I'm working with a dataset about migration across the country with the following columns:
i birth gender race region urban wage year educ
1 58 2 3 1 1 4620 1979 12
1 58 2 3 1 1 4620 1980 12
1 58 2 3 2 1 4620 1981 12
1 58 2 3 2 1 4700 1982 12
.....
i birth gender race region urban wage year educ
45 65 2 3 3 1 NA 1979 10
45 65 2 3 3 1 NA 1980 10
45 65 2 3 4 2 11500 1981 10
45 65 2 3 1 1 11500 1982 10
i = individual id. They follow a large group of people for 25 years and record changes in 'region' (categorical variables, 1-4) , 'urban' (dummy), 'wage' and 'educ'.
How do I count the aggregate number of times 'region' or 'urban' has changed (eg: from region 1 to region 3 or from urban 0 to 1) during the observation period (25 year period) within each subject? I also have some NA's in the data (which should be ignored)
A simplified version of expected output:
i changes in region
1 1
...
45 2
i changes in urban
1 0
...
45 2
I would then like to sum up the number of changes for region and urban.
I came across these answers: Count number of changes in categorical variables during repeated measurements and Identify change in categorical data across datapoints in R but I still don't get it.
Here's a part of the data for i=4.
i birth gender race region urban wage year educ
4 62 2 3 1 1 NA 1979 9
4 62 2 3 NA NA NA 1980 9
4 62 2 3 4 1 0 1981 9
4 62 2 3 4 1 1086 1982 9
4 62 2 3 1 1 70 1983 9
4 62 2 3 1 1 0 1984 9
4 62 2 3 1 1 0 1985 9
4 62 2 3 1 1 7000 1986 9
4 62 2 3 1 1 17500 1987 9
4 62 2 3 1 1 21320 1988 9
4 62 2 3 1 1 21760 1989 9
4 62 2 3 1 1 0 1990 9
4 62 2 3 1 1 0 1991 9
4 62 2 3 1 1 30500 1992 9
4 62 2 3 1 1 33000 1993 9
4 62 2 3 NA NA NA 1994 9
4 62 2 3 4 1 35000 1996 9
Here, output should be:
i change_reg change_urban
4 3 0
Here is something I hope will get your closer to what you need.
First you group by i. Then, you can then create a column that will indicate a 1 for each change in region. This compares the current value for the region with the previous value (using lag). Note if the previous value is NA (when looking at the first value for a given i), it will be considered no change.
Same approach is taken for urban. Then, summarize totaling up all the changes for each i. I left in these temporary variables so you can examine if you are getting the results desired.
Edit: If you wish to remove rows that have NA for region or urban you can add drop_na first.
library(dplyr)
library(tidyr)
df_tot <- df %>%
drop_na(region, urban) %>%
group_by(i) %>%
mutate(reg_change = ifelse(region == lag(region) | is.na(lag(region)), 0, 1),
urban_change = ifelse(urban == lag(urban) | is.na(lag(urban)), 0, 1)) %>%
summarize(tot_region = sum(reg_change),
tot_urban = sum(urban_change))
# A tibble: 3 x 3
i tot_region tot_urban
<int> <dbl> <dbl>
1 1 1 0
2 4 3 0
3 45 2 2
Edit: Afterwards, to get a grand total for both tot_region and tot_urban columns, you can use colSums. (Store your earlier result as df_tot as above.)
colSums(df_tot[-1])
tot_region tot_urban
6 2
The typical preparation steps for mstate involve converting "wide" format data (1x row per 'patient') into "multi-state" format data (multiple rows per 'patient' for each possible transition in the multi-state model).
