I am really new in r and stackoverflow. Apologies in advance for this novice question.
I have a panel data set like the following table.
ID Choice
1 1
1 1
1 2
1 5
1 1
2 1
2 1
2 5
2 1
2 1
3 3
3 1
3 1
3 2
3 4
I want to add another column like the following table when choice is 1. This is basically, sequencing the choice 1 within ID.
ID Choice BUS
1 1 0 (The first 1 will be considered as 0)
1 1 1
1 2 1
1 5 1
1 1 2
2 1 0
2 1 1
2 5 1
2 1 2
2 1 3
3 3 0
3 1 0
3 1 1
3 2 1
3 4 1
with(df, ave(Choice == 1, ID, FUN = cumsum))
Almost gives you what you want but as you want to consider first 1 as 0 it needs some modification.
df$BUS <- with(df, ave(Choice == 1, ID, FUN = function(x) {
inds = cumsum(x)
ifelse(inds > 0, inds - 1, inds)
}))
df
# ID Choice BUS
#1 1 1 0
#2 1 1 1
#3 1 2 1
#4 1 5 1
#5 1 1 2
#6 2 1 0
#7 2 1 1
#8 2 5 1
#9 2 1 2
#10 2 1 3
#11 3 3 0
#12 3 1 0
#13 3 1 1
#14 3 2 1
#15 3 4 1
Here we subtract 1 from cumulative sum from the first 1.
Using the same logic in dplyr
library(dplyr)
df %>%
group_by(ID) %>%
mutate(inds = cumsum(Choice == 1),
BUS = ifelse(inds > 0, inds - 1, inds)) %>%
select(-inds)
We can also use data.table
library(data.table)
setDT(df1)[, BUS := pmax(0, cumsum(Choice == 1)-1), ID]
df1
# ID Choice BUS
# 1: 1 1 0
# 2: 1 1 1
# 3: 1 2 1
# 4: 1 5 1
# 5: 1 1 2
# 6: 2 1 0
# 7: 2 1 1
# 8: 2 5 1
# 9: 2 1 2
#10: 2 1 3
#11: 3 3 0
#12: 3 1 0
#13: 3 1 1
#14: 3 2 1
#15: 3 4 1
data
df1 <- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L,
3L, 3L, 3L, 3L, 3L), Choice = c(1L, 1L, 2L, 5L, 1L, 1L, 1L, 5L,
1L, 1L, 3L, 1L, 1L, 2L, 4L)), class = "data.frame", row.names = c(NA,
-15L))
Related
Im trying to use str_detect to mutate only if the column "RedColor" is "1".
I have a dataset test which looks like this:
# id RedColor Color_Number
#1 1 1 1
#2 2 0 1
#3 3 1 3
#4 4 1 2
#5 6 0 2
#6 8 1 6
I tried the filter function but it returns me only a filtered dataset with all other cases with RedColor = "0" removed.
test <- test %>%
filter(RedColor==TRUE) %>%
mutate(DarkRed = str_detect(Color_Number, "1|2"))
Im expecting an output with the new column DarkRed = "1" in all cases with RedColor = 1 and 1 or 2 in column Color_Number.
# id RedColor Color_Number Dark_Red
#1 1 1 1 1
#2 2 0 1 0
#3 3 1 3 0
#4 4 1 2 1
#5 6 0 2 0
#6 8 1 6 0
Thank you!
Using base R
transform(df, Dark_Red = +(RedColor == 1& Color_Number %in% 1:2))
id RedColor Color_Number Dark_Red
1 1 1 1 1
2 2 0 1 0
3 3 1 3 0
4 4 1 2 1
5 6 0 2 0
6 8 1 6 0
data
df <- structure(list(id = c(1L, 2L, 3L, 4L, 6L, 8L), RedColor = c(1L,
0L, 1L, 1L, 0L, 1L), Color_Number = c(1L, 1L, 3L, 2L, 2L, 6L)), row.names = c(NA,
-6L), class = "data.frame")
Update on OP's request (see comments):
With this dataframe:
id RedColor Color_Number
1 1 1 one
2 2 0 one
3 3 1 three
4 4 1 two
5 6 0 two
6 8 1 six
you could use this code:
library(dplyr)
df %>%
mutate(Dark_Red = ifelse(
RedColor == 1 & Color_Number == "one" | Color_Number == "two", 1, 0))
Output:
id RedColor Color_Number Dark_Red
1 1 1 one 1
2 2 0 one 0
3 3 1 three 0
4 4 1 two 1
5 6 0 two 1
6 8 1 six 0
First answer:
We could use ifelse
str_detect is not appropriate as Ronak already explained:
library(dplyr)
df %>%
mutate(Dark_Red = ifelse(
RedColor == 1 & Color_Number == 1 | Color_Number == 2, 1, 0))
Output:
id RedColor Color_Number Dark_Red
1 1 1 1 1
2 2 0 1 0
3 3 1 3 0
4 4 1 2 1
5 6 0 2 1
6 8 1 6 0
For exact matches don't perform regex match. str_detect is used for pattern matching. Use %in% to match multiple values.
