This question already has answers here:
Count consecutive elements in a same length vector
(3 answers)
Closed 3 years ago.
Example
x<- c("a","a","b","c","c","c","d")
I like frequencies to be:
> a a b c c c d
> 2 2 1 3 3 3 1
and not
> a b c d
> 2 1 3 1
We can use table as a named vector
table(x)[x]
x
# a a b c c c d
# 2 2 1 3 3 3 1
or with ave
setNames(ave(seq_along(x), x, FUN = length), x)
#a a b c c c d
#2 2 1 3 3 3 1
You can also use rle():
with(rle(x), setNames(rep(lengths, lengths), rep(values, lengths)))
a a b c c c d
2 2 1 3 3 3 1
However, it assumes one run per group (as indicated by letters). Additionally, it could also work properly if you have several runs per group, but the vector could be sorted:
x <- c("a","b","c","c","c","d","a")
with(rle(sort(x)), setNames(rep(lengths, lengths), rep(values, lengths)))
a a b c c c d
2 2 1 3 3 3 1
If there are several runs per group and the vector could not be sorted, then this solution is not feasible.
Related
How does one count the characters based on the order they appear in a single length string. Below is an minimal example:
x <- "abbccdddaab"
First thought was this but it only counts them irrespective of order:
table(unlist(strsplit(x, "\\b")))
a b c d
3 3 2 3
But the desired output is:
a b c d a b
1 2 2 3 2 1
I would imagine the solution would require a for loop?
We can use rle instead of table as rle returns the output as a list of values and lengths based on checking whether the adjacent elements are same or not
out <- rle(strsplit(x, "\\b")[[1]])
setNames(out$lengths, out$values)
# a b c d a b
# 1 2 2 3 2 1
Using data.table::rleid :
x <- "abbccdddaab"
tmp <- strsplit(x, "\\b")[[1]]
table(data.table::rleid(tmp))
#1 2 3 4 5 6
#1 2 2 3 2 1
This question already has answers here:
Repeat each row of data.frame the number of times specified in a column
(10 answers)
Closed 3 years ago.
I have a data.frame with n rows and I would like to repeat this rows according to the observation of another variable
This is an example for a data.frame
df <- data.frame(a=1:3, b=letters[1:2])
df
a b
1 1 a
2 2 b
3 3 c
And this one is an example for a variable
df1 <- data.frame(x=1:3)
df1
x
1 1
2 2
3 3
In the next step I would like to repeat every row from the df with the observation of df1
So that it would look like this
a b
1 1 a
2 2 b
3 2 b
4 3 c
5 3 c
6 3 c
If you have any idea how to solve this problem, I would be very thankful
You simply can repeat the index like:
df[rep(1:3,df1$x),]
# a b
#1 1 a
#2 2 b
#2.1 2 b
#3 3 c
#3.1 3 c
#3.2 3 c
or not fixed to size 3
df[rep(seq_along(df1$x),df1$x),]
I have a dataset with several variables, but I want to keep the rows that are the same based on two columns. Here is an example of what I want to do:
a <- c(rep('A',3), rep('B', 3), rep('C',3))
b <- c(1,1,2,4,4,4,5,5,5)
df <- data.frame(a,b)
a b
1 A 1
2 A 1
3 A 2
4 B 4
5 B 4
6 B 4
7 C 5
8 C 5
9 C 5
I know that if I use the duplicated function I can get:
df[!duplicated(df),]
a b
1 A 1
3 A 2
4 B 4
7 C 5
But since the level 'A' on column a does not have a unique value in b, I want to drop both observations to get a new data.frame as this:
a b
4 B 4
7 C 5
I don't mind to have repeated values across b, as long as for every same level on a there is the same value in b.
Is there a way to do this? Thanks!
