data=data.frame("x1"=c(1:10),
"x2"=c(1:4,4,6:10),
"x3"=c(1:3,2:5,5:7),
"x4"=c(21:30),
"x5"=c(35:44))
recode=c("x1","x2","x3")
data <- data[recode %in% c(4,5)] <- NA
I want to store a specific set of variables for example above I store x1,x2,x3 in 'recode'. Then I want to change all the values for all variables in recode such that any value of 4 or 5 is set to NA.
We need to use replace with lapply
data[recode] <- lapply(data[recode], function(x) replace(x, x %in% 4:5, NA))
data
# x1 x2 x3 x4 x5
#1 1 1 1 21 35
#2 2 2 2 22 36
#3 3 3 3 23 37
#4 NA NA 2 24 38
#5 NA NA 3 25 39
#6 6 6 NA 26 40
#7 7 7 NA 27 41
#8 8 8 NA 28 42
#9 9 9 6 29 43
#10 10 10 7 30 44
Or with dplyr
library(dplyr)
data %>%
mutate_at(vars(recode), ~ na_if(., 4)) %>%
mutate_at(vars(recode), ~ na_if(., 5))
# x1 x2 x3 x4 x5
#1 1 1 1 21 35
#2 2 2 2 22 36
#3 3 3 3 23 37
#4 NA NA 2 24 38
#5 NA NA 3 25 39
#6 6 6 NA 26 40
#7 7 7 NA 27 41
#8 8 8 NA 28 42
#9 9 9 6 29 43
#10 10 10 7 30 44
One dplyr possibility could be:
data %>%
mutate_at(vars(recode), ~ replace(., . %in% 4:5, NA))
x1 x2 x3 x4 x5
1 1 1 1 21 35
2 2 2 2 22 36
3 3 3 3 23 37
4 NA NA NA 24 38
5 NA NA NA 25 39
6 6 6 4 26 40
7 7 7 5 27 41
8 8 8 5 28 42
9 9 9 6 29 43
10 10 10 7 30 44
Use Map().
data[recode] <- Map(function(x) ifelse(x %in% c(4, 5), NA, x), data[recode])
data
# x1 x2 x3 x4 x5
# 1 1 1 1 21 35
# 2 2 2 2 22 36
# 3 3 3 3 23 37
# 4 NA NA 2 24 38
# 5 NA NA 3 25 39
# 6 6 6 NA 26 40
# 7 7 7 NA 27 41
# 8 8 8 NA 28 42
# 9 9 9 6 29 43
# 10 10 10 7 30 44
Related
I have a long format dataset of 100,000+ individuals, capturing clinic visits at 5 different time points (not chronological). I've included an example dataset below that replicates the formatting of my data:
ID: participant ID visit_number: order of the clinic visits in the
original dataset
age_visit: age at the time of the visit
clinic_number: the identifier for the specific clinic location
age_sorted: For each ID, age sorted in ascending order across the 5
clinic visits
age_sorted_index: For each ID, the visit number
corresponding to the sorted age
I would like to create a new column (clinic_number_extracted) that fetches the clinic identifier (clinic_number) corresponding to each sorted age (age_sorted value) for each participant. I was thinking that it might be possible to use the age_sorted_index and visit_number variables to do so (generating key-value pairs?), but am not quite sure how to do this outside of data.table. A tidyverse solution would be preferred.
I've looked on R community and stack exchange for clues, but haven't been able to find exactly what I'm looking for (likely not using the correct search terms). I tried to play around with group_by(across()) and with_order(order_by()) functions without much success. I can potentially create a new variable with a few case_when() conditions but might run into issues if there are repeated age_assessment_sorted values.
