This question is similar to one already answered: R: Splitting dataframe into subgroups consisting of every consecutive 2 groups
However, rather than splitting into subgroups that have a type in common, I need to split into subgroups that contain two consecutive types and are distinct. The groups in my actual data have differing numbers of rows as well.
df <- data.frame(ID=c('1','1','1','1','1','1','1'), Type=c('a','a','b','c','c','d','d'), value=c(10,2,5,3,7,3,9))
ID Type value
1 1 a 10
2 1 a 2
3 1 b 5
4 1 c 3
5 1 c 7
6 1 d 3
7 1 d 9
So subgroup 1 would be Type a and b:
ID Type value
1 1 a 10
2 1 a 2
3 1 b 5
And subgroup 2 would be Type c and d:
ID Type value
4 1 c 3
5 1 c 7
6 1 d 3
7 1 d 9
I have tried manipulating the code from this previous example, but I can't figure out how to make this happen without having overlapping Types in each group. Any help would be greatly appreciated - thanks!
EDIT: thanks for pointing out I didn't actually include the correct link.
We can do a little manipulation of a dense_rank of the Type variable to make an appropriate grouping variable:
library(dplyr)
df %>%
group_by(g = (dense_rank(match(Type, Type)) - 1) %/% 2) %>%
group_split()
# [[1]]
# # A tibble: 3 × 4
# ID Type value g
# <chr> <chr> <dbl> <dbl>
# 1 1 a 10 0
# 2 1 a 2 0
# 3 1 b 5 0
#
# [[2]]
# # A tibble: 4 × 4
# ID Type value g
# <chr> <chr> <dbl> <dbl>
# 1 1 c 3 1
# 2 1 c 7 1
# 3 1 d 3 1
# 4 1 d 9 1
Explanation: match(Type, Type) converts Type into integers ordered by number of appearance - but not dense. dense_rank() makes that dense (no gaps). We then subtract 1 to make it start at 0 and %/% 2 to see how many 2s go into it, effectively grouping by pairs.
Here is a rle way, written as a function. Pass the data.frame and the split column name as a character string.
df <- data.frame(ID=c('1','1','1','1','1','1','1'),
Type=c('a','a','b','c','c','d','d'),
value=c(10,2,5,3,7,3,9))
split_two <- function(x, col) {
r <- rle(x[[col]])
r$values[c(FALSE, TRUE)] <- r$values[c(TRUE, FALSE)]
split(x, inverse.rle(r))
}
split_two(df, "Type")
#> $a
#> ID Type value
#> 1 1 a 10
#> 2 1 a 2
#> 3 1 b 5
#>
#> $c
#> ID Type value
#> 4 1 c 3
#> 5 1 c 7
#> 6 1 d 3
#> 7 1 d 9
Created on 2023-02-09 with reprex v2.0.2
My dataframe looks like this
data = data.frame(ID=c(1,2,3,4,5,6,7,8,9,10),
Gender=c('Male','Female','Female','Female','Male','Female','Male','Male','Female','Female'))
And I have a reference list that looks like this -
ref=list(Male=1,Female=2)
I'd like to replace values in the Gender column using this reference list, without adding a new column to my dataframe.
Here's my attempt
do.call(dplyr::recode, c(list(data), ref))
Which gives me the following error -
no applicable method for 'recode' applied to an object of class
"data.frame"
Any inputs would be greatly appreciated
An option would be do a left_join after stacking the 'ref' list to a two column data.frame
library(dplyr)
left_join(data, stack(ref), by = c('Gender' = 'ind')) %>%
select(ID, Gender = values)
A base R approach would be
unname(unlist(ref)[as.character(data$Gender)])
#[1] 1 2 2 2 1 2 1 1 2 2
In base R:
data$Gender = sapply(data$Gender, function(x) ref[[x]])
You can use factor, i.e.