For example, data in wide format:
library(mstate)
data(ebmt4)
ebmt <- ebmt4
> head(ebmt)
id rec rec.s ae ae.s recae recae.s rel rel.s srv srv.s year agecl proph match
1 1 22 1 995 0 995 0 995 0 995 0 1995-1998 20-40 no no gender mismatch
2 2 29 1 12 1 29 1 422 1 579 1 1995-1998 20-40 no no gender mismatch
3 3 1264 0 27 1 1264 0 1264 0 1264 0 1995-1998 20-40 no no gender mismatch
4 4 50 1 42 1 50 1 84 1 117 1 1995-1998 20-40 no gender mismatch
5 5 22 1 1133 0 1133 0 114 1 1133 0 1995-1998 >40 no gender mismatch
6 6 33 1 27 1 33 1 1427 0 1427 0 1995-1998 20-40 no no gender mismatch
Is converted to multi-state format:
tmat <- transMat(x = list(c(2, 3, 5, 6), c(4, 5, 6), c(4, 5, 6), c(5, 6), c(), c()), names = c("Tx", "Rec", "AE", "Rec+AE", "Rel", "Death"))
msebmt <- msprep(data = ebmt, trans = tmat, time = c(NA, "rec", "ae", "recae", "rel", "srv"), status = c(NA, "rec.s", "ae.s", "recae.s", "rel.s", "srv.s"), keep = c("match", "proph", "year", "agecl"))
> head(msebmt)
An object of class 'msdata'
Data:
id from to trans Tstart Tstop time status match proph year agecl
1 1 1 2 1 0 22 22 1 no gender mismatch no 1995-1998 20-40
2 1 1 3 2 0 22 22 0 no gender mismatch no 1995-1998 20-40
3 1 1 5 3 0 22 22 0 no gender mismatch no 1995-1998 20-40
4 1 1 6 4 0 22 22 0 no gender mismatch no 1995-1998 20-40
5 1 2 4 5 22 995 973 0 no gender mismatch no 1995-1998 20-40
6 1 2 5 6 22 995 973 0 no gender mismatch no 1995-1998 20-40
But what if my original dataset has time-varying covariates (i.e. long format) and I want to format the data into multi-state mode? All of the tutorials I have found online are only for converting initially wide data to multi-state data (not initially long data); for example the mstate package vignette.
So, let's say I have the below data df, where id is for a 'patient', (start,stop] tell us the time periods, state is the state the patient is in at the end of the time period, and tv.cov is their time-varying covariate (assumed constant over the time period). Note that only patient id=5 has 3x entries and that person's tv.cov changes.
id start stop state tv.cov
1 0 1 1 1
2 0 4 1 2
3 0 7 1 1
4 0 10 1 5
5 0 6 1 4
5 6 10 2 10
5 10 15 3 12
Assuming the basic "illness-death" transition model:
tmat <- mstate::trans.illdeath(names = c("healthy", "sick", "death"))
> tmat
to
from healthy sick death
healthy NA 1 2
sick NA NA 3
death NA NA NA
How can I prep df into multi-state format?
As a hack, should I setup the data in "wide" format, format the data into "multi-state" format using msprep and then join another frame onto it which contains the time-varying covariates for each patient at each time interval?
I really need to speed some R code up. I have a large dataset from a particular sport. Each row in the data frame represents some type of action in the game. For each game (game_id) we have two teams (team_id) that take part in the game. time_ref in the data frame are the actions in chronological order for each game. type_id is the type of action in the game. player_off is set as TRUE or FALSE and is linked to action_id=3. action_id=3 represents a player getting a card and player_off is set to TRUE/FALSE if the player was sent off when they got that card. Example data.frame:
> df
game_id team_id action_id player_off time_ref
100 10 1 NA 1000
100 10 1 NA 1001
100 10 1 NA 1002
100 11 1 NA 1003
100 11 2 NA 1004
100 11 1 NA 1005
100 10 3 1 1006
100 11 1 NA 1007
100 10 1 NA 1008
100 10 1 NA 1009
101 12 3 0 1000
101 12 1 NA 1001
101 12 1 NA 1002
101 13 2 NA 1003
101 13 3 1 1004
101 12 1 NA 1005
101 13 1 NA 1006
101 13 1 NA 1007
101 12 1 NA 1008
101 12 1 NA 1009
What I need is another column in the data frame that gives me TRUE or FALSE on whether both teams had an equal/unequal number of players on the field while each action (row) took place.
