This is my data frame
df <- data.frame(
id = 1:14,
group_id = c(rep(1:2, each = 3), rep(3:4, each = 4)),
type = rep("A", 14), stringsAsFactors = FALSE)
df[c(2,4,8,12),"type"] <- "B"
id group_id type
1 1 1 A
2 2 1 B
3 3 1 A
4 4 2 B
5 5 2 A
6 6 2 A
7 7 3 A
8 8 3 B
9 9 3 A
10 10 3 A
11 11 4 A
12 12 4 B
13 13 4 A
14 14 4 A
I'd like to keep all rows with type B as well as the following row.
I could do...
B <- which(df$type=="B")
afterB <- B+1
df_sel <- df[c(B, afterB), ]
df_sel <- df_sel[order(df_sel$id),]
df_sel
...to get what I want.
id group_id type
2 2 1 B
3 3 1 A
4 4 2 B
5 5 2 A
8 8 3 B
9 9 3 A
12 12 4 B
13 13 4 A
How can this be done in a more generic way.
Another way, very similar to what you do but in one step and without the need to reorder:
df_sel <- df[rep(which(df$type=="B"), e=2)+c(0, 1), ]
df_sel
# id group_id type
# 2 2 1 B
# 3 3 1 A
# 4 4 2 B
# 5 5 2 A
# 8 8 3 B
# 9 9 3 A
# 12 12 4 B
# 13 13 4 A
Using lag from dplyr
library(dplyr)
df[df$type == "B" | lag(df$type == "B", default = FALSE), ]
# id group_id type
#2 2 1 B
#3 3 1 A
#4 4 2 B
#5 5 2 A
#8 8 3 B
#9 9 3 A
#12 12 4 B
#13 13 4 A
using grep will provide a row index of all instances of B - rows; concatenate (c()) this with rows + 1 to select from df will work.
rows <- grep("B", df[, "type"])
df[sort(c(rows, rows + 1)), ]
gives:
id group_id type
2 2 1 B
3 3 1 A
4 4 2 B
5 5 2 A
8 8 3 B
9 9 3 A
12 12 4 B
13 13 4 A
Related
I like to create two columns with cumulative frequency of "A" and "B" in the assignment columns.
df = data.frame(id = 1:10, assignment= c("B","A","B","B","B","A","B","B","A","B"))
id assignment
1 1 B
2 2 A
3 3 B
4 4 B
5 5 B
6 6 A
7 7 B
8 8 B
9 9 A
10 10 B
The resulting table would have this format
id assignment A B
1 1 B 0 1
2 2 A 1 1
3 3 B 1 2
4 4 B 1 3
5 5 B 1 4
6 6 A 2 4
7 7 B 2 5
8 8 B 2 6
9 9 A 3 6
10 10 B 3 7
How to generalize the codes for more than 2 categories (say for "A","B",C")?
Thanks
Use lapply over unique values in assignment to create new columns.
vals <- sort(unique(df$assignment))
df[vals] <- lapply(vals, function(x) cumsum(df$assignment == x))
df
# id assignment A B
#1 1 B 0 1
#2 2 A 1 1
#3 3 B 1 2
#4 4 B 1 3
#5 5 B 1 4
#6 6 A 2 4
#7 7 B 2 5
#8 8 B 2 6
#9 9 A 3 6
#10 10 B 3 7
We can use model.matrix with colCumsums
library(matrixStats)
cbind(df, colCumsums(model.matrix(~ assignment - 1, df[-1])))
A base R option
transform(
df,
A = cumsum(assignment == "A"),
B = cumsum(assignment == "B")
)
gives
id assignment A B
1 1 B 0 1
2 2 A 1 1
3 3 B 1 2
4 4 B 1 3
5 5 B 1 4
6 6 A 2 4
7 7 B 2 5
8 8 B 2 6
9 9 A 3 6
10 10 B 3 7
how do I extract specific row of data when the column has repetitive value? my data looks like this: I want to extract the row of the end of each repeat of x (A 3 10, A 2 3 etc) or the index of the last value
Name X M
A 1 1
A 2 9
A 3 10
A 1 1
A 2 3
A 1 5
A 2 6
A 3 4
A 4 5
A 5 3
B 1 1
B 2 9
B 3 10
B 1 1
B 2 3
Expected output
Index Name X M
3 A 3 10
5 A 2 3
10 A 5 3
13 B 3 10
15 B 2 3
Using base R duplicated and cumsum:
dups <- !duplicated(cumsum(dat$X == 1), fromLast=TRUE)
cbind(dat[dups,], Index=which(dups))
# Name X M Index
#3 A 3 10 3
#5 A 2 3 5
#10 A 5 3 10
#13 B 3 10 13
#15 B 2 3 15
A solution using dplyr.
library(dplyr)
df2 <- df %>%
mutate(Flag = ifelse(lead(X) < X, 1, 0)) %>%
mutate(Index = 1:n()) %>%
filter(Flag == 1 | is.na(Flag)) %>%
select(Index, X, M)
df2
# Index X M
# 1 3 3 10
# 2 5 2 3
# 3 10 5 3
# 4 13 3 10
# 5 15 2 3
Flag is a column showing if the next number in A is smaller than the previous number. If TRUE, Flag is 1, otherwise is 0. We can then filter for Flag == 1 or where Flag is NA, which is the last row. df2 is the final filtered data frame.