library(dplyr)
df <- df %>% mutate(Dark_Red = as.integer(RedColor == 1 & Color_Number %in% 1:2))
df
# id RedColor Color_Number Dark_Red
#1 1 1 1 1
#2 2 0 1 0
#3 3 1 3 0
#4 4 1 2 1
#5 6 0 2 0
#6 8 1 6 0
If you want to write this in base R use transform -
df <- transform(df, Dark_Red = as.integer(RedColor == 1 & Color_Number %in% 1:2))
data
df <- structure(list(id = c(1L, 2L, 3L, 4L, 6L, 8L), RedColor = c(1L,
0L, 1L, 1L, 0L, 1L), Color_Number = c(1L, 1L, 3L, 2L, 2L, 6L)),
row.names = c(NA, -6L), class = "data.frame")
you can use ifelse inside the mutate call instead of filtering:
test <- test %>%
mutate(Darkred=ifelse((RedColor==TRUE & Color_Number %in% 1:2), 1,0))
> test
# A tibble: 10 × 4
id RedColor Color_Number Darkred
<int> <int> <int> <dbl>
1 1 1 2 1
2 2 1 2 1
3 3 1 3 0
4 4 1 3 0
5 5 0 4 0
6 6 0 2 0
7 7 1 3 0
8 8 1 4 0
9 9 0 5 0
10 10 0 3 0
Data:
test<-data_frame(id=1:10,
RedColor=rbinom(10,1,0.5),
Color_Number=sample(1:5,10,TRUE,rep(.2,5)))
I have a data based 3 groups : SAMPN,PERNO,loop
there are 2 columns, mode1 and mode2. and a column called int.
SAMPN PERNO loop mode1 mode2 int
1 1 1 1 2 NA
1 1 1 2 1 NA
1 1 1 3 2 0
1 2 1 3 2 NA
1 2 1 1 1 2
2 2 1 3 2 NA
2 2 1 1 3 NA
2 2 1 3 1 0
2 2 2 1 2 NA
2 2 2 3 1 2
SAMPN is family index, PERNO is index of persons in each family and loop is tour of each person. the last row of each loop for each person is 0 or 2 and and rest of loop is NA. in each family and for each person and each loop I want copy the column mode 1 in int if the last row of loop is 0 and copy mode2 if the last row of loo is 2.
output
SAMPN PERNO loop mode1 mode2 int
1 1 1 1 2 1
1 1 1 2 1 2
1 1 1 3 2 3
1 2 1 3 2 2
1 2 1 1 1 1
2 2 1 3 2 3
2 2 1 1 3 1
2 2 1 3 1 3
2 2 2 1 2 2
2 2 2 3 1 1
the first 3 rows is loop of first person in the first family, I filled that loop by mode1 because the third row was 0. and so on
Here's a way using dplyr
df <- read.table(h=T,text="SAMPN PERNO loop mode1 mode2 int
1 1 1 1 2 NA
1 1 1 2 1 NA
1 1 1 3 2 0
1 2 1 3 2 NA
1 2 1 1 1 2
2 2 1 3 2 NA
2 2 1 1 3 NA
2 2 1 3 1 0
2 2 2 1 2 NA
2 2 2 3 1 2")
library(dplyr)
df %>%
group_by(loop, SAMPN, PERNO) %>%
mutate(int = if(last(int) == 0) mode1 else mode2) %>%
ungroup()
#> # A tibble: 10 x 6
#> SAMPN PERNO loop mode1 mode2 int
#> <int> <int> <int> <int> <int> <int>
#> 1 1 1 1 1 2 1
#> 2 1 1 1 2 1 2
#> 3 1 1 1 3 2 3
#> 4 1 2 1 3 2 2
#> 5 1 2 1 1 1 1
#> 6 2 2 1 3 2 3
#> 7 2 2 1 1 3 1
#> 8 2 2 1 3 1 3
#> 9 2 2 2 1 2 2
#> 10 2 2 2 3 1 1
If you have more values than 0 or 2, switch could be a good alternative :
df %>%
group_by(loop, SAMPN, PERNO) %>%
mutate(int = switch(
as.character(last(int)),
`0` = mode1,
`2` = mode2)) %>%
ungroup()