So game_id=100 had an action_id=3 & player_off=1 for team_id=10 at time_ref=1006. So we know the teams were equal with number of players on the field up to that point but unequal for the rest of the game (time_ref>1006). The same thing occurred in game_id=101 also.
This an example of the data frame with an extra column I would like to have for the dataset.
>df
game_id team_id action_id player_off time_ref is_even
100 10 1 NA 1000 1
100 10 1 NA 1001 1
100 10 1 NA 1002 1
100 11 1 NA 1003 1
100 11 2 NA 1004 1
100 11 1 NA 1005 1
100 10 3 1 1006 1
100 11 1 NA 1007 0
100 10 1 NA 1008 0
100 10 1 NA 1009 0
101 12 3 0 1000 1
101 12 1 NA 1001 1
101 12 1 NA 1002 1
101 13 2 NA 1003 1
101 13 3 1 1004 1
101 12 1 NA 1005 0
101 13 1 NA 1006 0
101 13 1 NA 1007 0
101 12 1 NA 1008 0
101 12 1 NA 1009 0
So you can see that in game_id=100 a player was sent off at time_ref=1006 so all previous rows were marked as is_even=1 and subsequent marked as uneven or 0. Similar for game_id=101 at time_ref=1004.
What is the most efficient way of achieving this extra column? Preferably not using for loops.
For some vector
x = c(0, NA, NA, NA, 1, NA, NA, NA)
write a function to standardize the data (0 or 1 player lost), calculate the cumulative number of players lost, and compare this to zero,
fun0 = function(x) {
x[is.na(x)] = 0
cumsum(x) == 0
}
For several groups, use ave() with a grouping variable
x = c(x, rev(x))
grp = rep(1:2, each = length(x) / 2)
ave(x, grp, FUN = fun0)
For the data in the question, try
df$is_even = ave(df$player_off, df$game_id, FUN = fun)
Semantically, it seems likely that fun0() is more complicated than implied in this solution, specifically that if each team loses a player, they are again even, as #SunLisa says. If so, clean the data
df$player_off[is.na(df$player_off)] = 0
and change fun0(), e.g.,
fun1 <- function(x, team) {
is_team_1 <- team == head(team, 1) # is 'team' the first team?
x1 <- x & is_team_1 # lost player & team 1
x2 <- x & !is_team_1 # lost player & team 2
cumsum(x1) == cumsum(x2) # same total number of players?
}
(it doesn't seem like a good idea to coerce the logical return value to an integer). This could be applied by group with
df$is_even = ave(seq_len(nrow(df)), df$game_id, FUN = function(i) {
fun1(df$player_off[i], df$team_id[i])
})
or
split(df$is_even, df$game_id) <-
Map(fun1,
split(df$player_off, df$game_id),
split(df$team_id, df$game_id)
)
The implementation of ave() is useful to look at, the important line being
split(x, g) <- lapply(split(x, g), FUN)
The right-hand side splits x by group g, then applies FUN() to each group. The left-hand side split<-() is a tricky operation, using the group indexes to update the original vector x.
Comments
The original question asked for 'no for loops', but actually lapply() (in ave()) and Map() are exactly that; ave() is relatively efficient because of the split-apply-combine strategy it adopts, rather than what the OP probably implemented, which was likely to iterate through games, subset the data frame, then update the data.frame for each game. The subsetting would have duplicated subsets of the entire data set, and the update in particular would have copied at least the entire result column on each assignment; this copying would have slowed the execution down alot. It's also possible that the OP was struggling with fun0(); it would help to clarify the question, especially title, to identify that as the problem.
There are faster ways, especially using the data.table package, but the principle is the same -- identify a function that operates on a vector the way you'd like, and apply it by group.