This one maybe?
ag <- aggregate(b~a, df, unique)
ag[lengths(ag$b)==1,]
# a b
#2 B 4
#3 C 5
Maybe something like this:
> ind <- apply(sapply(with(df, split(b,a)), diff), 2, function(x) all(x==0) )
> out <- df[!duplicated(df),]
> out[out$a %in% names(ind)[ind], ]
a b
4 B 4
7 C 5
Here is another option with data.table
library(data.table)
setDT(df)[, if(uniqueN(b)==1) .SD[1L], by = a]
# a b
#1: B 4
#2: C 5
Given the following df:
a=c('a','b','c')
b=c(1,2,5)
c=c(2,3,4)
d=c(2,1,6)
df=data.frame(a,b,c,d)
a b c d
1 a 1 2 2
2 b 2 3 1
3 c 5 4 6
I'd like to apply a function that normally takes a vector (and returns a vector) like cummax row by row to the columns in position b to d.
Then, I'd like to have the output back in the df, either as a vector in a new column of the df, or replacing the original data.
I'd like to avoid writing it as a for loop that would iterate every row, pull out the content of the cells into a vector, do its thing and put it back.
Is there a more efficient way? I've given the apply family functions a go, but I'm struggling to first get a good way to vectorise content of columns by row and get the right output.
the final output could look something like that (imagining I've applied a cummax() function).
a b c d
1 a 1 2 2
2 b 2 3 3
3 c 5 5 6
or
a b c d output
1 a 1 2 2 (1,2,2)
2 b 2 3 1 (2,3,3)
3 c 5 4 6 (5,5,6)
where output is a vector.
Seems this would just be a simple apply problem that you want to cbind to df:
> cbind(df, apply(df[ , 4:2] # work with columns in reverse order
, 1, # do it row-by-row
cummax) )
a b c d 1 2 3
d a 1 2 2 2 1 6
c b 2 3 1 2 3 6
b c 5 4 6 2 3 6
Ouch. Bitten by failing to notice that this would be returned in a column oriented matrix and need to transpose that result; Such a newbie mistake. But it does show the value of having a question with a reproducible dataset I suppose.
> cbind(df, t(apply(df[ , 4:2] , 1, cummax) ) )
a b c d d c b
1 a 1 2 2 2 2 2
2 b 2 3 1 1 3 3
3 c 5 4 6 6 6 6
To destructively assign the result to df you would just use:
df <- # .... that code.
This does the concatenation with commas (and as a result no longer needs to be transposed:
> cbind(df, output=apply(df[ , 4:2] , 1, function(x) paste( cummax(x), collapse=",") ) )
a b c d output
1 a 1 2 2 2,2,2
2 b 2 3 1 1,3,3
3 c 5 4 6 6,6,6
I have a matrix that contains following:
A B C D
a 1 3 2 5
b 3 2 5 8
a 2 1 0 9
a 4 2 1 3
c 4 3 1 1
b 2 5 1 9
A, B, C, D are column names and
a, b, c, d are row names.
I want to make it look like
A B C D
a 4 3 2 9
b 3 5 5 9
c 4 3 1 1
using R, Which is to
1) order the row in alphabetical order,
2) and then if there are redundant rows (i.e. there are other rows with the same row name), pick a maximum value among the redundant rows for each column and delete the others.
I first used python to do this process, but I was wondering if there is
more convenient way for this job in R.
I would appreciate any help.
You can use data.table
dt_in <- data.table(matrix_in)
dt_in[, name := rownames(matrix_in)]
dt_max <- dt_in[, list(A = max(A), B = max(B), C = max(C), D = max(D)), by = "name"]
as.matrix(data.frame(dt_max))
Here's a one liner using data.table you can keep the rows while converting to data.table and then apply max function over all columns using lapply(.SD,...) by the rn variable (the saved row names)
library(data.table)
data.table(m, keep.rownames = TRUE)[, lapply(.SD, max), by = rn]
# rn A B C D
# 1: a 4 3 2 9
# 2: b 3 5 5 9
# 3: c 4 3 1 1
You can simply use aggregate function:
aggregate(matrix ~ rownames(matrix), matrix, max)