# same output!
We can also use case_when
library(dplyr)
df %>%
group_by(loop, SAMPN, PERNO) %>%
mutate(int = case_when(rep(last(int) == 0, n()) ~ mode1, TRUE ~mode2))
# A tibble: 10 x 6
# Groups: loop, SAMPN, PERNO [4]
# SAMPN PERNO loop mode1 mode2 int
# <int> <int> <int> <int> <int> <int>
# 1 1 1 1 1 2 1
# 2 1 1 1 2 1 2
# 3 1 1 1 3 2 3
# 4 1 2 1 3 2 2
# 5 1 2 1 1 1 1
# 6 2 2 1 3 2 3
# 7 2 2 1 1 3 1
# 8 2 2 1 3 1 3
#9 2 2 2 1 2 2
#10 2 2 2 3 1 1
data
df <- structure(list(SAMPN = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L), PERNO = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), loop = c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L), mode1 = c(1L, 2L, 3L, 3L,
1L, 3L, 1L, 3L, 1L, 3L), mode2 = c(2L, 1L, 2L, 2L, 1L, 2L, 3L,
1L, 2L, 1L), int = c(NA, NA, 0L, NA, 2L, NA, NA, 0L, NA, 2L)),
class = "data.frame", row.names = c(NA,
-10L))
I have a data.frame as below and I want to add a variable describing the longest consecutive count of 1 in the VALUE variable observed in the group (i.e. longest consecutive rows with 1 in VALUE per group).
GROUP_ID VALUE
1 0
1 1
1 1
1 1
1 1
1 0
2 1
2 1
2 0
2 1
2 1
2 1
3 1
3 0
3 1
3 0
So the output would look like this:
GROUP_ID VALUE CONSECUTIVE
1 0 4
1 1 4
1 1 4
1 1 4
1 1 4
1 0 4
2 1 3
2 1 3
2 0 3
2 1 3
2 1 3
2 1 3
3 1 1
3 0 1
3 1 1
3 0 1
Any help would be greatly appreciated!
Using dplyr:
library(dplyr)
dat %>%
group_by(GROUP_ID) %>%
mutate(CONSECUTIVE = {rl <- rle(VALUE); max(rl$lengths[rl$values == 1])})
which gives:
# A tibble: 16 x 3
# Groups: GROUP_ID [3]
GROUP_ID VALUE CONSECUTIVE
<int> <int> <int>
1 1 0 4
2 1 1 4
3 1 1 4
4 1 1 4
5 1 1 4
6 1 0 4
7 2 1 3
8 2 1 3
9 2 0 3
10 2 1 3
11 2 1 3
12 2 1 3
13 3 1 1
14 3 0 1
15 3 1 1
16 3 0 1
Or with data.table:
library(data.table)
setDT(dat) # convert to a 'data.table'
dat[, CONSECUTIVE := {rl <- rle(VALUE); max(rl$lengths[rl$values == 1])}
, by = GROUP_ID][]
We can use ave with rle and get maximum occurrence of consecutive 1's for each group. (GROUP_ID)
df$Consecutive <- ave(df$VALUE, df$GROUP_ID, FUN = function(x) {
y <- rle(x == 1)
max(y$lengths[y$values])
})
df
# GROUP_ID VALUE Consecutive
#1 1 0 4
#2 1 1 4
#3 1 1 4
#4 1 1 4
#5 1 1 4
#6 1 0 4
#7 2 1 3
#8 2 1 3
#9 2 0 3
#10 2 1 3
#11 2 1 3
#12 2 1 3
#13 3 1 1
#14 3 0 1
#15 3 1 1
#16 3 0 1
Here is another option with data.table