DATA
df <- read.table(text = "Name X M
A 1 1
A 2 9
A 3 10
A 1 1
A 2 3
A 1 5
A 2 6
A 3 4
A 4 5
A 5 3
B 1 1
B 2 9
B 3 10
B 1 1
B 2 3",
header = TRUE, stringsAsFactors = FALSE)
This question already has answers here:
Using dplyr to get cumulative count by group
(3 answers)
Closed 5 years ago.
I come to an issue with numbering the duplicated rows in data.frame and could not find a similar post.
Let's say we have a data like this
df <- data.frame(gr=gl(7,2),x=c("a","a","b","b","c","c","a","a","c","c","d","d","a","a"))
> df
gr x
1 1 a
2 1 a
3 2 b
4 2 b
5 3 c
6 3 c
7 4 a
8 4 a
9 5 c
10 5 c
11 6 d
12 6 d
13 7 a
14 7 a
and want to add new column called x_dupl to show that first occurrence of x values is numbered as 1 and second time 2 and third time 3 and so on..
thanks in advance!
The expected output
> df
gr x x_dupl
1 1 a 1
2 1 a 1
3 2 b 1
4 2 b 1
5 3 c 1
6 3 c 1
7 4 a 2
8 4 a 2
9 5 c 2
10 5 c 2
11 6 d 1
12 6 d 1
13 7 a 3
14 7 a 3
Your example data (plus rows where gr = 7 as in your output), and named df1, not df:
df1 <- data.frame(gr = gl(7,2),
x = c("a","a","b","b","c","c","a","a","c","c","d","d","a","a"))
library(dplyr)
df1 %>%
group_by(x) %>%
mutate(x_dupl = dense_rank(gr)) %>%
ungroup()
# A tibble: 14 x 3
gr x x_dupl
<fctr> <fctr> <int>
1 1 a 1
2 1 a 1
3 2 b 1
4 2 b 1
5 3 c 1
6 3 c 1
7 4 a 2
8 4 a 2
9 5 c 2
10 5 c 2
11 6 d 1
12 6 d 1
13 7 a 3
14 7 a 3
A base R solution:
df <- data.frame(gr=gl(7,2),x=c("a","a","b","b","c","c","a","a","c","c","d","d","a","a"))
x <- rle(as.numeric(df$x))
x$values <- ave(x$values, x$values, FUN = seq_along)
df$x_dupl <- inverse.rle(x)
# gr x x_dupl
# 1 1 a 1
# 2 1 a 1
# 3 2 b 1
# 4 2 b 1
# 5 3 c 1
# 6 3 c 1
# 7 4 a 2
# 8 4 a 2
# 9 5 c 2
# 10 5 c 2
# 11 6 d 1
# 12 6 d 1
# 13 7 a 3
# 14 7 a 3
My data looks like this:
mydata <- data.frame(id = c(1,1,1,2,2,3,3,3,3),
subid = c(1,2,3,1,2,1,2,3,4),
time = c(16, 18, 20, 10, 11, 7, 9, 10, 11))
id subid time
1 1 1 16
2 1 2 18
3 1 3 20
4 2 1 10
5 2 2 11
6 3 1 7
7 3 2 9
8 3 3 10
9 3 4 11
My goal is to transform the data to:
newdata <- data.frame(id = c(1,1,1,2,3,3,3,3,3,3),
subid.1 = c(1,1,2,1,1,1,1,2,2,3),
subid.2 = c(2,3,3,2,2,3,4,3,4,4),
time.1 = c(16,16,18,10,7,7,7,9,9,10),
time.2 = c(18,20,20,11,9,10,11,10,11,11))
id subid.1 subid.2 time.1 time.2
1 1 1 2 16 18
2 1 1 3 16 20
3 1 2 3 18 20
4 2 1 2 10 11
5 3 1 2 7 9
6 3 1 3 7 10
7 3 1 4 7 11
8 3 2 3 9 10
9 3 2 4 9 11
10 3 3 4 10 11
So it's not a simple reshape from long-to-wide procedure: The idea is, within groups defined by id, to take all possible combinations of
subid's and their corresponding time values, and get those into a wide format.
