How do I convert 2 columns from a data.frame onto 2 different columns?
I.E:
Data
A B C D
1 3 5 7
2 4 6 8
to
Data
A B
1 3
2 4
5 7
6 8
You can use rbind
rbind(df[,1:2], data.frame(A = df$C, B = df$D))
You can use a fast version of rbind, rbindlist from data.table:
library(data.table)
rbindlist(lapply(seq(1, ncol(df), 2), function(i) df[,i:(i+1)]))
Here is my solution but it requires to change names of the columns.
names(dat) <- c("A", "B", "A", "B")
merge(dat[1:2], dat[3:4], all = T)
A B
1 1 3
2 2 4
3 5 7
4 6 8
And here is another solution more easy.
dat[3:4, ] <- dat[ ,3:4]
dat <- dat[1:2]
dat
A B
1 1 3
2 2 4
3 5 7
4 6 8
For scalability, a solution that will halve any even size data frame and append the rows:
half <- function(df) {m <- as.matrix(df)
dim(m) <- c(nrow(df)*2,ncol(df)/2)
nd <- as.data.frame(m)
names(nd) <- names(df[(1:dim(nd)[2])]);nd}
half(Data)
A B
1 1 5
2 2 6
3 3 7
4 4 8
Related
I have to subset the data of 6 rows every time. How to do that in R?
data:
col1 : 1,2,3,4,5,6,7,8,9,10
col2 : a1,a2,a3,a4,a5,a6,a7,a8,a9,a10
I want to do subset of 6 rows every time. First subset of the rows will have 1:6 ,next subset of the rows will have 7:nrow(data). I have tried using seq function.
seqData <- seq(1,nrow(data),6)
output: It is giving 1 and 7th row but I want 1 to 6 rows first, next onwards 7 to nrow(data).
How to get output like that.
Will this work:
set.seed(1)
dat <- data.frame(c1 = sample(1:5,12,T),
c2 = sample(1:5,12,T))
dat
c1 c2
1 1 2
2 4 2
3 1 1
4 2 5
5 5 5
6 3 1
7 2 1
8 3 5
9 3 5
10 1 2
11 5 2
12 5 1
split(dat, rep(1:ceiling(nrow(dat)/6), each = 6))
$`1`
c1 c2
1 1 2
2 4 2
3 1 1
4 2 5
5 5 5
6 3 1
$`2`
c1 c2
7 2 1
8 3 5
9 3 5
10 1 2
11 5 2
12 5 1
The function below creates a numeric vector with integers increasing by 1 unit every n rows. And uses this vector to split the data as needed.
data <- data.frame(col1 = 1:10, col2 = paste0("a", 1:10))
split_nrows <- function(x, n){
f <- c(1, rep(0, n - 1))
f <- rep(f, length.out = NROW(x))
f <- cumsum(f)
split(x, f)
}
split_nrows(data, 6)
Here's a simple example with mtcars that yields a list of 6 subset dfs.
nrows <- nrow(mtcars)
breaks <- seq(1, nrows, 6)
listdfs <- lapply(breaks, function(x) mtcars[x:(x+5), ]) # increment by 5 not 6
listdfs[[6]] <- listdfs[[6]][1:2, ] #last df: remove 4 NA rows (36 - 32)
Given a two column data.frame with one containing group labels and a second containing integer values ordered from smallest to largest. How can the data be expanded creating pairs of combinations of the integer column?
Not sure the best way to state this. I'm not interested in all possible combinations but instead all unique combinations starting from the lowest value.