library(data.table)
library(dplyr)
setDT(df1)[, CONSECUTIVE := max(table(na_if(rleid(VALUE)*VALUE, 0))), .(GROUP_ID)]
df1
# GROUP_ID VALUE CONSECUTIVE
# 1: 1 0 4
# 2: 1 1 4
# 3: 1 1 4
# 4: 1 1 4
# 5: 1 1 4
# 6: 1 0 4
# 7: 2 1 3
# 8: 2 1 3
# 9: 2 0 3
#10: 2 1 3
#11: 2 1 3
#12: 2 1 3
#13: 3 1 1
#14: 3 0 1
#15: 3 1 1
#16: 3 0 1
data
df1 <- structure(list(GROUP_ID = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L, 3L, 3L, 3L, 3L), VALUE = c(0L, 1L, 1L, 1L, 1L, 0L,
1L, 1L, 0L, 1L, 1L, 1L, 1L, 0L, 1L, 0L)), class = "data.frame", row.names = c(NA,
-16L))
I am trying to identify which trials, within a long form dataset, are repeated but only within certain blocks per participant. My data is structured something like this:
sub block trial item
1 1 1 A
1 1 2 B
1 2 1 A
1 2 2 B
1 3 1 B
1 3 2 C
2 1 1 A
2 1 2 B
2 2 1 A
2 2 2 B
2 3 1 B
2 3 2 C
What I would like to create is a new column that indicates for each participant, which items are repeating and another new column with a new trial code, but only if the items are repeated in blocks 2 and 3. So it would look something like this:
sub block trial item dup newtrial
1 1 1 A FALSE 1
1 1 2 B FALSE 2
1 2 1 A FALSE 1
1 2 2 B FALSE 2
1 3 1 C FALSE 1
1 3 2 B TRUE 102
2 1 1 A FALSE 1
2 1 2 B FALSE 2
2 2 1 A FALSE 1
2 2 2 B FALSE 2
2 3 1 C FALSE 1
2 3 2 B TRUE 102
I have been able to identify duplicates across the whole dataset and add 100 to each trial number using the following code:
data$dup<-duplicated(data$item)
data$newtrial<-NA
data<-transform(data,
item=make.unique(as.character(item)),
newtrial=ifelse(duplicated(item),trial+100, trial))
What I have not been able to figure out is how to constrain the function to each individual subject and only certain blocks within each subject number.
Thanks!
another option using data.table:
library(data.table)
xt <- fread("sub block trial item
1 1 1 A
1 1 2 B
1 2 1 A
1 2 2 B
1 3 1 B
1 3 2 B
2 1 1 A
2 1 2 B
2 2 1 A
2 2 2 B
2 3 1 B
2 3 2 B")
xt[,
c("dup","ntrial") := {
dup <- duplicated(item)
tt <- ifelse(dup,trial+100L,trial)
list(dup,tt)
},"sub,block"]
You can do this using dplyr grouping the observations by sub and block:
library(dplyr)
res <- data %>% group_by(sub,block) %>%
mutate(dup=duplicated(item)) %>%
ungroup %>%
mutate(newtrial=ifelse(dup,trial+100,trial))
We use mutate to create new columns dup and newtrial.
