How to populate values of one row conditional of another row in R? - r

Question

I inherited a data set coded in an unusual way. I would like to learn a less verbose way of reshaping it. The data frame looks like this:
# Input.
participant = c(rep("John",6), rep("Mary",6))
day = c(rep(1,3), rep(2,3), rep(1,3), rep(2,3))
likes = c("apples", "apples", "18", "apples", "apples", "7", "bananas", "bananas", "24", "bananas", "bananas", "3")
question = rep(c(1,1,0),4)
number = c(rep(18,3), rep(7,3), rep(24,3), rep(3,3))
df = data.frame(participant, day, question, likes)
participant day question likes
1 John 1 1 apples
2 John 1 1 apples
3 John 1 0 18
4 John 2 1 apples
5 John 2 1 apples
6 John 2 0 7
7 Mary 1 1 bananas
8 Mary 1 1 bananas
9 Mary 1 0 24
10 Mary 2 1 bananas
11 Mary 2 1 bananas
12 Mary 2 0 3
As you can see, the column likes is heterogeneous. When question equals 0, likes conveys a number chosen by the participants, not their preferred fruit. So I would like to re-code it in a new column as follows:
participant day question likes number
1 John 1 1 apples 18
2 John 1 1 apples 18
3 John 1 0 18 18
4 John 2 1 apples 7
5 John 2 1 apples 7
6 John 2 0 7 7
7 Mary 1 1 bananas 24
8 Mary 1 1 bananas 24
9 Mary 1 0 24 24
10 Mary 2 1 bananas 3
11 Mary 2 1 bananas 3
12 Mary 2 0 3 3
My current solution with base R involves subsetting the initial data frame, creating a lookup table, changing the column names and then merging the lookup table with the original data frame. But this involves several steps and I worry that there should be a simpler solution. I think that tidyr might be the answer, but I don't know how to use it to spread values in one column (likes) conditional other columns (day and question).
Do you have any suggestions? Thanks a lot!

Answer 1

Using the data set above, you can try the following. You group your data by participant and day and look for a row with question == 0 for each group.
library(dplyr)
group_by(df, participant, day) %>%
mutate(age = as.numeric(as.character(likes[which(question == 0)])))
Or as alistaire suggested, you can use grep() too.
group_by(df, participant, day) %>%
mutate(age = as.numeric(grep('\\d+', likes, value = TRUE)))
# participant day question likes age
# (fctr) (dbl) (dbl) (fctr) (dbl)
#1 John 1 1 apples 18
#2 John 1 1 apples 18
#3 John 1 0 18 18
#4 John 2 1 apples 7
#5 John 2 1 apples 7
#6 John 2 0 7 7
#7 Mary 1 1 bananas 24
#8 Mary 1 1 bananas 24
#9 Mary 1 0 24 24
#10 Mary 2 1 bananas 3
#11 Mary 2 1 bananas 3
#12 Mary 2 0 3 3
If you want to use data.table, you can do:
library(data.table)
setDT(df)[, age := as.numeric(as.character(likes[which(question == 0)])),
by = list(participant, day)]
NOTE
The present data set is a new one. Jota's answer works for the deleted data set.

Answer 2

Addressing the new example data:
# create a key column, overwrite it later
df$number <- paste0(df$participant, df$day) # use as a key
# create lookup table
lookup <- df[!is.na(as.numeric(as.character(df$likes))), c("number", "likes")]
# use lookup to overwrite df$number with the appropriate number
df$number <- lookup$likes[match(df$number, lookup$number)]
# participant day question likes number
#1 John 1 1 apples 18
#2 John 1 1 apples 18
#3 John 1 0 18 18
#4 John 2 1 apples 7
#5 John 2 1 apples 7
#6 John 2 0 7 7
#7 Mary 1 1 bananas 24
#8 Mary 1 1 bananas 24
#9 Mary 1 0 24 24
#10 Mary 2 1 bananas 3
#11 Mary 2 1 bananas 3
#12 Mary 2 0 3 3
The warning about NAs be introduced by coercion is expected due to converting characters to numeric (as.numeric(as.character(df$likes))),.
If you're data are ordered like in the example, you can use na.locf from the zoo package:
library(zoo)
df$age <- na.locf(as.numeric(as.character(df$likes)), fromLast = TRUE)