I have the following data frame
id<-c(1,1,1,1,1,1,1,1,2,2,2,2,3,3,3,3)
time<-c(0,1,2,3,4,5,6,7,0,1,2,3,0,1,2,3)
value<-c(1,1,6,1,2,0,0,1,2,6,2,2,1,1,6,1)
d<-data.frame(id, time, value)
The value 6 appears only once for each id. For every id, i would like to remove all rows after the line with the value 6 per id except the first two lines coming after.
I've searched and found a similar problem, but i couldnt adapt it myself. I therefore use the code of this thread
In the above case the final data frame should be
id time value
1 0 1
1 1 1
1 2 6
1 3 1
1 4 2
2 0 2
2 1 6
2 2 2
2 3 2
3 0 1
3 1 1
3 2 6
3 3 1
On of the solution given seems getting very close to what i need. But i didn't manage to adapt it. Could u help me?
library(plyr)
ddply(d, "id",
function(x) {
if (any(x$value == 6)) {
subset(x, time <= x[x$value == 6, "time"])
} else {
x
}
}
)
Thank you very much.
We could use data.table. Convert the 'data.frame' to 'data.table' (setDT(d)). Grouped by the 'id' column, we get the position of 'value' that is equal to 6. Add 2 to it. Find the min of the number of elements for that group (.N) and the position, get the seq, and use that to subset the dataset. We can also add an if/else condition to check whether there are any 6 in the 'value' column or else to return .SD without any subsetting.
library(data.table)
setDT(d)[, if(any(value==6)) .SD[seq(min(c(which(value==6) + 2, .N)))]
else .SD, by = id]
# id time value
# 1: 1 0 1
# 2: 1 1 1
# 3: 1 2 6
# 4: 1 3 1
# 5: 1 4 2
# 6: 2 0 2
# 7: 2 1 6
# 8: 2 2 2
# 9: 2 3 2
#10: 3 0 1
#11: 3 1 1
#12: 3 2 6
#13: 3 3 1
#14: 4 0 1
#15: 4 1 2
#16: 4 2 5
Or as #Arun mentioned in the comments, we can use the ?head to subset, which would be faster
setDT(d)[, if(any(value==6)) head(.SD, which(value==6L)+2L) else .SD, by = id]
Or using dplyr, we group by 'id', get the position of 'value' 6 with which, add 2, get the seq and use that numeric index within slice to extract the rows.
library(dplyr)
d %>%
group_by(id) %>%
slice(seq(which(value==6)+2))
# id time value
#1 1 0 1
#2 1 1 1
#3 1 2 6
#4 1 3 1
#5 1 4 2
#6 2 0 2
#7 2 1 6
#8 2 2 2
#9 2 3 2
#10 3 0 1
#11 3 1 1
#12 3 2 6
#13 3 3 1
#14 4 0 1
#15 4 1 2
#16 4 2 5
data
d <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 4L, 4L, 4L), time = c(0L, 1L, 2L, 3L, 4L, 0L, 1L,
2L, 3L, 0L, 1L, 2L, 3L, 0L, 1L, 2L), value = c(1L, 1L, 6L, 1L,
2L, 2L, 6L, 2L, 2L, 1L, 1L, 6L, 1L, 1L, 2L, 5L)), .Names = c("id",
"time", "value"), class = "data.frame", row.names = c(NA, -16L))
Related
I´m having a data.frame of the following form:
ID Var1
1 1
1 1
1 3
1 4
1 1
1 0
2 2
2 2
2 6
2 7
2 8
2 0
3 0
3 2
3 1
3 3
3 2
3 4
and I would like to get there:
ID Var1 X
1 1 0
1 1 0
1 3 0
1 4 5
1 1 5
1 0 5
2 2 0
2 2 0
2 6 0
2 7 10
2 8 10
2 0 10
3 0 0
3 2 0
3 1 0
3 3 3
3 2 3
3 4 3
so in words: I´d like to calculate the sum of the variable in a window = 3, and then report the results obtained in the previous window. This should happen with respect to the IDs and thus the first three observations on every ID should be returned with 0, as there is no previous time period that could be reported.