In r, the combn function gives the desired output not considering groups, for example:
t(combn(seq(1:4),2))
[,1] [,2]
[1,] 1 2
[2,] 1 3
[3,] 1 4
[4,] 2 3
[5,] 2 4
[6,] 3 4
Since the first values is 1 we get the unique combination of (1,2) and not the additional combination of (2,1) which I don't need. How would one then apply a similar method by groups?
for example given a data.frame
test <- data.frame(Group = rep(c("A","B"),each=4),
Val = c(1,3,6,8,2,4,5,7))
test
Group Val
1 A 1
2 A 3
3 A 6
4 A 8
5 B 2
6 B 4
7 B 5
8 B 7
I was able to come up with this solution that gives the desired output:
test <- data.frame(Group = rep(c("A","B"),each=4),
Val = c(1,3,6,8,2,4,5,7))
j=1
for(i in unique(test$Group)){
if(j==1){
one <- filter(test,i == Group)
two <- data.frame(t(combn(one$Val,2)))
test1 <- data.frame(Group = i,Val1=two$X1,Val2=two$X2)
j=j+1
}else{
one <- filter(test,i == Group)
two <- data.frame(t(combn(one$Val,2)))
test2 <- data.frame(Group = i,Val1=two$X1,Val2=two$X2)
test1 <- rbind(test1,test2)
}
}
test1
Group Val1 Val2
1 A 1 3
2 A 1 6
3 A 1 8
4 A 3 6
5 A 3 8
6 A 6 8
7 B 2 4
8 B 2 5
9 B 2 7
10 B 4 5
11 B 4 7
12 B 5 7
However, this is not elegant and is really slow as the number of groups and length of each group become large. It seems like there should be a more elegant and efficient solution but so far I have not come across anything on SO.
I would appreciate any ideas!
here is a data.table approach
library( data.table )
#make test a data.table
setDT(test)
#split by group
L <- split( test, by = "Group")
#get unique combinations of 2 Vals
L2 <- lapply( L, function(x) {
as.data.table( t( combn( x$Val, m = 2, simplify = TRUE ) ) )
})
#merge them back together
data.table::rbindlist( L2, idcol = "Group" )
# Group V1 V2
# 1: A 1 3
# 2: A 1 6
# 3: A 1 8
# 4: A 3 6
# 5: A 3 8
# 6: A 6 8
# 7: B 2 4
# 8: B 2 5
# 9: B 2 7
#10: B 4 5
#11: B 4 7
#12: B 5 7
You can set simplify = F in combn() and then use unnest_wider() in dplyr.
library(dplyr)
library(tidyr)
test %>%
group_by(Group) %>%
summarise(Val = combn(Val, 2, simplify = F)) %>%
unnest_wider(Val, names_sep = "_")
# Group Val_1 Val_2
# <chr> <dbl> <dbl>
# 1 A 1 3
# 2 A 1 6
# 3 A 1 8
# 4 A 3 6
# 5 A 3 8
# 6 A 6 8
# 7 B 2 4
# 8 B 2 5
# 9 B 2 7
# 10 B 4 5
# 11 B 4 7
# 12 B 5 7
library(tidyverse)
df2 <- split(df$Val, df$Group) %>%
map(~gtools::combinations(n = 4, r = 2, v = .x)) %>%
map(~as_tibble(.x, .name_repair = "unique")) %>%
bind_rows(.id = "Group")
I have a dataframe in the following format
> x <- data.frame("a" = c(1,1),"b" = c(2,2),"c" = c(3,4))
> x
a b c
1 1 2 3
2 1 2 4
I'd like to add 3 new columns which is a cumulative product of the columns a b c, however I need a reverse cumulative product i.e. the output should be
row 1:
result_d = 1*2*3 = 6 , result_e = 2*3 = 6, result_f = 3
and similarly for row 2
The end result will be
a b c result_d result_e result_f
1 1 2 3 6 6 3
2 1 2 4 8 8 4
the column names do not matter this is just an example. Does anyone have any idea how to do this?
as per my comment, is it possible to do this on a subset of columns? e.g. only for columns b and c to return:
a b c results_e results_f
1 1 2 3 6 3
2 1 2 4 8 4
so that column "a" is effectively ignored?