An alternative, fully-vectorized, solution follows this suggestion to calculate a cumulative sum by group. For fun0(), standardize x to be the number of players leaving the game at a particular timepoint, without NAs
x[is.na(x)] = 0
For the equivalent of fun(), calculate the cumulative sum of players leaving the game, irrespective of group
cs = cumsum(x)
Correct this for the group that the cumulative sum applies to
in_game = cs - (grp - 1)
and set this to 'TRUE' when 0 players have left the game
is_even = (in_game == 0)
This relies on grp indexing from 1 to the number of groups; for the data here one might grp = match(df$game_id, unique(df$game_id)). A similar solution exists for fun1().
Here's a dplyr + tidyr solution to the problem, with the summary of what was done:
Manipulate the data by converting all NAs in player_off to 0 for easier summing and assigning the smaller team_num (assuming there are only 2) to team1 and the other to team2
"Tally" the player_offs using spread and fill the invalid combinations in the data with 0 -- for example, in game_id = 100, there's no team_id = 11 for time_ref = 1000
Take the cumulative sum of the lagged team1 and team2 vectors (and of course fill NAs with 0)
Code below:
require(dplyr)
require(tidyr)
df %>%
group_by(game_id) %>%
mutate(
player_off = player_off %>% replace(list = is.na(.), values = 0),
team_num = if_else(team_id == min(team_id), "team1", "team2")
) %>%
spread(key = team_num, value = player_off, fill = 0) %>%
arrange(game_id, time_ref) %>%
mutate(
team1_cum = cumsum(lag(team1, default = 0)),
team2_cum = cumsum(lag(team2, default = 0)),
is_even = as.integer(team1_cum == team2_cum)
) %>%
ungroup() %>%
select(-team1, -team2, -team1_cum, -team2_cum)
Output:
# A tibble: 20 x 5
game_id team_id action_id time_ref is_even
<int> <int> <int> <int> <int>
1 100 10 1 1000 1
2 100 10 1 1001 1
3 100 10 1 1002 1
4 100 11 1 1003 1
5 100 11 2 1004 1
6 100 11 1 1005 1
7 100 10 3 1006 1
8 100 11 1 1007 0
9 100 10 1 1008 0
10 100 10 1 1009 0
11 101 12 3 1000 1
12 101 12 1 1001 1
13 101 12 1 1002 1
14 101 13 2 1003 1
15 101 13 3 1004 1
16 101 12 1 1005 0
17 101 13 1 1006 0
18 101 13 1 1007 0
19 101 12 1 1008 0
20 101 12 1 1009 0
Here's my think:
data.table is going to work well, especially when you are working with large data sets. It's faster. We just need to group it, cumsum 2 team's layoff, and see if they equal.
First I have to say:
(problem solved by Martin Morgan, his updated answer no longer has this error)
I don't think #Martin Morgan 's answer is right. Let's imagine a certain case:
when team 1 had one player off, after which team 2 had another player off, then 2 teams should be even, but #Martin Morgan's output would be FALSE.
I'll make an example with this dataset, where player_off of record 19 was modified to 1, which means that in game 101, after team 13 had had 1 player off at 1004, team 12 had 1 player off at 1008, which would make 2 teams even at 1009.