For understanding: In the actual dataset each row corresponds to one week and the window = 7. So X is supposed to give information on the sum of Var1 in the previous week.
I have tried using some rollapply stuff, but always ended in an error and also the window would be a rolling window if I got that right, which is specifically not what I need.
Thanks for your answers!
In rollapply, the width argument can be a list which provides the offsets to use. In this case we want to use the points 3, 2 and 1 back for the first point, 4, 3 and 2 back for the second, 5, 4 and 3 back for the third and then recycle. That is, for a window width of k = 3 we would want the following list of offset vectors:
w <- list(-(3:1), -(4:2), -(5:3))
In general we can write w below in terms of the window width k. ave then invokes rollapply with that width list for each ID.
library(zoo)
k <- 3
w <- lapply(1:k, function(x) seq(to = -x, length = k))
transform(DF, X = ave(Var1, ID, FUN = function(x) rollapply(x, w, sum, fill = 0)))
giving:
ID Var1 X
1 1 1 0
2 1 1 0
3 1 3 0
4 1 4 5
5 1 1 5
6 1 0 5
7 2 2 0
8 2 2 0
9 2 6 0
10 2 7 10
11 2 8 10
12 2 0 10
13 3 0 0
14 3 2 0
15 3 1 0
16 3 3 3
17 3 2 3
18 3 4 3
Note
The input DF in reproducible form is:
DF <- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L), Var1 = c(1L, 1L, 3L, 4L, 1L,
0L, 2L, 2L, 6L, 7L, 8L, 0L, 0L, 2L, 1L, 3L, 2L, 4L)),
class = "data.frame", row.names = c(NA, -18L))
We could group by 'ID', create a new grouping column with window size of 3 using gl, then get the summarized output by taking the sum of 'Var1' and placing the 'Var1' in a list, get the lag of 'X' and unnest
library(dplyr) #1.0.0
library(tidyr)
df1 %>%
# // grouping by ID
group_by(ID) %>%
# // create another group added with gl
group_by(grp = as.integer(gl(n(), 3, n())), .add = TRUE) %>%
# // get the sum of Var1, while changing the Var1 in a list
summarise(X = sum(Var1), Var1 = list(Var1)) %>%
# // get the lag of X
mutate(X = lag(X, default = 0)) %>%
# // unnest the list column
unnest(c(Var1)) %>%
select(names(df1), X)
# A tibble: 18 x 3
# Groups: ID [3]
# ID Var1 X
# <int> <int> <dbl>
# 1 1 1 0
# 2 1 1 0
# 3 1 3 0
# 4 1 4 5
# 5 1 1 5
# 6 1 0 5
# 7 2 2 0
# 8 2 2 0
# 9 2 6 0
#10 2 7 10
#11 2 8 10
#12 2 0 10
#13 3 0 0
#14 3 2 0
#15 3 1 0
#16 3 3 3
#17 3 2 3
#18 3 4 3
data
df1 <- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L), Var1 = c(1L, 1L, 3L, 4L, 1L,
0L, 2L, 2L, 6L, 7L, 8L, 0L, 0L, 2L, 1L, 3L, 2L, 4L)), class = "data.frame",
row.names = c(NA,
-18L))
I am quite a beginner in R but thanks to the community of Stackoverflow I am improving!
However, I am stuck with a problem:
I have a dataset with 5 variables:
id_house represents the id for each household
id_ind is an id which values 1 for the first individual in the household, 2 for the next, 3 for the third...
Indicator_tb_men which indicates if the first person has answered to the survey (1 = yes, 0 = no). All the other members of the household take the value 0.
id_house id_ind indicator_tb_men
1 1 1
1 2 0
2 1 1
3 1 0
3 2 0
3 3 0
4 1 1
5 1 0
I would like to delete all members of households where the first individual has not answered the survey.