One option is to loop through the rows and apply cumprod over the reverse of elements and then do the reverse
nm1 <- paste0("result_", c("d", "e", "f"))
x[nm1] <- t(apply(x, 1,
function(x) rev(cumprod(rev(x)))))
x
# a b c result_d result_e result_f
#1 1 2 3 6 6 3
#2 1 2 4 8 8 4
Or a vectorized option is rowCumprods
library(matrixStats)
x[nm1] <- rowCumprods(as.matrix(x[ncol(x):1]))[,ncol(x):1]
temp = data.frame(Reduce("*", x[NCOL(x):1], accumulate = TRUE))
setNames(cbind(x, temp[NCOL(temp):1]),
c(names(x), c("res_d", "res_e", "res_f")))
# a b c res_d res_e res_f
#1 1 2 3 6 6 3
#2 1 2 4 8 8 4
Say I have a data frame df
df <- data.frame( a1 = 1:10, b1 = 2:11, c2 = 3:12 )
I wish to subset the columns, but with a pattern
df1 <- subset( df, select= (pattern = "1") )
To get
> df1
a1 b1
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 10
10 10 11
Is this possible?
It is possible to do this via
subset(df, select = grepl("1", names(df)))
For automating this as a function, one can use use [ to do the subsetting. Couple that with one of R's regular expression functions and you have all you need.
By way of an example, here is a custom function implementing the ideas I mentioned above.
Subset <- function(df, pattern) {
ind <- grepl(pattern, names(df))
df[, ind]
}
Note this does not error checking etc and just relies upon grepl to return a logical vector indicating which columns match pattern, which is then passed to [ to subset by columns. Applied to your df this gives:
> Subset(df, pattern = "1")
a1 b1
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 10
10 10 11
Same same but different:
df2 <- df[, grep("1", names(df))]
a1 b1
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 10
10 10 11
Base R now has a convenience function endsWith():
df[, endsWith(names(df), "1")]
In data.table you can do:
library(data.table)
setDT(df)
df[, .SD, .SDcols = patterns("1")]
# Or more precisely
df[, .SD, .SDcols = patterns("1$")]
In dplyr:
library(dplyr)
select(df, ends_with("1"))
Hoping there's a simple answer here but I can't find it anywhere.
I have a numeric matrix with row names and column names:
# 1 2 3 4
# a 6 7 8 9
# b 8 7 5 7
# c 8 5 4 1
# d 1 6 3 2
I want to melt the matrix to a long format, with the values in one column and matrix row and column names in one column each. The result could be a data.table or data.frame like this:
# col row value
# 1 a 6
# 1 b 8
# 1 c 8
# 1 d 1
# 2 a 7
# 2 c 5
# 2 d 6
...
Any tips appreciated.
Use melt from reshape2:
library(reshape2)
#Fake data
x <- matrix(1:12, ncol = 3)
colnames(x) <- letters[1:3]
rownames(x) <- 1:4
x.m <- melt(x)
x.m
Var1 Var2 value
1 1 a 1
2 2 a 2
3 3 a 3
4 4 a 4
...
The as.table and as.data.frame functions together will do this:
> m <- matrix( sample(1:12), nrow=4 )
> dimnames(m) <- list( One=letters[1:4], Two=LETTERS[1:3] )
> as.data.frame( as.table(m) )
One Two Freq
1 a A 7
2 b A 2
3 c A 1
4 d A 5
5 a B 9
6 b B 6
7 c B 8
8 d B 10
9 a C 11
10 b C 12
11 c C 3
12 d C 4
Assuming 'm' is your matrix...
data.frame(col = rep(colnames(m), each = nrow(m)),
row = rep(rownames(m), ncol(m)),
value = as.vector(m))
This executes extremely fast on a large matrix and also shows you a bit about how a matrix is made, how to access things in it, and how to construct your own vectors.
A modification that doesn't require you to know anything about the storage structure, and that easily extends to high dimensional arrays if you use the dimnames, and slice.index functions:
data.frame(row=rownames(m)[as.vector(row(m))],
col=colnames(m)[as.vector(col(m))],
value=as.vector(m))