factor(data$Gender, levels = names(ref), labels = ref)
#[1] 1 2 2 2 1 2 1 1 2 2
You can unlist ref to give you a named vector of codes, and then index this with your data:
transform(data,Gender=unlist(ref)[as.character(Gender)])
ID Gender
1 1 1
2 2 2
3 3 2
4 4 2
5 5 1
6 6 2
7 7 1
8 8 1
9 9 2
10 10 2
Surprisingly, that one works as well:
data$Gender <- ref[as.character(data$Gender)]
#> data
# ID Gender
# 1 1 1
# 2 2 2
# 3 3 2
# 4 4 2
# 5 5 1
# 6 6 2
# 7 7 1
# 8 8 1
# 9 9 2
# 10 10 2
I have a dataset in R, which contains observations by time. For each subject, I have up to 4 rows, and a variable of ID along with a variable of Time and a variable called X, which is numerical (but can also be categorical for the sake of the question). I wish to compute the change from baseline for each row, by ID. Until now, I did this in SAS, and this was my SAS code:
data want;
retain baseline;
set have;
if (first.ID) then baseline = .;
if (first.ID) then baseline = X;
else baseline = baseline;
by ID;
Change = X-baseline;
run;
My question is: How do I do this in R ?
Thank you in advance.
Dataset Example (in SAS, I don't know how to do it in R).
data have;
input ID, Time, X;
datalines;
1 1 5
1 2 6
1 3 8
1 4 9
2 1 2
2 2 2
2 3 7
2 4 0
3 1 1
3 2 4
3 3 5
;
run;
Generate some example data:
dta <- data.frame(id = rep(1:3, each=4), time = rep(1:4, 3), x = rnorm(12))
# > dta
# id time x
# 1 1 1 -0.232313499
# 2 1 2 1.116983376
# 3 1 3 -0.682125947
# 4 1 4 -0.398029820
# 5 2 1 0.440525082
# 6 2 2 0.952058966
# 7 2 3 0.690180586
# 8 2 4 -0.995872696
# 9 3 1 0.009735667
# 10 3 2 0.556254340
# 11 3 3 -0.064571775
# 12 3 4 -1.003582676
I use the package dplyr for this. This package is not installed by default, so, you'll have to install it first if it isn't already.
The steps are: group the data by id (following operations are done per group), sort the data to make sure it is ordered on time (that the first record is the baseline), then calculate a new column which is the difference between x and the first value of x. The result is stored in a new data.frame, but can of course also be assigned back to dta.
library(dplyr)
dta_new <- dta %>% group_by(id) %>% arrange(id, time) %>%
mutate(change = x - first(x))
# > dta_new
# Source: local data frame [12 x 4]
# Groups: id [3]
#
# id time x change
# <int> <int> <dbl> <dbl>
# 1 1 1 -0.232313499 0.00000000
# 2 1 2 1.116983376 1.34929688
# 3 1 3 -0.682125947 -0.44981245
# 4 1 4 -0.398029820 -0.16571632
# 5 2 1 0.440525082 0.00000000
# 6 2 2 0.952058966 0.51153388
# 7 2 3 0.690180586 0.24965550
# 8 2 4 -0.995872696 -1.43639778
# 9 3 1 0.009735667 0.00000000
# 10 3 2 0.556254340 0.54651867
# 11 3 3 -0.064571775 -0.07430744
# 12 3 4 -1.003582676 -1.01331834
I have a data frame in R which is similar to the follows. Actually my real ’df’ dataframe is much bigger than this one here but I really do not want to confuse anybody so that is why I try to simplify things as much as possible.
So here’s the data frame.
id <-c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3)
a <-c(3,1,3,3,1,3,3,3,3,1,3,2,1,2,1,3,3,2,1,1,1,3,1,3,3,3,2,1,1,3)
b <-c(3,2,1,1,1,1,1,1,1,1,1,2,1,3,2,1,1,1,2,1,3,1,2,2,1,3,3,2,3,2)
c <-c(1,3,2,3,2,1,2,3,3,2,2,3,1,2,3,3,3,1,1,2,3,3,1,2,2,3,2,2,3,2)
d <-c(3,3,3,1,3,2,2,1,2,3,2,2,2,1,3,1,2,2,3,2,3,2,3,2,1,1,1,1,1,2)
e <-c(2,3,1,2,1,2,3,3,1,1,2,1,1,3,3,2,1,1,3,3,2,2,3,3,3,2,3,2,1,3)
df <-data.frame(id,a,b,c,d,e)
df
Basically what I would like to do is to get the occurrences of numbers for each column (a,b,c,d,e) and for each id group (1,2,3) (for this latter grouping see my column ’id’).