So it would give:
id_house id_ind indicator_tb_men
1 1 1
1 2 0
2 1 1
4 1 1
Using dplyr here is one way :
library(dplyr)
df %>%
arrange(id_house, id_ind) %>%
group_by(id_house) %>%
filter(first(indicator_tb_men) != 0)
# id_house id_ind indicator_tb_men
# <int> <int> <int>
#1 1 1 1
#2 1 2 NA
#3 2 1 1
#4 4 1 1
data
df <- structure(list(id_house = c(1L, 1L, 2L, 3L, 3L, 3L, 4L, 5L),
id_ind = c(1L, 2L, 1L, 1L, 2L, 3L, 1L, 1L), indicator_tb_men = c(1L,
NA, 1L, 0L, NA, NA, 1L, 0L)), class = "data.frame", row.names = c(NA, -8L))
in base we can use nested logic
df[df$id_house %in% df$id_house[df$id_ind == 1 & df$indicator_tb_men == 1],]
id_house id_ind indicator_tb_men
1 1 1 1
2 1 2 NA
3 2 1 1
7 4 1 1
Data: Using Ronak Shah's data
ID Number Var
1 2 6
1 2 7
1 1 8
1 2 9
1 2 10
2 2 3
2 2 4
2 1 5
2 2 6
Each person has several records.
There is only one record of a person whose Number is 1, the rest is 2.
The variable Var has different values for the same person.
When the Number equals to 1, the corresponding Var (we call it P) is different for different persons.
Now, I want to delete the rows whose Var > P for every person.
At the end, I want this
ID Number Var
1 2 6
1 2 7
1 1 8
2 2 3
2 2 4
2 1 5
You can use dplyr::first where Num==1 to get the first Var value
library(dplyr)
df %>% group_by(ID) %>% mutate(Flag=first(Var[Number==1])) %>%
filter(Var <= Flag) %>% select(-Flag)
#short version and you sure there is a one Num==1
df %>% group_by(ID) %>% filter(Var <= Var[Number==1])
Here is a solution with data.table:
library(data.table)
dt <- fread(
"ID Number Var
1 2 6
1 2 7
1 1 8
1 2 9
1 2 10
2 2 3
2 2 4
2 1 5
2 2 6")
dt[, .SD[Var <= Var[Number==1]], ID]
# ID Number Var
# 1: 1 2 6
# 2: 1 2 7
# 3: 1 1 8
# 4: 2 2 3
# 5: 2 2 4
# 6: 2 1 5
A base R option would be
df1[with(df1, Var <= ave(Var * (Number == 1), ID, FUN = function(x) x[x!=0])),]
# ID Number Var
#1 1 2 6
#2 1 2 7
#3 1 1 8
#6 2 2 3
#7 2 2 4
#8 2 1 5
data
df1 <- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), Number = c(2L,
2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L), Var = c(6L, 7L, 8L, 9L, 10L,
3L, 4L, 5L, 6L)), row.names = c(NA, -9L), class = "data.frame")
Suppose we have the following data with column names "id", "time" and "x":
df<-
structure(
list(
id = c(1L, 1L, 1L, 2L, 2L, 3L, 3L),
time = c(20L, 6L, 7L, 11L, 13L, 2L, 6L),
x = c(1L, 1L, 0L, 1L, 1L, 1L, 0L)
),
.Names = c("id", "time", "x"),
class = "data.frame",
row.names = c(NA,-7L)
)
Each id has multiple observations for time and x. I want to extract the last observation for each id and form a new dataframe which repeats these observations according to the number of observations per each id in the original data. I am able to extract the last observations for each id using the following codes
library(dplyr)
df<-df%>%
group_by(id) %>%
filter( ((x)==0 & row_number()==n())| ((x)==1 & row_number()==n()))
What is left unresolved is the repetition aspect. The expected output would look like
df <-
structure(
list(
id = c(1L, 1L, 1L, 2L, 2L, 3L, 3L),
time = c(7L, 7L, 7L, 13L, 13L, 6L, 6L),
x = c(0L, 0L, 0L, 1L, 1L, 0L, 0L)
),
.Names = c("id", "time", "x"),
class = "data.frame",
row.names = c(NA,-7L)
)
Thanks for your help in advance.
