Given the following first two columns(id and time_diff), i want to generate the 'block' column
test
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5
The data is already sorted by id and time. The time_diff was computed based on the difference of the previous time and the time value for the row, given the same id. I want to create a block id which is an auto-increment value and increases when a new ID or a time_diff of >10 with the same id is encountered.
How can I achieve this in R?
Importing your data as a data frame with something like:
df = read.table(text='
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5')
You can do a one-liner like this to get occurrences satisfying your two conditions:
> new_col = as.vector(cumsum(
na.exclude(
c(F,diff(as.numeric(as.factor(df$id)))) | # change of id OR
df$time_diff > 10 # time_diff greater than 10
)
))
> new_col
[1] 0 0 0 0 0 1 2 2 2 2 3 3 4 4 4
And finally append this new column to your dataframe with cbind:
> cbind(df, block = c(0,new_col))
id time_diff block block
1 a NA 1 0
2 a 1 1 0
3 a 1 1 0
4 a 1 1 0
5 a 3 1 0
6 a 3 1 0
7 b NA 2 1
8 b 11 3 2
9 b 1 3 2
10 b 1 3 2
11 b 1 3 2
12 b 12 4 3
13 b 1 4 3
14 c NA 5 4
15 c 4 5 4
16 c 7 5 4
You will notice an offset between your wanted block variable and mine: correcting it is easy and can be done at several different step, I will leave it to you :)
Another variation of #Jealie's method would be:
with(test, cumsum(c(TRUE,id[-1]!=id[-nrow(test)])|time_diff>10))
#[1] 1 1 1 1 1 1 2 3 3 3 3 4 4 5 5 5
After learning from Jealie and akrun, I came up with this idea.
mydf %>%
mutate(group = cumsum(time_diff > 10 |!duplicated(id)))
# id time_diff block group
#1 a NA 1 1
#2 a 1 1 1
#3 a 1 1 1
#4 a 1 1 1
#5 a 3 1 1
#6 a 3 1 1
#7 b NA 2 2
#8 b 11 3 3
#9 b 1 3 3
#10 b 1 3 3
#11 b 1 3 3
#12 b 12 4 4
#13 b 1 4 4
#14 c NA 5 5
#15 c 4 5 5
#16 c 7 5 5
Here is an approach using dplyr:
require(dplyr)
set.seed(999)
test <- data.frame(
id = rep(letters[1:4], each = 3),
time_diff = sample(4:15)
)
test %>%
mutate(
b = as.integer(id) - lag(as.integer(id)),
more10 = time_diff > 10,
increment = pmax(b, more10, na.rm = TRUE),
increment = ifelse(row_number() == 1, 1, increment),
block = cumsum(increment)
) %>%
select(id, time_diff, block)
Try:
> df
id time_diff
1 a NA
2 a 1
3 a 1
4 a 1
5 a 3
6 a 3
7 b NA
8 b 11
9 b 1
10 b 1
11 b 1
12 b 12
13 b 1
14 c NA
15 c 4
16 c 7
block= c(1)
for(i in 2:nrow(df))
block[i] = ifelse(df$time_diff[i]>10 || df$id[i]!=df$id[i-1],
block[i-1]+1,
block[i-1])
df$block = block
df
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5
Related
I'm trying to add a column with repeating sequence but one that changes for each group. In the example data, the group is the id column.
data <- tibble::expand_grid(id = 1:12, condition = c("a", "b", "c"))
data
id condition
1 a
1 b
1 c
2 a
2 b
2 c
3 a
3 b
3 c
... and so on
I'd like to add a column called order to repeat various combinations like 1 2 3 2 3 1 3 1 2 1 3 2 2 1 3 3 2 1 for each id.
In the end, the desired output will look like this
id condition order
1 a 1
1 b 2
1 c 3
2 a 2
2 b 3
2 c 1
3 a 3
3 b 1
3 c 2
... and so on
I'm looking for a simple mutate solution or base R solution. I tried generating a list of combinations but I'm not sure how to create a variable from that.
