/3/1968 | 6/3/2000 |
| 25 | Antony | 6/6/2009 | 4/23/2010 |
| 25 | Anthony | 2/20/2010 | 7/21/2016 |
| 25 | Anthony | 9/12/2014 | 11/26/2019 |
How can I achieve an output as shown above? I went through the substitute and parse functions, but was unable to understand how they apply to this problem.
My dataset would be:
df=data.frame(Id=c("10","10","25","25","25","25"),Name=c("Mark","","","","Anthony",""),
Start_Date=c("4/2/1999", "1/1/2000","5/3/1968","6/6/2009","2/20/2010","9/12/2014"),
End_Date=c("7/5/2018","9/24/2018","6/3/2000","4/23/2010","7/21/2016","11/26/2019"))

Answer 1

We can change the blanks ("") to NA and use fill to replace the NA elements with the previous non-NA element
library(dplyr)
library(tidyr)
df1 %>%
mutate(Name = na_if(Name, "")) %>%
group_by(Id) %>%
fill(Name, .direction = "down") %>%
fill(Name, .direction = "up)
# A tibble: 6 x 4
# Groups: Id [2]
# Id Name Start_Date End_Date
# <chr> <chr> <chr> <chr>
#1 10 Mark 4/2/1999 7/5/2018
#2 10 Mark 1/1/2000 9/24/2018
#3 25 Anthony 5/3/1968 6/3/2000
#4 25 Anthony 6/6/2009 4/23/2010
#5 25 Anthony 2/20/2010 7/21/2016
#6 25 Anthony 9/12/2014 11/26/2019
In the devel version of tidyr (‘0.8.3.9000’), this can be done in a single fill statement as .direction = "downup" is also an option
df1 %>%
mutate(Name = na_if(Name, "")) %>%
group_by(Id) %>%
fill(Name, .direction = "downup")
Or another option is to group by 'Id', and mutate the 'Name' as the first non-blank element
df1 %>%
group_by(Id) %>%
mutate(Name = first(Name[Name!=""]))
# A tibble: 6 x 4
# Groups: Id [2]
# Id Name Start_Date End_Date
# <chr> <chr> <chr> <chr>
#1 10 Mark 4/2/1999 7/5/2018
#2 10 Mark 1/1/2000 9/24/2018
#3 25 Anthony 5/3/1968 6/3/2000
#4 25 Anthony 6/6/2009 4/23/2010
#5 25 Anthony 2/20/2010 7/21/2016
#6 25 Anthony 9/12/2014 11/26/2019
data
df1 <- structure(list(Id = c("10", "10", "25", "25", "25", "25"), Name = c("Mark",
"", "", "", "Anthony", ""), Start_Date = c("4/2/1999", "1/1/2000",
"5/3/1968", "6/6/2009", "2/20/2010", "9/12/2014"), End_Date = c("7/5/2018",
"9/24/2018", "6/3/2000", "4/23/2010", "7/21/2016", "11/26/2019"
)), class = "data.frame", row.names = c(NA, -6L))

Answer 2

Using DF defined reproducibly in the Note at the end, replace each zero-length element of Name with NA and then use na.omit to get the unique non-NA to use to fill. We have assumed that there is only one non-NA per Id which is the case in the question. If not we could replace na.omit with function(x) unique(na.omit(x)) assuming that the non-NAs are all the same within Id. No packages are used.
transform(DF, Name = ave(replace(Name, !nzchar(Name), NA), Id, FUN = na.omit))
giving:
Id Name Start_Date End_Date
1 10 Mark 4/2/1999 7/5/2018
2 10 Mark 1/1/2000 9/24/2018
3 25 Anthony 5/3/1968 6/3/2000
4 25 Anthony 6/6/2009 4/23/2010
5 25 Anthony 2/20/2010 7/21/2016
6 25 Anthony 9/12/2014 11/26/2019
na.strings
We can simplify this slightly if we make sure that the zero length elements of Name are NA in the first place. We replace the read.table line in the Note with the first line below. Then it is just a matter of using na.locf0.
DF <- read.table(text = Lines, header = TRUE, as.is = TRUE, sep = "|",
strip.white = TRUE, na.strings = "")
transform(DF, Name = ave(Name, Id, FUN = na.omit))
Note
The input in reproducible form:
Lines <- "
Id | Name | Start_Date | End_Date
10 | Mark | 4/2/1999 | 7/5/2018
10 | | 1/1/2000 | 9/24/2018
25 | | 5/3/1968 | 6/3/2000
25 | | 6/6/2009 | 4/23/2010
25 | Anthony | 2/20/2010 | 7/21/2016
25 | | 9/12/2014 | 11/26/2019"
DF <- read.table(text = Lines, header = TRUE, as.is = TRUE, sep = "|", strip.white = TRUE)