We can use ave to find the max row number for each ID and subset it from the data frame.
df[ave(1:nrow(df), df$id, FUN = max), ]
# id time x
#3 1 7 0
#3.1 1 7 0
#3.2 1 7 0
#5 2 13 1
#5.1 2 13 1
#7 3 6 0
#7.1 3 6 0
You can do this by using last() to grab the last row within each id.
df %>%
group_by(id) %>%
mutate(time = last(time),
x = last(x))
Because last(x) returns a single value, it gets expanded out to fill all the rows in the mutate() call.
This can also be applied to an arbitrary number of variables using mutate_at:
df %>%
group_by(id) %>%
mutate_at(vars(-id), ~ last(.))
slice will be your friend in the tidyverse I reckon:
df %>%
group_by(id) %>%
slice(rep(n(),n()))
## A tibble: 7 x 3
## Groups: id [3]
# id time x
# <int> <int> <int>
#1 1 7 0
#2 1 7 0
#3 1 7 0
#4 2 13 1
#5 2 13 1
#6 3 6 0
#7 3 6 0
In data.table, you could also use the mult= argument of a join:
library(data.table)
setDT(df)
df[df[,.(id)], on="id", mult="last"]
# id time x
#1: 1 7 0
#2: 1 7 0
#3: 1 7 0
#4: 2 13 1
#5: 2 13 1
#6: 3 6 0
#7: 3 6 0
And in base R, a merge will get you there too:
merge(df["id"], df[!duplicated(df$id, fromLast=TRUE),])
# id time x
#1 1 7 0
#2 1 7 0
#3 1 7 0
#4 2 13 1
#5 2 13 1
#6 3 6 0
#7 3 6 0
Using data.table you can try
library(data.table)
setDT(df)[,.(time=rep(time[.N],.N), x=rep(x[.N],.N)), by=id]
id time x
1: 1 7 0
2: 1 7 0
3: 1 7 0
4: 2 13 1
5: 2 13 1
6: 3 6 0
7: 3 6 0
Following #thelatemai, to avoid name the columns you can also try
df[, .SD[rep(.N,.N)], by=id]
id time x
1: 1 7 0
2: 1 7 0
3: 1 7 0
4: 2 13 1
5: 2 13 1
6: 3 6 0
7: 3 6 0
I would like to create variable "Time" which basically indicates the number of times variable ID showed up within each day minus 1. In other words, the count is lagged by 1 and the first time ID showed up in a day should be left blank. Second time the same ID shows up on a given day should be 1.
Basically, I want to create the "Time" variable in the example below.
ID Day Time Value
1 1 0
1 1 1 0
1 1 2 0
1 2 0
1 2 1 0
1 2 2 0
1 2 3 1
2 1 0
2 1 1 0
2 1 2 0
Below is the code I am working on. Have not been successful with it.
data$time<-data.frame(data$ID,count=ave(data$ID==data$ID, data$Day, FUN=cumsum))
We can do this with data.table. Convert the 'data.frame' to 'data.table' (setDT(df1)), grouped by 'ID', 'Day', we get the lag of sequence of rows (shift(seq_len(.N))) and assign (:=) it as "Time" column.
library(data.table)
setDT(df1)[, Time := shift(seq_len(.N)), .(ID, Day)]
df1
# ID Day Value Time
# 1: 1 1 0 NA
# 2: 1 1 0 1
# 3: 1 1 0 2
# 4: 1 2 0 NA
# 5: 1 2 0 1
# 6: 1 2 0 2
# 7: 1 2 1 3
# 8: 2 1 0 NA
# 9: 2 1 0 1
#10: 2 1 0 2
Or with base R
with(df1, ave(Day, Day, ID, FUN= function(x)
ifelse(seq_along(x)!=1, seq_along(x)-1, NA)))
#[1] NA 1 2 NA 1 2 3 NA 1 2
Or without the ifelse
with(df1, ave(Day, Day, ID, FUN= function(x)
NA^(seq_along(x)==1)*(seq_along(x)-1)))
#[1] NA 1 2 NA 1 2 3 NA 1 2
data
df1 <- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L),
Day = c(1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 1L), Value = c(0L,
0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L)), .Names = c("ID", "Day",
"Value"), row.names = c(NA, -10L), class = "data.frame")