> dt.1
game_id team_id action_id player_off time_ref
1 100 10 1 NA 1000
2 100 10 1 NA 1001
3 100 10 1 NA 1002
4 100 11 1 NA 1003
5 100 11 2 NA 1004
6 100 11 1 NA 1005
7 100 10 3 1 1006
8 100 11 1 NA 1007
9 100 10 1 NA 1008
10 100 10 1 NA 1009
11 101 12 3 0 1000
12 101 12 1 NA 1001
13 101 12 1 NA 1002
14 101 13 2 NA 1003
15 101 13 3 1 1004
16 101 12 1 NA 1005
17 101 13 1 NA 1006
18 101 13 1 NA 1007
19 101 12 1 1 1008
20 101 12 1 NA 1009
But #Martin Morgan 's function would produce this output:
> dt.1$is_even = ave(df$player_off, df$game_id, FUN = fun)
> dt.1
game_id team_id action_id player_off time_ref is_even
1 100 10 1 NA 1000 1
2 100 10 1 NA 1001 1
3 100 10 1 NA 1002 1
4 100 11 1 NA 1003 1
5 100 11 2 NA 1004 1
6 100 11 1 NA 1005 1
7 100 10 3 1 1006 1
8 100 11 1 NA 1007 0
9 100 10 1 NA 1008 0
10 100 10 1 NA 1009 0
11 101 12 3 0 1000 1
12 101 12 1 NA 1001 1
13 101 12 1 NA 1002 1
14 101 13 2 NA 1003 1
15 101 13 3 1 1004 1
16 101 12 1 NA 1005 0
17 101 13 1 NA 1006 0
18 101 13 1 NA 1007 0
19 101 12 1 1 1008 0
20 101 12 1 NA 1009 0
Notice how at line 19 and line 20, is.even=0. Which is not what op wants.
My code does not process NAs, so I am going to transform NA to 0 first.
> dt.1<-as.data.table(dt.1)
> dt.1[is.na(dt.1)]<-0
My code would produce the correct output, at time 1008 and 1009, where both team 12 and team 13 had 1 off, two teams are even.
> dt.1[,.(action_id,team2_off=(team_id==max(team_id))*player_off,team1_off=(team_id==min(team_id))*player_off,team_id,time_ref,player_off),by=game_id][order(game_id,time_ref)][,.(team_id,time_ref,action_id,player_off,even=as.numeric(cumsum(team2_off)==cumsum(team1_off))),by=game_id]
game_id team_id time_ref action_id player_off even
1: 100 10 1000 1 0 1
2: 100 10 1001 1 0 1
3: 100 10 1002 1 0 1
4: 100 11 1003 1 0 1
5: 100 11 1004 2 0 1
6: 100 11 1005 1 0 1
7: 100 10 1006 3 1 0
8: 100 11 1007 1 0 0
9: 100 10 1008 1 0 0
10: 100 10 1009 1 0 0
11: 101 12 1000 3 0 1
12: 101 12 1001 1 0 1
13: 101 12 1002 1 0 1
14: 101 13 1003 2 0 1
15: 101 13 1004 3 1 0
16: 101 12 1005 1 0 0
17: 101 13 1006 1 0 0
18: 101 13 1007 1 0 0
19: 101 12 1008 1 1 1
20: 101 12 1009 1 0 1
I understand it is a messy looking chunk of data.table code, let me explain step by step.
dt[, .(
action_id,
team2_off = (team_id == max(team_id)) * player_off,
team1_off = (team_id == min(team_id)) * player_off,
team_id,
time_ref,
player_off
), by = game_id][order(game_id, time_ref)][, .(team_id,
time_ref,
action_id,
player_off,
even = cumsum(team2_off) == cumsum(team1_off)), by = game_id]
first, we take data.table dt, group by game_id, and does this calculation:
team2_off = (team_id == max(team_id)) * player_off,
team1_off = (team_id == min(team_id)) * player_off
data.table has some problem taking 2 grouping at once (group by game_id and team_id), but it handles logical expression inside of each group well. In this way, we effectively get team1_off and team2_off, by multiplying a logical output of team_id == max/min(team_id) with player_off. When both are 1, the output would be 1, which means, 1 player was off in the selected team.
