Using spread with duplicate identifiers for rows
Since tidyr 1.0.0 pivot_wider is the recommended replacement of spread and you could do the following :
jj <- data.frame(month=rep(1:3,4),
student=rep(c("Amy", "Bob"), each=6),
A=c(9, 7, 6, 8, 6, 9, 3, 2, 1, 5, 6, 5),
B=c(6, 7, 8, 5, 6, 7, 5, 4, 6, 3, 1, 5))
library(tidyr)
pivot_wider(
jj,
names_from = "student",
values_from = c("A","B"),
names_sep = ".",
values_fn = list(A= list, B= list)) %>%
unchop(everything())
#> # A tibble: 6 x 5
#> month A.Amy A.Bob B.Amy B.Bob
#> <int> <dbl> <dbl> <dbl> <dbl>
#> 1 1 9 3 6 5
#> 2 1 8 5 5 3
#> 3 2 7 2 7 4
#> 4 2 6 6 6 1
#> 5 3 6 1 8 6
#> 6 3 9 5 7 5
Created on 2019-09-14 by the reprex package (v0.3.0)
The twist in this problem is that month is not unique by student, to solve this :
values_fn = list(A= list, B= list))puts the multiple values in a listunchop(everything())unnest the lists vertically, you can useunnestas well here
The issue is the two columns for both A and B. If we can make that one value column, we can spread the data as you would like. Take a look at the output for jj_melt when you use the code below.
library(reshape2)
jj_melt <- melt(jj, id=c("month", "student"))
jj_spread <- dcast(jj_melt, month ~ student + variable, value.var="value", fun=sum)
# month Amy_A Amy_B Bob_A Bob_B
# 1 1 17 11 8 8
# 2 2 13 13 8 5
# 3 3 15 15 6 11
I won't mark this as a duplicate since the other question did not summarize by sum, but the data.table answer could help with one additional argument, fun=sum:
library(data.table)
dcast(setDT(jj), month ~ student, value.var=c("A", "B"), fun=sum)
# month A_sum_Amy A_sum_Bob B_sum_Amy B_sum_Bob
# 1: 1 17 8 11 8
# 2: 2 13 8 13 5
# 3: 3 15 6 15 11
If you would like to use the tidyr solution, combine it with dcast to summarize by sum.
as.data.frame(jj)
library(tidyr)
jj %>%
gather(variable, value, -(month:student)) %>%
unite(temp, student, variable) %>%
dcast(month ~ temp, fun=sum)
# month Amy_A Amy_B Bob_A Bob_B
# 1 1 17 11 8 8
# 2 2 13 13 8 5
# 3 3 15 15 6 11
Edit
Based on your new requirements, I have added an activity column.
library(dplyr)
jj %>% group_by(month, student) %>%
mutate(id=1:n()) %>%
melt(id=c("month", "id", "student")) %>%
dcast(... ~ student + variable, value.var="value")
# month id Amy_A Amy_B Bob_A Bob_B
# 1 1 1 9 6 3 5
# 2 1 2 8 5 5 3
# 3 2 1 7 7 2 4
# 4 2 2 6 6 6 1
# 5 3 1 6 8 1 6
# 6 3 2 9 7 5 5
The other solutions can also be used. Here I added an optional expression to arrange the final output by activity number:
library(tidyr)
jj %>%
gather(variable, value, -(month:student)) %>%
unite(temp, student, variable) %>%
group_by(temp) %>%
mutate(id=1:n()) %>%
dcast(... ~ temp) %>%
arrange(id)
# month id Amy_A Amy_B Bob_A Bob_B
# 1 1 1 9 6 3 5
# 2 2 2 7 7 2 4
# 3 3 3 6 8 1 6
# 4 1 4 8 5 5 3
# 5 2 5 6 6 6 1
# 6 3 6 9 7 5 5
The data.table syntax is compact because it allows for multiple value.var columns and will take care of the spread for us. We can then skip the melt -> cast process.
library(data.table)
setDT(jj)[, activityID := rowid(student)]
dcast(jj, ... ~ student, value.var=c("A", "B"))
# month activityID A_Amy A_Bob B_Amy B_Bob
# 1: 1 1 9 3 6 5
# 2: 1 4 8 5 5 3
# 3: 2 2 7 2 7 4
# 4: 2 5 6 6 6 1
# 5: 3 3 6 1 8 6
# 6: 3 6 9 5 7 5
Your answer was missing mutate id! Here is the solution using dplyr packge only.
jj %>%
gather(variable, value, -(month:student)) %>%
unite(temp, student, variable) %>%
group_by(temp) %>%
mutate(id=1:n()) %>%
spread(temp, value)
# A tibble: 6 x 6
# month id Amy_A Amy_B Bob_A Bob_B
# * <int> <int> <dbl> <dbl> <dbl> <dbl>
# 1 1 1 9 6 3 5
# 2 1 4 8 5 5 3
# 3 2 2 7 7 2 4
# 4 2 5 6 6 6 1
# 5 3 3 6 8 1 6
# 6 3 6 9 7 5 5