set.seed(42)
# Beginning dataset
das <- data.frame(id = rep(letters[1:3], each = 5),
visit_number = rep(1:5, times = 3),
age_visit = c(50, rep(NA_real_, times = 7), 34, 40, 72, rep(NA_real_, times = 3), 87),
clinic_number = sample(30:50, 15, replace=TRUE),
age_sorted = c(50, rep(NA_real_, times = 4), 34, 40,rep(NA_real_, times = 3), 72, 87, rep(NA_real_, times = 3)),
age_sorted_index = c(rep(1:5), 4, 5, rep(1:3), 1, 5, 2, 3, 4))
# Print out dataset
das
#> id visit_number age_visit clinic_number age_sorted age_sorted_index
#> 1 a 1 50 46 50 1
#> 2 a 2 NA 34 NA 2
#> 3 a 3 NA 30 NA 3
#> 4 a 4 NA 39 NA 4
#> 5 a 5 NA 33 NA 5
#> 6 b 1 NA 47 34 4
#> 7 b 2 NA 46 40 5
#> 8 b 3 NA 44 NA 1
#> 9 b 4 34 36 NA 2
#> 10 b 5 40 33 NA 3
#> 11 c 1 72 34 72 1
#> 12 c 2 NA 43 87 5
#> 13 c 3 NA 49 NA 2
#> 14 c 4 NA 47 NA 3
#> 15 c 5 87 44 NA 4
Desired data:
das_final <- cbind(das,
clinic_number_extracted = c(46, rep(NA_real_, times = 4), 36, 33, rep(NA_real_, times = 3), 34, 44, rep(NA_real_, times = 3)))
# Print out final dataset
das_final
#> id visit_number age_visit clinic_number age_sorted age_sorted_index
#> 1 a 1 50 46 50 1
#> 2 a 2 NA 34 NA 2
#> 3 a 3 NA 30 NA 3
#> 4 a 4 NA 39 NA 4
#> 5 a 5 NA 33 NA 5
#> 6 b 1 NA 47 34 4
#> 7 b 2 NA 46 40 5
#> 8 b 3 NA 44 NA 1
#> 9 b 4 34 36 NA 2
#> 10 b 5 40 33 NA 3
#> 11 c 1 72 34 72 1
#> 12 c 2 NA 43 87 5
#> 13 c 3 NA 49 NA 2
#> 14 c 4 NA 47 NA 3
#> 15 c 5 87 44 NA 4
#> clinic_number_extracted
#> 1 46
#> 2 NA
#> 3 NA
#> 4 NA
#> 5 NA
#> 6 36
#> 7 33
#> 8 NA
#> 9 NA
#> 10 NA
#> 11 34
#> 12 44
#> 13 NA
#> 14 NA
#> 15 NA
Created on 2022-05-06 by the reprex package (v2.0.1)
We may use match between the 'age_sorted' and 'age_visit', use that index to subset the 'clinic_number' and replace the elements where they are NA in 'age_sorted' to NA
library(dplyr)
das_final %>%
group_by(id) %>%
mutate(clinic_new = clinic_number[match(age_sorted, age_visit)] *
NA^is.na(age_sorted)) %>%
ungroup
-output
# A tibble: 15 × 8
id visit_number age_visit clinic_number age_sorted age_sorted_index clinic_number_extracted clinic_new
<chr> <int> <dbl> <int> <dbl> <dbl> <dbl> <dbl>
1 a 1 50 46 50 1 46 46
2 a 2 NA 34 NA 2 NA NA
3 a 3 NA 30 NA 3 NA NA
4 a 4 NA 39 NA 4 NA NA
5 a 5 NA 33 NA 5 NA NA
6 b 1 NA 47 34 4 36 36
7 b 2 NA 46 40 5 33 33
8 b 3 NA 44 NA 1 NA NA
9 b 4 34 36 NA 2 NA NA
10 b 5 40 33 NA 3 NA NA
11 c 1 72 34 72 1 34 34
12 c 2 NA 43 87 5 44 44
13 c 3 NA 49 NA 2 NA NA
14 c 4 NA 47 NA 3 NA NA
15 c 5 87 44 NA 4 NA NA
Could you just do this without needing to group? It seems to solve the issue in the sample data, but perhaps I am missing something:
das$clinic_number_extracted <- ifelse(!is.na(das$age_sorted), das$clinic_number, NA)
Output:
# id visit_number age_visit clinic_number age_sorted age_sorted_index clinic_number_extracted
1 a 1 50 46 50 1 46
#2 a 2 NA 34 NA 2 NA
#3 a 3 NA 30 NA 3 NA
#4 a 4 NA 39 NA 4 NA
#5 a 5 NA 33 NA 5 NA
#6 b 1 NA 47 34 4 47
#7 b 2 NA 46 40 5 46
#8 b 3 NA 44 NA 1 NA
#9 b 4 34 36 NA 2 NA
#10 b 5 40 33 NA 3 NA
#11 c 1 72 34 72 1 34
#12 c 2 NA 43 87 5 43
#13 c 3 NA 49 NA 2 NA
#14 c 4 NA 47 NA 3 NA
#15 c 5 87 44 NA 4 NA
Testing against desired data provided:
all.equal(das[,"clinic_number_extracted"], das_final[,"clinic_number_extracted"])
#[1] TRUE
Using reduce(bind_cols), the list elements of same dimension may be combined. However, I would like to know how to combine only same dimension (may be specified dimesion in some way) elements from a list which may have elements of different dimension.