Now we have a data table of:
> dt.1[,.(action_id,team2_off=(team_id==max(team_id))*player_off,team1_off=(team_id==min(team_id))*player_off,team_id,time_ref,player_off),by=game_id]
game_id action_id team2_off team1_off team_id time_ref player_off
1: 100 1 0 0 10 1000 0
2: 100 1 0 0 10 1001 0
3: 100 1 0 0 10 1002 0
4: 100 1 0 0 11 1003 0
5: 100 2 0 0 11 1004 0
6: 100 1 0 0 11 1005 0
7: 100 3 0 1 10 1006 1
8: 100 1 0 0 11 1007 0
9: 100 1 0 0 10 1008 0
10: 100 1 0 0 10 1009 0
11: 101 3 0 0 12 1000 0
12: 101 1 0 0 12 1001 0
13: 101 1 0 0 12 1002 0
14: 101 2 0 0 13 1003 0
15: 101 3 1 0 13 1004 1
16: 101 1 0 0 12 1005 0
17: 101 1 0 0 13 1006 0
18: 101 1 0 0 13 1007 0
19: 101 1 0 1 12 1008 1
20: 101 1 0 0 12 1009 0
Now we no longer need to group by two groups (team_id, game_id), we can just do cumsum by game_id, and compare if cumsum(team1_off)==cumsum(team2_off), also, order it by game_id and time_ref, so the result would have the correct order.
I understand that NAs may have different meanings than 0 in this scenario. If you really care that much, just create a dummy column of player_off.
> dt$dummy<-dt$player_off
> dt$dummy[is.na(dt$dummy)]<-0
> dt<-as.data.table(dt)
> dt[, .(
+ action_id,
+ team2_off = (team_id == max(team_id)) * dummy,
+ team1_off = (team_id == min(team_id)) * dummy,
+ team_id,
+ time_ref,
+ player_off
+ ), by = game_id][order(game_id, time_ref)][, .(team_id,
+ time_ref,
+ action_id,
+ player_off,
+ even = as.numeric(cumsum(team2_off) == cumsum(team1_off))), by = game_id]
game_id team_id time_ref action_id player_off even
1: 100 10 1000 1 NA 1
2: 100 10 1001 1 NA 1
3: 100 10 1002 1 NA 1
4: 100 11 1003 1 NA 1
5: 100 11 1004 2 NA 1
6: 100 11 1005 1 NA 1
7: 100 10 1006 3 1 0
8: 100 11 1007 1 NA 0
9: 100 10 1008 1 NA 0
10: 100 10 1009 1 NA 0
11: 101 12 1000 3 0 1
12: 101 12 1001 1 NA 1
13: 101 12 1002 1 NA 1
14: 101 13 1003 2 NA 1
15: 101 13 1004 3 1 0
16: 101 12 1005 1 NA 0
17: 101 13 1006 1 NA 0
18: 101 13 1007 1 NA 0
19: 101 12 1008 1 NA 0
20: 101 12 1009 1 NA 0
I really think you question is very interesting, and I was dedicated to solve this using data.table. It took me few hours and I almost gave up on data.table, thinking that data.table just can't process two grouping at a time. I eventually solved it with a logical multiplication.
Great fun I had
team1_off = (team_id == min(team_id)) * dummy
team2_off = (team_id == max(team_id)) * dummy
Take the following patient data example from a hospital.
YEAR <- sample(1980:1995,15, replace=T)
Pat_ID <- sample(1:100,15)
sex <- c(1,0,1,0,1,0,0,1,0,0,0,0,1,0,0)
df1 <- data.frame(Pat_ID,YEAR,sex)
I want to introduce a dummy variable $PAIR_IDENTIFIER that takes a new value each time a new sex==1 appears. The problem is there is no constant patern to the sex variable.
You see sometimes the succeeding 1 appears in the ith+2 position and then ith+3 position etc.
so $PAIR_IDENTIFIER <- c(1,1,2,2,3,3,3,4,4,4,4,4 .....)