You can use perms from package pracma to generate all permutations, e.g.,
data %>%
cbind(order = c(t(pracma::perms(1:3))))
which gives
id condition order
1 1 a 3
2 1 b 2
3 1 c 1
4 2 a 3
5 2 b 1
6 2 c 2
7 3 a 2
8 3 b 3
9 3 c 1
10 4 a 2
11 4 b 1
12 4 c 3
13 5 a 1
14 5 b 2
15 5 c 3
16 6 a 1
17 6 b 3
18 6 c 2
19 7 a 3
20 7 b 2
21 7 c 1
22 8 a 3
23 8 b 1
24 8 c 2
25 9 a 2
26 9 b 3
27 9 c 1
28 10 a 2
29 10 b 1
30 10 c 3
31 11 a 1
32 11 b 2
33 11 c 3
34 12 a 1
35 12 b 3
36 12 c 2
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
I have a data.table with firm information.
library(data.table)
DT <- fread("
iso Firm GDP year
A 1 1 1
A 2 1 1
A 3 1 1
A 4 1 1
A 5 3 2
A 6 3 2
A 7 3 2
A 8 3 2
B 9 2 1
B 10 2 1
B 11 2 1
B 12 2 1
B 13 4 1
B 14 4 1
B 15 4 1
B 16 4 1",
header = TRUE)
I want to calculate GDPgrowth (per country) from one year to the other and add it to the dataset ((N-O)/O). However, if I do:
DT <- DT[,GDPgrowth :=((GDP- shift(GDP))/shift(GDP)), by=iso]
the outcome will be zero because it subtracts the firm observations from each other.
How can I make sure it calculates for the whole group of firms belonging to the country together?
Desired output:
library(data.table)
DT <- fread("
iso Firm GDP GDPgrowth year
A 1 1 NA 1
A 2 1 NA 1
A 3 1 NA 1
A 4 1 NA 1
A 5 3 2 2
A 6 3 2 2
A 7 3 2 2
A 8 3 2 2
B 9 2 NA 1
B 10 2 NA 1
B 11 2 NA 1
B 12 2 NA 1
B 13 4 1 1
B 14 4 1 1
B 15 4 1 1
B 16 4 1 1",
header = TRUE)
Here is one way continuing from your current approach :
library(data.table)
DT[,GDPgrowth :=((GDP- shift(GDP))/shift(GDP)), by=iso]
DT[GDPgrowth == 0, GDPgrowth := NA]
DT[, GDPgrowth:= zoo::na.locf(GDPgrowth, na.rm = FALSE), .(iso, year)]
DT
# iso Firm GDP year GDPgrowth
# 1: A 1 1 1 NA
# 2: A 2 1 1 NA
# 3: A 3 1 1 NA
# 4: A 4 1 1 NA
# 5: A 5 3 2 2
# 6: A 6 3 2 2
# 7: A 7 3 2 2
# 8: A 8 3 2 2
# 9: B 9 2 1 NA
#10: B 10 2 1 NA
#11: B 11 2 1 NA
#12: B 12 2 1 NA
#13: B 13 4 1 1
#14: B 14 4 1 1
#15: B 15 4 1 1
#16: B 16 4 1 1
Using dplyr and tidyr::fill it can be done as
library(dplyr)
DT %>%
group_by(iso) %>%
mutate(GDPgrowth = (GDP - lag(GDP))/lag(GDP),
GDPgrowth = replace(GDPgrowth, GDPgrowth == 0, NA)) %>%
group_by(iso, year) %>%
tidyr::fill(GDPgrowth)
So, expand.grid returns a df of all the combinations of the vectors passed.
df <- expand.grid(1:3, 1:3)
df <- expand.grid(1:3, 1:3, 1:3)
What I would like is a generalized function that takes 1 parameter (number of vectors) and returns the appropriate data frame.
combinations <- function(n) {
return(expand.grid(0, 1, ... n))
}
Such that
combinations(2) returns(expand.grid(1:3, 1:3))
combinations(3) returns(expand.grid(1:3, 1:3, 1:3))
combinations(4) returns(expand.grid(1:3, 1:3, 1:3, 1:3))
etc.