library(tidyverse)
df1 <- data.frame(A1 = 1:10, A2 = 10:1)
df2 <- data.frame(B = 11:30)
df3 <- data.frame(C = 31:40)
ls1 <- list(df1, df3)
ls1
[[1]]
A1 A2
1 1 10
2 2 9
3 3 8
4 4 7
5 5 6
6 6 5
7 7 4
8 8 3
9 9 2
10 10 1
[[2]]
C
1 31
2 32
3 33
4 34
5 35
6 36
7 37
8 38
9 39
10 40
ls1 %>%
reduce(bind_cols)
A1 A2 C
1 1 10 31
2 2 9 32
3 3 8 33
4 4 7 34
5 5 6 35
6 6 5 36
7 7 4 37
8 8 3 38
9 9 2 39
10 10 1 40
ls2 <- list(df1, df2, df3)
ls2
[[1]]
A1 A2
1 1 10
2 2 9
3 3 8
4 4 7
5 5 6
6 6 5
7 7 4
8 8 3
9 9 2
10 10 1
[[2]]
B
1 11
2 12
3 13
4 14
5 15
6 16
7 17
8 18
9 19
10 20
11 21
12 22
13 23
14 24
15 25
16 26
17 27
18 28
19 29
20 30
[[3]]
C
1 31
2 32
3 33
4 34
5 35
6 36
7 37
8 38
9 39
10 40
ls2 %>%
reduce(bind_cols)
Error: Can't recycle `..1` (size 10) to match `..2` (size 20).
Run `rlang::last_error()` to see where the error occurred.
Question
Looking for a function to combine all data.frames in a list with an argument of number of rows.
One option could be:
map(split(lst, map_int(lst, NROW)), bind_cols)
$`10`
A1 A2 C
1 1 10 31
2 2 9 32
3 3 8 33
4 4 7 34
5 5 6 35
6 6 5 36
7 7 4 37
8 8 3 38
9 9 2 39
10 10 1 40
$`20`
B
1 11
2 12
3 13
4 14
5 15
6 16
7 17
8 18
9 19
10 20
11 21
12 22
13 23
14 24
15 25
16 26
17 27
18 28
19 29
20 30
You can use -
n <- 1:max(sapply(ls2, nrow))
res <- do.call(cbind, lapply(ls2, `[`, n, ,drop = FALSE))
res
# A1 A2 B C
#1 1 10 11 31
#2 2 9 12 32
#3 3 8 13 33
#4 4 7 14 34
#5 5 6 15 35
#6 6 5 16 36
#7 7 4 17 37
#8 8 3 18 38
#9 9 2 19 39
#10 10 1 20 40
#NA NA NA 21 NA
#NA.1 NA NA 22 NA
#NA.2 NA NA 23 NA
#NA.3 NA NA 24 NA
#NA.4 NA NA 25 NA
#NA.5 NA NA 26 NA
#NA.6 NA NA 27 NA
#NA.7 NA NA 28 NA
#NA.8 NA NA 29 NA
#NA.9 NA NA 30 NA
A little-bit shorter with purrr::map_dfc
purrr::map_dfc(ls2, `[`, n, , drop = FALSE)
We can use cbind.fill from rowr
library(rowr)
do.call(cbind.fill, c(ls2, fill = NA))
A base R option using tapply + sapply
tapply(
ls2,
sapply(ls2, nrow),
function(x) do.call(cbind, x)
)
gives
$`10`
A1 A2 C
1 1 10 31
2 2 9 32
3 3 8 33
4 4 7 34
5 5 6 35
6 6 5 36
7 7 4 37
8 8 3 38
9 9 2 39
10 10 1 40
$`20`
B
1 11
2 12
3 13
4 14
5 15
6 16
7 17
8 18
9 19
10 20
11 21
12 22
13 23
14 24
15 25
16 26
17 27
18 28
19 29
20 30
You may also use if inside reduce if you want to combine similar elements of list (case: when first item in list has priority)
df1 <- data.frame(A1 = 1:10, A2 = 10:1)
df2 <- data.frame(B = 11:30)
df3 <- data.frame(C = 31:40)
ls1 <- list(df1, df3)
ls2 <- list(df1, df2, df3)
library(tidyverse)
reduce(ls2, ~if(nrow(.x) == nrow(.y)){bind_cols(.x, .y)} else {.x})
#> A1 A2 C
#> 1 1 10 31
#> 2 2 9 32
#> 3 3 8 33
#> 4 4 7 34
#> 5 5 6 35
#> 6 6 5 36
#> 7 7 4 37
#> 8 8 3 38
#> 9 9 2 39
#> 10 10 1 40
Created on 2021-06-09 by the reprex package (v2.0.0)
Here's another tidyverse option.