Data: Modifying your data slightly to introduce duplicate item for sub=1, block=3 and sub=2, block=3:
data <- structure(list(sub = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 2L), block = c(1L, 1L, 2L, 2L, 3L, 3L, 1L, 1L, 2L, 2L, 3L,
3L), trial = c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L
), item = structure(c(1L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 2L,
2L, 2L), .Label = c("A", "B"), class = "factor")), .Names = c("sub",
"block", "trial", "item"), class = "data.frame", row.names = c(NA,
-12L))
## sub block trial item
##1 1 1 1 A
##2 1 1 2 B
##3 1 2 1 A
##4 1 2 2 B
##5 1 3 1 B
##6 1 3 2 B
##7 2 1 1 A
##8 2 1 2 B
##9 2 2 1 A
##10 2 2 2 B
##11 2 3 1 B
##12 2 3 2 B
Using this data:
print(res)
### A tibble: 12 x 6
## sub block trial item dup newtrial
## <int> <int> <int> <fctr> <lgl> <dbl>
##1 1 1 1 A FALSE 1
##2 1 1 2 B FALSE 2
##3 1 2 1 A FALSE 1
##4 1 2 2 B FALSE 2
##5 1 3 1 B FALSE 1
##6 1 3 2 B TRUE 102
##7 2 1 1 A FALSE 1
##8 2 1 2 B FALSE 2
##9 2 2 1 A FALSE 1
##10 2 2 2 B FALSE 2
##11 2 3 1 B FALSE 1
##12 2 3 2 B TRUE 102
I am trying to append rows to an R data.frame. Here is an example of a data.frame "foo":
A B C D
1 1 1 200
1 1 2 50
1 1 3 15
1 2 1 150
1 2 4 50
1 3 1 300
2 1 2 40
2 1 4 90
2 3 2 80
For every A, there are 3 possible values of B, and for every B, there are 4 possible values of C. However, the initial df only contains non-zero values of D. I'd like to manipulate the df so that zeros are included for both B and C. Thus, the df would show 0's in D for any value of B/C that was 0. I have seen questions that address this with one column, but couldn't find a question addressing it with multiple columns. The final df would look like this:
A B C D
1 1 1 200
1 1 2 50
1 1 3 15
1 1 4 0
1 2 1 150
1 2 2 0
1 2 3 0
1 2 4 50
1 3 1 300
1 3 2 0
1 3 3 0
1 3 4 0
2 1 1 0
2 1 2 40
2 1 3 0
2 1 4 90
2 2 1 0
2 2 2 0
2 2 3 0
2 2 4 0
2 3 1 0
2 3 2 80
2 3 3 0
2 3 4 0
I first tried creating a dummy data frame that then merged with the initial df, but something isn't working right. Here's the current code, which I know is wrong because this code only generates rows based on A. I think I want to make the dummy frame based on A and B but I don't know how - could an if/else function work here?:
# create dummy df
dummy <- as.data.frame(
cbind(
sort(rep(unique(foo$A), 12)),
rep(1:3,length(unique(foo$A)))))
colnames(dummy) <- c("A","B")
foo$A <- as.numeric(foo$A)
foo$B <- as.numeric(foo$C)
# merge with foo
mergedummy <- merge(dummy,foo,all.x=T)
Any insight is greatly appreciated - thanks!
A one liner:
merge(dat, data.frame(table(dat[1:3]))[-4],all.y=TRUE)
# A B C D
#1 1 1 1 200
#2 1 1 2 50
#3 1 1 3 15
#4 1 1 4 NA
#...
Or maybe less complicated:
out <- data.frame(xtabs(D ~ ., data=dat))
out[do.call(order,out[1:3]),]
# A B C Freq
#1 1 1 1 200
#7 1 1 2 50
#13 1 1 3 15
#19 1 1 4 0
#...
Where dat is:
dat <- structure(list(A = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), B = c(1L,
1L, 1L, 2L, 2L, 3L, 1L, 1L, 3L), C = c(1L, 2L, 3L, 1L, 4L, 1L,
2L, 4L, 2L), D = c(200L, 50L, 15L, 150L, 50L, 300L, 40L, 90L,
80L)), .Names = c("A", "B", "C", "D"), class = "data.frame", row.names = c(NA,
-9L))
I created a master data frame which includes all combinations of A, B, and C as you describe in the expected outcome. Then, I merge the master data frame and your data frame. Finally, I replaced NA with 0.
master <- data.frame(A = rep(1:2, each = 12),
B = rep(1:3, each = 4),
C = rep(1:4, times = 6))
library(dplyr)
master %>%
left_join(., mydf) %>%
mutate(D = ifelse(D %in% NA, 0, D))
# A B C D
#1 1 1 1 200
#2 1 1 2 50
#3 1 1 3 15
#4 1 1 4 0
#5 1 2 1 150
#6 1 2 2 0
#7 1 2 3 0
#8 1 2 4 50
#9 1 3 1 300
#10 1 3 2 0
#11 1 3 3 0
#12 1 3 4 0
#13 2 1 1 0
#14 2 1 2 40
#15 2 1 3 0
#16 2 1 4 90
#17 2 2 1 0
#18 2 2 2 0
#19 2 2 3 0
#20 2 2 4 0
#21 2 3 1 0
#22 2 3 2 80
#23 2 3 3 0
#24 2 3 4 0
Here is one solution:
foo <- merge(expand.grid(lapply(foo[,1:3], unique)), foo, all=TRUE, sort=TRUE)
foo[is.na(foo)] <- 0