I know I can get all possible combinations using, for example gtools::combinations. The first group consists of 3 rows, so
gtools::combinations(n=3, r=2)
gives me the matrix of the new subid.1 and subid.2 pair for group id==1:
[,1] [,2]
[1,] 1 2
[2,] 1 3
[3,] 2 3
But then I don't know how to proceed (neither to reshape the group with id==1 to this format, nor how to do that separately for each group). Thank you!
with base R:
subset(merge(mydata, mydata, by="id", suffix=c(".1",".2")), subid.1 < subid.2)
# id subid.1 time.1 subid.2 time.2
# 1 1 1 16 2 18
# 2 1 1 16 3 20
# 3 1 2 18 3 20
# 4 2 1 10 2 11
# 5 3 1 7 2 9
# 6 3 1 7 3 10
# 7 3 1 7 4 11
# 8 3 2 9 3 10
# 9 3 2 9 4 11
# 10 3 3 10 4 11
dplyr version:
mydata %>% inner_join(.,.,by="id",suffix=c(".1",".2")) %>% filter(subid.1 < subid.2)
data.table version :
setDT(mydata)
mydata[mydata, on="id", allow.cartesian=TRUE][subid < i.subid]
# id subid time i.subid i.time
# 1: 1 1 16 2 18
# 2: 1 1 16 3 20
# 3: 1 2 18 3 20
# 4: 2 1 10 2 11
# 5: 3 1 7 2 9
# 6: 3 1 7 3 10
# 7: 3 2 9 3 10
# 8: 3 1 7 4 11
# 9: 3 2 9 4 11
# 10: 3 3 10 4 11
or to get your column names right, but it kills the fun of a short solution :).
merge(mydata, mydata, by="id", suffix=c(".1",".2"), allow.cartesian=TRUE)[subid.1 < subid.2]
Forgot to state that I came up with this rather lame 4-step solution:
step1 <- lapply(unique(mydata$id), function(x) {
nrows <- nrow(mydata[which(mydata$id == x), ])
combos <- gtools::combinations(n=nrows, r=2)
return(as.data.frame(cbind(x, combos)))
})
step2 <- dplyr::bind_rows(step1)
step3a <- merge(step2, mydata, by.x = c("x", "V2"), by.y = c("id", "subid"))
step3b <- merge(step3a, mydata, by.x = c("x", "V3"), by.y = c("id", "subid"))
step4 <- step3b[, c(1, 3, 2, 4, 5)]
names(step4) <- c("id", "subid.1", "subid.2", "time.1", "time.2")
It's ugly but works.
Using the data.table-package:
library(data.table)
setDT(mydata)[, .(subid = c(t(combn(subid, 2)))), by = id
][, grp := rep(1:2, each = .N/2), by = id
][mydata, on = .(id, subid), time := time
][, dcast(.SD, id + rowid(grp) ~ grp, value.var = list('subid','time'), sep = '.')]
which gives you:
id grp subid.1 subid.2 time.1 time.2
1: 1 1 1 2 16 18
2: 1 2 1 3 16 20
3: 1 3 2 3 18 20
4: 2 4 1 2 10 11
5: 3 5 1 2 7 9
6: 3 6 1 3 7 10
7: 3 7 1 4 7 11
8: 3 8 2 3 9 10
9: 3 9 2 4 9 11
10: 3 10 3 4 10 11
I have this data.frame:
a <- c(rep("1", 3), rep("2", 3), rep("3",3), rep("4",3), rep("5",3))
b <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)
df <-data.frame(a,b)
a b
1 1 1
2 1 2
3 1 3
4 2 4
5 2 5
6 2 6
7 3 7
8 3 8
9 3 9
10 4 10
11 4 11
12 4 12
13 5 13
14 5 14
15 5 15
I want to have something like this:
a <- c(rep("2", 3), rep("3", 3))
b <- c(4,5,6,7,8,9)
dffinal<-data.frame(a,b)
a b
1 2 4
2 2 5
3 2 6
4 3 7
5 3 8
6 3 9
I could use the "subset" function, but its not working
sub <- subset(df,c(2,3) == a )
a b
5 2 5
8 3 8
This command only takes one row of "2" and "3" in column "a".
Any Help?
You're confusing == with %in%:
subset(df, a %in% c(2,3))
# a b
# 4 2 4
# 5 2 5
# 6 2 6
# 7 3 7
# 8 3 8
# 9 3 9
what about this?
library(dplyr)
df %>% filter(a == 2 | a==3)
a b
1 2 4
2 2 5
3 2 6
4 3 7
5 3 8
6 3 9
We can use data.table. We convert the 'data.frame' to 'data.table' (setDT(df)), and set the 'key' as column 'a', then we subset the rows.
library(data.table)
setDT(df, key= 'a')[c('2','3')]
# a b
#1: 2 4
#2: 2 5
#3: 2 6
#4: 3 7
#5: 3 8
#6: 3 9