We're creating a dummy ID in each data.frame based on the row_number(), then joining all data.frames by the dummy ID, and then dropping the dummy ID.
ls2 %>%
map(., ~mutate(.x, id = row_number())) %>%
reduce(full_join, by = "id") %>%
select(-id)
This gives us:
A1 A2 B C
1 1 10 11 31
2 2 9 12 32
3 3 8 13 33
4 4 7 14 34
5 5 6 15 35
6 6 5 16 36
7 7 4 17 37
8 8 3 18 38
9 9 2 19 39
10 10 1 20 40
11 NA NA 21 NA
12 NA NA 22 NA
13 NA NA 23 NA
14 NA NA 24 NA
15 NA NA 25 NA
16 NA NA 26 NA
17 NA NA 27 NA
18 NA NA 28 NA
19 NA NA 29 NA
20 NA NA 30 NA
We can also use Reduce function from base R:
lst <- list(df1, df2, df3)
# First we create id number for each underlying data set
lst |>
lapply(\(x) {x$id <- 1:nrow(x);
x
}
) -> ls2
Reduce(function(x, y) if(nrow(x) == nrow(y)){
merge(x, y, by = "id")
} else {
x
}, ls2)
id A1 A2 C
1 1 1 10 31
2 2 2 9 32
3 3 3 8 33
4 4 4 7 34
5 5 5 6 35
6 6 6 5 36
7 7 7 4 37
8 8 8 3 38
9 9 9 2 39
10 10 10 1 40
This is my data, and I want to replace NA with "No". I can replace missing values one by one. However, I need to replace NAs in s_1:s_4 in the code. Just as a reminder, all of the variables are factor levels.
id x s_0 s_1 s_2 s_3
1 5 75 A 4 110
2 9 36 NA NA 921
3 11 13 B 7 769
4 11 34 C 2 912
5 11 NA C NA 835
6 13 39 NA 4 NA
7 14 45 B 4 577
8 19 42 D 6 NA
9 20 4 NA 7 577
10 13 28 NA 3 573
If these are already existing factors, you can use forcats::fct_explicit_na():
library(dplyr)
library(forcats)
# Make sample data vars factors
dat <- dat %>%
mutate(across(starts_with("s_"), as.factor))
# Add 'No' as factor level
dat %>%
mutate(across(starts_with("s_"), fct_explicit_na, "No"))
# A tibble: 10 x 6
id x s_0 s_1 s_2 s_3
<dbl> <dbl> <fct> <fct> <fct> <fct>
1 1 5 75 A 4 110
2 2 9 36 No No 921
3 3 11 13 B 7 769
4 4 11 34 C 2 912
5 5 11 No C No 835
6 6 13 39 No 4 No
7 7 14 45 B 4 577
8 8 19 42 D 6 No
9 9 20 4 No 7 577
10 10 13 28 No 3 573
In base R, you need to include "No" as factor level before turning NA's to "No".