You can do this by simply using the cumsum,
df1$PAIR_IDENTIFIER <- cumsum(df1$sex)
df1
# Pat_ID YEAR sex PAIR_IDENTIFIER
#1 54 1991 1 1
#2 100 1992 0 1
#3 6 1995 1 2
#4 99 1994 0 2
#5 42 1988 1 3
#6 65 1990 0 3
#7 53 1994 0 3
#8 96 1987 1 4
I have an R script that allows me to select a sample size and take fifty individual random samples with replacement. Below is an example of this code:
## Creates data frame
df = as.data.table(data)
## Select sample size
sample.size = 5
## Creates Sample 1 (Size 5)
Sample.1<-df[,
Dollars[sample(.N, size=sample.size, replace=TRUE)], by = Num]
Sample.1$Sample <- c("01")
According to the R script above, I first created a data frame. I then select my sample size, which in this case is 5. This represents just one sample. Due to my lack of experience with R, I repeat this code 49 more times. The last piece of code looks like this:
## Creates Sample 50 (Size 5)
Sample.50<-df[,
Dollars[sample(.N, size=sample.size, replace=TRUE)], by = Num]
Sample.50$Sample <- c("50")
The sample output would look something like this (Sample Range 1 - 50):
Num Dollars Sample
1 85000 01
1 4900 01
1 18000 01
1 6900 01
1 11000 01
1 8800 50
1 3800 50
1 10400 50
1 2200 50
1 29000 50
It should be noted that varaible 'Num' was created for grouping purposes and has little to no influence on my overall question (which is posted below).
Instead of repeating this code fifty times, to get me fifty individual samples (with a size of 5), is there a loop I can create to help me limit my code? I have been recently asked to create ten thousand random samples, each of a size of 5. I obviously cannot repeat this code ten thousand times so I need some sort of loop.
A sample of my final output should look something like this (Sample Range 1 - 10,000):
Num Dollars Sample
1 85000 01
1 4900 01
1 18000 01
1 6900 01
1 11000 01
1 9900 10000
1 8300 10000
1 10700 10000
1 6800 10000
1 31000 10000
Thank you all in advance for your help, its greatly appreciated.
Here is some sample code if needed:
Num Dollars
1 31002
1 13728
1 23526
1 80068
1 86244
1 9330
1 27169
1 13694
1 4781
1 9742
1 20060
1 35230
1 15546
1 7618
1 21604
1 8738
1 5299
1 12081
1 7652
1 16779
A very simple method would be to use a for loop and store the results in a list:
lst <- list()
for(i in seq_len(3)){
lst[[i]] <- df[sample(seq_len(nrow(df)), 5, replace = TRUE),]
lst[[i]]["Sample"] <- i
}
> lst
[[1]]
Num Dollars Sample
20 1 16779 1
1 1 31002 1
12 1 35230 1
14 1 7618 1
14.1 1 7618 1
[[2]]
Num Dollars Sample
9 1 4781 2
13 1 15546 2
12 1 35230 2
17 1 5299 2
12.1 1 35230 2
[[3]]
Num Dollars Sample
1 1 31002 3
7 1 27169 3
17 1 5299 3
5 1 86244 3
6 1 9330 3
Then, to create a single data.frame, use do.call to rbind the list elements together:
do.call(rbind, lst)
Num Dollars Sample
20 1 16779 1
1 1 31002 1
12 1 35230 1
14 1 7618 1
14.1 1 7618 1
9 1 4781 2
13 1 15546 2
121 1 35230 2
17 1 5299 2
12.1 1 35230 2
11 1 31002 3
7 1 27169 3
171 1 5299 3
5 1 86244 3
6 1 9330 3
It's worth noting that if you're sampling with replacement, then drawing 50 (or 10,000) samples of size 5 is equivalent to drawing one sample of size 250 (or 50,000). Thus I would do it like this (you'll see I stole a line from #beginneR's answer):
df = as.data.table(data)
## Select sample size
sample.size = 5
n.samples = 10000
# Sample and assign groups
draws <- df[sample(seq_len(nrow(df)), sample.size * n.samples, replace = TRUE), ]
draws[, Sample := rep(1:n.samples, each = sample.size)]