combinations <- function(n)
expand.grid(rep(list(1:3),n))
> combinations(2)
Var1 Var2
1 1 1
2 2 1
3 3 1
4 1 2
5 2 2
6 3 2
7 1 3
8 2 3
9 3 3
> combinations(3)
Var1 Var2 Var3
1 1 1 1
2 2 1 1
3 3 1 1
4 1 2 1
5 2 2 1
6 3 2 1
7 1 3 1
8 2 3 1
9 3 3 1
10 1 1 2
11 2 1 2
12 3 1 2
13 1 2 2
14 2 2 2
15 3 2 2
16 1 3 2
17 2 3 2
18 3 3 2
19 1 1 3
20 2 1 3
21 3 1 3
22 1 2 3
23 2 2 3
24 3 2 3
25 1 3 3
26 2 3 3
27 3 3 3
I know how to delete columns in R, but I am not sure how to delete them based on the following set of conditions.
Suppose a data frame such as:
DF <- data.frame(L = c(2,4,5,1,NA,4,5,6,4,3), J= c(3,4,5,6,NA,3,6,4,3,6), K= c(0,1,1,0,NA,1,1,1,1,1),D = c(1,1,1,1,NA,1,1,1,1,1))
DF
L J K D
1 2 3 0 1
2 4 4 1 1
3 5 5 1 1
4 1 6 0 1
5 NA NA NA NA
6 4 3 1 1
7 5 6 1 1
8 6 4 1 1
9 4 3 1 1
10 3 6 1 1
The data frame has to be set up in this fashion. Column K corresponds to column L, and column D, corresponds to column J. Because column D has values that are all equal to one, I would like to delete column D, and the corresponding column J yielding a dataframe that looks like:
DF
L K
1 2 0
2 4 1
3 5 1
4 1 0
5 NA NA
6 4 1
7 5 1
8 6 1
9 4 1
10 3 1
I know there has got to be a simple command to do so, I just can't think of any. And if it makes any difference, the NA's must be retained.
Additional helpful information, in my real data frame there are a total of 20 columns, so there are 10 columns like L and J, and another 10 that are like K and D, I need a function that can recognize the correspondence between these two groups and delete columns accordingly if necessary
Thank you in advance!
Okey, assuming the column-number based correspondence, here is an example:
> n <- 10
>
> # sample data
> d <- data.frame(lapply(1:n, function(x)sample(n)), lapply(1:n, function(x)sample(2, n, T, c(0.1, 0.9))-1))
> names(d) <- c(LETTERS[1:n], letters[1:n])
> head(d)
A B C D E F G H I J a b c d e f g h i j
1 5 5 2 7 4 3 4 3 5 8 0 1 1 1 1 1 1 1 1 1
2 9 8 4 6 7 8 8 2 10 5 1 1 1 1 1 1 1 1 1 1
3 6 6 10 3 5 6 2 1 8 6 1 1 1 1 1 1 1 1 1 1
4 1 7 5 5 1 10 10 4 2 4 1 1 1 1 1 1 1 1 1 1
5 10 9 6 2 9 5 6 9 9 9 1 1 0 1 1 1 1 1 1 1
6 2 1 1 4 6 1 5 8 4 10 1 1 1 1 1 1 1 1 1 1
>
> # find the column that should be left.
> idx <- which(colMeans(d[(n+1):(2*n)], na.rm = TRUE) != 1)
>
> # filter the data
> d[, c(idx, idx+n)]
A B C D F a b c d f
1 5 5 2 7 3 0 1 1 1 1
2 9 8 4 6 8 1 1 1 1 1
3 6 6 10 3 6 1 1 1 1 1
4 1 7 5 5 10 1 1 1 1 1
5 10 9 6 2 5 1 1 0 1 1
6 2 1 1 4 1 1 1 1 1 1
7 8 4 7 10 2 1 1 1 1 0
8 7 3 9 9 4 1 0 1 0 1
9 3 10 3 1 9 1 1 0 1 1
10 4 2 8 8 7 1 0 1 1 1
I basically agree with koshke (whose SO work is excellent), but would suggest that the test to use is colSums(d[(n+1):(2*n)], na.rm=TRUE) == NROW(d) , since a paired 0 and 2 or -1 and 3 could throw off the colMeans test.