cols <- grep('s_\\d+', names(df))
df[cols] <- lapply(df[cols], function(x) {
levels(x) <- c(levels(x), 'No')
x[is.na(x)] <- 'No'
x
})
df
# id x s_0 s_1 s_2 s_3
#1 1 5 75 A 4 110
#2 2 9 36 No No 921
#3 3 11 13 B 7 769
#4 4 11 34 C 2 912
#5 5 11 No C No 835
#6 6 13 39 No 4 No
#7 7 14 45 B 4 577
#8 8 19 42 D 6 No
#9 9 20 4 No 7 577
#10 10 13 28 No 3 573
I would like to repeat the first two rows for each id two times. I don't know how to do that. Does anyone have a suggestion?
id <- rep(1:4,each=6)
scored <- c(12,13,NA,NA,NA,NA,14,20,NA,NA,NA,NA,23,56,NA,NA,NA,NA, 45,78,NA,NA,NA,NA)
df <- data.frame(id,scored)
df
id scored
1 1 12
2 1 13
3 1 NA
4 1 NA
5 1 NA
6 1 NA
7 2 14
8 2 20
9 2 NA
10 2 NA
11 2 NA
12 2 NA
13 3 23
14 3 56
15 3 NA
16 3 NA
17 3 NA
18 3 NA
19 4 45
20 4 78
21 4 NA
22 4 NA
23 4 NA
24 4 NA
>
I want it to look like:
df
id score
1 1 12
2 1 13
3 1 12
4 1 13
5 1 12
6 1 13
7 2 14
8 2 20
9 2 14
10 2 20
11 2 14
12 2 20
13 3 23
14 3 56
15 3 23
16 3 56
17 3 23
18 3 56
19 4 45
20 4 78
21 4 45
22 4 78
23 4 45
24 4 78
>
..................................................
..................................................
..................................................
We can do a group by rep on the non-NA elements of 'scored'
library(dplyr)
df %>%
group_by(id) %>%
mutate(scored = rep(scored[!is.na(scored)], length.out = n()))
# A tibble: 24 x 2
# Groups: id [4]
# id scored
# <int> <dbl>
# 1 1 12
# 2 1 13
# 3 1 12
# 4 1 13
# 5 1 12
# 6 1 13
# 7 2 14
# 8 2 20
# 9 2 14
#10 2 20
# … with 14 more rows
I have data that look something like this:
id <- c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,8,8,8,9,9,9)
yr <- c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3,1,2,3)
gr <- c(3,4,5,3,4,5,3,4,5,4,5,6,4,5,6,4,5,6,5,6,7,5,6,7,5,6,7)
x <- c(33,48,31,41,31,36,25,38,28,17,39,53,60,60,19,39,34,47,20,28,38,15,17,49,48,45,39)
df <- data.frame(id,yr,gr,x)
id yr gr x
1 1 1 3 33
2 1 2 4 48
3 1 3 5 31
4 2 1 3 41
5 2 2 4 31
6 2 3 5 36
7 3 1 3 25
8 3 2 4 38
9 3 3 5 28
10 4 1 4 17
11 4 2 5 39
12 4 3 6 53
13 5 1 4 60
14 5 2 5 60
15 5 3 6 19
16 6 1 4 39
17 6 2 5 34
18 6 3 6 47
19 7 1 5 20
20 7 2 6 28
21 7 3 7 38
22 8 1 5 15
23 8 2 6 17
24 8 3 7 49
25 9 1 5 48
26 9 2 6 45
27 9 3 7 39
I would like to create a new variable in the data frame that contains the quantiles of "x" computed within each unique combination of "yr" and "gr". That is, rather than finding the quantiles of "x" based on all 27 rows of data in the example, I would like to compute the quantiles by two grouping variables: yr and gr. For instance, the quantiles of "x" when yr = 1 and gr = 3, yr = 1 and gr = 4, etc.
Once these values are computed, I would like them to be appended to the data frame as a single column, say "x_quant".
I am able to split the data into the separate groups I need, and I am know how to calculate quantiles, but I am having trouble combining the two steps in a way that is amenable to creating a new column in the existing data frame.
Any help y'all can provide would be greatly appretiated! Thank you much!
