CODEWITHSUNDEEP
rdplyr
Y.Olshanskiy
Y.Olshanskiy

Reputation: 135

R dplyr line-by-line execution

I am looking for a way to execute line-by-line statements using dplyr package in R, which is similar to execution by loop: we do something with the next line only when the previous lines are updated.

For instance,

X <- data.frame(a = c(1,NA,NA,NA))
for (i in 2:nrow(X)){
    X$a[i] = X$a[i-1] + 1     
}
X
  a
1 1
2 2
3 3
4 4

So line 3 takes values from previous lines only when line 2 has received value = 2 at the previous loop step.

If I try to do it by usual dplyr::mutate function then I have

library(dplyr)
X <- data.frame(a = c(1,NA,NA,NA))
X %>% mutate(a = if_else(row_number() == 1, a, lag(a) + 1) )
   a
1  1
2  2
3 NA
4 NA

Any ideas how to get the first output using dplyr?

Let me give more specific and complicated example:

> X <- data.frame(date_1 = c("2000-01-01", "2001-01-01", NA, NA, NA, "2007-01-01", NA, NA),
+                 date_2 = c("2002-01-01", "2002-01-01", "2002-01-01", "2002-01-01", "2003-01-01", "2008-01-01", "2010-01-01", "2010-01-01"),
+                 stringsAsFactors=FALSE)
> X
      date_1     date_2
1 2000-01-01 2002-01-01
2 2001-01-01 2002-01-01
3       <NA> 2002-01-01
4       <NA> 2002-01-01
5       <NA> 2003-01-01
6 2007-01-01 2008-01-01
7       <NA> 2010-01-01
8       <NA> 2010-01-01
>

and I want to fill it using the following loop:

> for (i in 2:nrow(X)){
+      X$date_1[i] <- if_else(!is.na(X$date_1[i]), X$date_1[i],
+                       if_else(X$date_2[i-1] == X$date_2[i], X$date_1[i-1],  X$date_2[i-1]))
+ }
> X
      date_1     date_2
1 2000-01-01 2002-01-01
2 2001-01-01 2002-01-01
3 2001-01-01 2002-01-01
4 2001-01-01 2002-01-01
5 2002-01-01 2003-01-01
6 2007-01-01 2008-01-01
7 2008-01-01 2010-01-01
8 2008-01-01 2010-01-01

dplyr version would look like:

> X %>% mutate( date_1 = if_else(row_number() == 1, date_1,
+                         if_else(!is.na(date_1), date_1,
+                          if_else(date_2 == lag(date_2), lag(date_1),
+                                  lag(date_2))))
+         )
      date_1     date_2
1 2000-01-01 2002-01-01
2 2001-01-01 2002-01-01
3 2001-01-01 2002-01-01
4       <NA> 2002-01-01
5 2002-01-01 2003-01-01
6 2007-01-01 2008-01-01
7 2008-01-01 2010-01-01
8       <NA> 2010-01-01

Upvotes: 1

Views: 736

Answers (1)

Abdou
Abdou

Reputation: 13274

Try:

library(tidyverse)


x %>%
    fill(a) %>%
    mutate(a = a+seq_along(a)-1)

or

x %>%
    fill(a) %>%
    mutate(a = a+which(!!a)-1)

That should yield:

#  a
#1 1
#2 2
#3 3
#4 4

Edit:

A solution for the latest example:

X <- data.frame(date_1 = c("2000-01-01", "2001-01-01", NA, NA, NA, "2007-01-01", NA, NA),
                date_2 = c("2002-01-01", "2002-01-01", "2002-01-01", "2002-01-01", "2003-01-01","2008-01-01", "2010-01-01", "2010-01-01"), stringsAsFactors=FALSE)

X %>%
    group_by(date_2) %>%
    fill(date_1) %>%
    ungroup() %>%
    mutate(date_3 = lag(date_2)) %>%
    group_by(date_1, date_2) %>%
    mutate(date_3 = if_else(is.na(date_1), head(date_3,1), date_3)) %>%
    ungroup() %>%
    mutate(date_1 = if_else(is.na(date_1), date_3, date_1)) %>%
    select(date_1, date_2)

Output:

 date_1     date_2    
 2000-01-01 2002-01-01
 2001-01-01 2002-01-01
 2001-01-01 2002-01-01
 2001-01-01 2002-01-01
 2002-01-01 2003-01-01
 2007-01-01 2008-01-01
 2008-01-01 2010-01-01
 2008-01-01 2010-01-01

I hope this helps.

Upvotes: 1

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