So, for column ’a’ and for id number ’1’ (for the latter see column ’id’) the code would be something like this:
as.numeric(table(df[1:10,2]))
##The results are:
[1] 3 7
Just to briefly explain my results: in column ’a’ (and regarding only those records which have number ’1’ in column ’id’) we can say that number '1' occured 3 times and number '3' occured 7 times.
Again, just to show you another example. For column ’a’ and for id number ’2’ (for the latter grouping see again column ’id’):
as.numeric(table(df[11:20,2]))
##After running the codes the results are:
[1] 4 3 3
Let me explain a little again: in column ’a’ and regarding only those observations which have number ’2’ in column ’id’) we can say that number '1' occured 4 times, number '2' occured 3 times and number '3' occured 3 times.
So this is what I would like to do. Calculating the occurrences of numbers for each custom-defined subsets (and then collecting these values into a data frame). I know it is not a difficult task but the PROBLEM is that I’m gonna have to change the input ’df’ dataframe on a regular basis and hence both the overall number of rows and columns might change over time…
What I have done so far is that I have separated the ’df’ dataframe by columns, like this:
for (z in (2:ncol(df))) assign(paste("df",z,sep="."),df[,z])
So df.2 will refer to df$a, df.3 will equal df$b, df.4 will equal df$c etc. But I’m really stuck now and I don’t know how to move forward…
Is there a proper, ”automatic” way to solve this problem?
How about -
> library(reshape)
> dftab <- table(melt(df,'id'))
> dftab
, , value = 1
variable
id a b c d e
1 3 8 2 2 4
2 4 6 3 2 4
3 4 2 1 5 1
, , value = 2
variable
id a b c d e
1 0 1 4 3 3
2 3 3 3 6 2
3 1 4 5 3 4
, , value = 3
variable
id a b c d e
1 7 1 4 5 3
2 3 1 4 2 4
3 5 4 4 2 5
So to get the number of '3's in column 'a' and group '1'
you could just do
> dftab[3,'a',1]
[1] 4
A combination of tapply and apply can create the data you want:
tapply(df$id,df$id,function(x) apply(df[id==x,-1],2,table))
However, when a grouping doesn't have all the elements in it, as in 1a, the result will be a list for that id group rather than a nice table (matrix).
$`1`
$`1`$a
1 3
3 7
$`1`$b
1 2 3
8 1 1
$`1`$c
1 2 3
2 4 4
$`1`$d
1 2 3
2 3 5
$`1`$e
1 2 3
4 3 3
$`2`
a b c d e
1 4 6 3 2 4
2 3 3 3 6 2
3 3 1 4 2 4
$`3`
a b c d e
1 4 2 1 5 1
2 1 4 5 3 4
3 5 4 4 2 5
I'm sure someone will have a more elegant solution than this, but you can cobble it together with a simple function and dlply from the plyr package.
ColTables <- function(df) {
counts <- list()
for(a in names(df)[names(df) != "id"]) {
counts[[a]] <- table(df[a])
}
return(counts)
}
results <- dlply(df, "id", ColTables)
This gets you back a list - the first "layer" of the list will be the id variable; the second the table results for each column for that id variable. For example:
> results[['2']]['a']
$a
1 2 3
4 3 3
For id variable = 2, column = a, per your above example.
A way to do it is using the aggregate function, but you have to add a column to your dataframe
> df$freq <- 0
> aggregate(freq~a+id,df,length)
a id freq
1 1 1 3
2 3 1 7
3 1 2 4
4 2 2 3
5 3 2 3
6 1 3 4
7 2 3 1
8 3 3 5
Of course you can write a function to do it, so it's easier to do it frequently, and you don't have to add a column to your actual data frame
> frequency <- function(df,groups) {
+ relevant <- df[,groups]
+ relevant$freq <- 0
+ aggregate(freq~.,relevant,length)
+ }
> frequency(df,c("b","id"))
b id freq
1 1 1 8
2 2 1 1
3 3 1 1
4 1 2 6
5 2 2 3
6 3 2 1
7 1 3 2
8 2 3 4
9 3 3 4
You didn't say how you'd like the data. The by function might give you the output you like.
by(df, df$id, function(x) lapply(x[,-1], table))