~kj
# turn "yr" and "gr" into sortable column
df$y <- paste(df$yr,"",df$gr)
df.ordered <- df[order(df$y),] #sort df based on group
grp <- split(df.ordered,df.ordered$y);grp
# get quantiles and turn results into string
q <- vector('list')
for (i in 1:length(grp)) {
a <- quantile(grp[[i]]$x)
q[i] <- paste(a[1],"",a[2],"",a[3],"",a[4],"",a[5])
}
x_quant <- unlist(sapply(q, `[`, 1))
x_quant <- rep(x_quant,each=3)
# append quantile back to data frame. Gave new column a more descriptive name
df.ordered$xq_0_25_50_75_100 <- x_quant
df.ordered$y <- NULL
df <- df.ordered;df </pre>
Output:
> # turn "yr" and "gr" into sortable column
> df$y <- paste(df$yr,"",df$gr)
> df.ordered <- df[order(df$y),] #sort df based on group
> grp <- split(df.ordered,df.ordered$y);grp
$`1 3`
id yr gr x y
1 1 1 3 33 1 3
4 2 1 3 41 1 3
7 3 1 3 25 1 3
$`1 4`
id yr gr x y
10 4 1 4 17 1 4
13 5 1 4 60 1 4
16 6 1 4 39 1 4
$`1 5`
id yr gr x y
19 7 1 5 20 1 5
22 8 1 5 15 1 5
25 9 1 5 48 1 5
$`2 4`
id yr gr x y
2 1 2 4 48 2 4
5 2 2 4 31 2 4
8 3 2 4 38 2 4
$`2 5`
id yr gr x y
11 4 2 5 39 2 5
14 5 2 5 60 2 5
17 6 2 5 34 2 5
$`2 6`
id yr gr x y
20 7 2 6 28 2 6
23 8 2 6 17 2 6
26 9 2 6 45 2 6
$`3 5`
id yr gr x y
3 1 3 5 31 3 5
6 2 3 5 36 3 5
9 3 3 5 28 3 5
$`3 6`
id yr gr x y
12 4 3 6 53 3 6
15 5 3 6 19 3 6
18 6 3 6 47 3 6
$`3 7`
id yr gr x y
21 7 3 7 38 3 7
24 8 3 7 49 3 7
27 9 3 7 39 3 7
> # get quantiles and turn results into string
> q <- vector('list')
> for (i in 1:length(grp)) {
+ a <- quantile(grp[[i]]$x)
+ q[i] <- paste(a[1],"",a[2],"",a[3],"",a[4],"",a[5])
+ }
> x_quant <- unlist(sapply(q, `[`, 1))
> x_quant <- rep(x_quant,each=3)
> # append quantile back to data frame
> df.ordered$xq_0_25_50_75_100 <- x_quant
> df.ordered$y <- NULL
> df <- df.ordered
> df
id yr gr x xq_0_25_50_75_100
1 1 1 3 33 25 29 33 37 41
4 2 1 3 41 25 29 33 37 41
7 3 1 3 25 25 29 33 37 41
10 4 1 4 17 17 28 39 49.5 60
13 5 1 4 60 17 28 39 49.5 60
16 6 1 4 39 17 28 39 49.5 60
19 7 1 5 20 15 17.5 20 34 48
22 8 1 5 15 15 17.5 20 34 48
25 9 1 5 48 15 17.5 20 34 48
2 1 2 4 48 31 34.5 38 43 48
5 2 2 4 31 31 34.5 38 43 48
8 3 2 4 38 31 34.5 38 43 48
11 4 2 5 39 34 36.5 39 49.5 60
14 5 2 5 60 34 36.5 39 49.5 60
17 6 2 5 34 34 36.5 39 49.5 60
20 7 2 6 28 17 22.5 28 36.5 45
23 8 2 6 17 17 22.5 28 36.5 45
26 9 2 6 45 17 22.5 28 36.5 45
3 1 3 5 31 28 29.5 31 33.5 36
6 2 3 5 36 28 29.5 31 33.5 36
9 3 3 5 28 28 29.5 31 33.5 36
12 4 3 6 53 19 33 47 50 53
15 5 3 6 19 19 33 47 50 53
18 6 3 6 47 19 33 47 50 53
21 7 3 7 38 38 38.5 39 44 49
24 8 3 7 49 38 38.5 39 44 49
27 9 3 7 39 38 38.5 39 44 49
>