I have a dataframe:
group_eg <- data.frame(
CID = c(091, 091, 091, 091,
101, 101, 101,
102, 102, 102,
103, 103,
104, 104, 104),
PID_A = c(1091, 1091, 1091, 1091,
1101, 1101, 1101,
1102, 1102, 1102,
1103, 1103,
1104, 1104, 1104),
PID_B = c(2091, 2091, 2091, 2091,
2101, 2101, 2101,
2102, 2102, 2102,
2103, 2103,
2104, 2104, 2104),
text = c("eg1", "eg2", "eg3", "eg3",
"eg4","eg5", "eg6",
"eg7", "eg8","eg9",
"eg10", "eg11",
"eg12", "eg13", "eg14")
)
I want to divide this dataframe into a list of smaller dataframes. Each of the smaller dataframes should only contain rows that belong to 2 unique CIDs in ascending order. If the number of CIDs is not divisible by 2, the last dataframe can contain the rows that belong to just 1 CID. Notice that there are different number of observations for each unique CID, so I'm a bit stuck.
Here are the example outputs:
output1 <- data.frame(
CID = c(091, 091, 091, 091,
101, 101, 101),
PID_A = c(1091, 1091, 1091, 1091,
1101, 1101, 1101),
PID_B = c(2091, 2091, 2091, 2091,
2101, 2101, 2101),
text = c("eg1", "eg2", "eg3", "eg3",
"eg4","eg5", "eg6")
)
output2 <- data.frame(
CID = c(102, 102, 102,
103, 103),
PID_A = c(1102, 1102, 1102,
1103, 1103),
PID_B = c(2102, 2102, 2102,
2103, 2103),
text = c( "eg7", "eg8","eg9",
"eg10", "eg11")
)
output3 <- data.frame(
CID = c(104, 104, 104),
PID_A = c(1104, 1104, 1104),
PID_B = c(2104, 2104, 2104),
text = c("eg12", "eg13", "eg14")
)
Does anyone know how to do this? Thank you!
Using dplyr::consecutive_id and integer division %/% you could do:
library(dplyr, warn = FALSE)
group_eg |>
group_by(group = (consecutive_id(CID) + 1) %/% 2) |>
group_split()
#> <list_of<
#> tbl_df<
#> CID : double
#> PID_A: double
#> PID_B: double
#> text : character
#> group: double
#> >
#> >[3]>
#> [[1]]
#> # A tibble: 7 × 5
#> CID PID_A PID_B text group
#> <dbl> <dbl> <dbl> <chr> <dbl>
#> 1 91 1091 2091 eg1 1
#> 2 91 1091 2091 eg2 1
#> 3 91 1091 2091 eg3 1
#> 4 91 1091 2091 eg3 1
#> 5 101 1101 2101 eg4 1
#> 6 101 1101 2101 eg5 1
#> 7 101 1101 2101 eg6 1
#>
#> [[2]]
#> # A tibble: 5 × 5
#> CID PID_A PID_B text group
#> <dbl> <dbl> <dbl> <chr> <dbl>
#> 1 102 1102 2102 eg7 2
#> 2 102 1102 2102 eg8 2
#> 3 102 1102 2102 eg9 2
#> 4 103 1103 2103 eg10 2
#> 5 103 1103 2103 eg11 2
#>
#> [[3]]
#> # A tibble: 3 × 5
#> CID PID_A PID_B text group
#> <dbl> <dbl> <dbl> <chr> <dbl>
#> 1 104 1104 2104 eg12 3
#> 2 104 1104 2104 eg13 3
#> 3 104 1104 2104 eg14 3