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
Related
Say we have a data frame with a factor (Group) that is a grouping variable for a list of IDs:
set.seed(123)
data <- data.frame(Group = factor(sample(5,10, replace = T)),
ID = c(1:10))
In this example, the ID's belong to one of 5 Groups, labeled 1:5. We simply want to replace 1:5 with A:E. In other words, if Group == 1, we want to change it to A, if Group == 2, we want to change it to B, and so on. What is the simplest way to achieve this?
You may assign new labels= in a names list using factor once again.
data$Group1 <- factor(data$Group, labels=list("1"="A", "2"="B", "3"="C", "4"="D", "5"="E"))
## more succinct:
data$Group2 <- factor(data$Group, labels=setNames(list("A", "B", "C", "D", "E"), 1:5))
data
# Group ID Group1 Group2 Group3
# 1 3 1 C C C
# 2 3 2 C C C
# 3 2 3 B B B
# 4 2 4 B B B
# 5 3 5 C C C
# 6 5 6 E E E
# 7 4 7 D D D
# 8 1 8 A A A
# 9 2 9 B B B
# 10 3 10 C C C
This for general, if indeed capital letters are wanted see #RonakShah's solution.
You can use the built-in constant in R LETTERS :
data$new_group <- LETTERS[data$Group]
data
# Group ID new_group
#1 3 1 C
#2 3 2 C
#3 2 3 B
#4 2 4 B
#5 3 5 C
#6 5 6 E
#7 4 7 D
#8 1 8 A
#9 2 9 B
#10 3 10 C
Created a new column (new_group) here for comparison purposes. You can overwrite the same column if you wish to.
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
Suppose I have a dataframe like so
a b c
1 2 3
1 3 4
1 4 5
2 5 6
2 6 7
3 7 8
4 8 9
What I want is the following:
a b c d
1 2 3 a
1 3 4 b
1 4 5 c
2 5 6 a
2 6 7 b
3 7 8 a
4 8 9 a
Essentially, I want to do a cycling, for each group by the column a, I want to create a new column which cycles the letters from a to z in order. Group 1 has three elements, so the letter goes from 'a' to 'c'. Group 3 and 4 has only 1 element, so the letter only gets assigned 'a'.
A data.table option is
library(data.table)
setDT(dd)[, d:= letters[seq_len(.N)], by = a]
One way to do this is with a split-apply-combine paradigm, as in plyr (or dplyr or data.table or ...
Create data:
dd <- data.frame(a=rep(1:4,c(3,2,1,1)),
b=2:8,c=3:9)
Use ddply to split the data frame by variable a, transforming each piece by adding an appropriate variable, then recombine:
library("plyr")
ddply(dd,"a",
transform,
d=letters[1:length(b)])
Or in dplyr:
library("dplyr")
dd %>% group_by(a) %>%
mutate(d=letters[1:n()])
Or in base R (thanks #thelatemail):
dd$d <- ave(rownames(dd), dd$a,
FUN=function(x) letters[seq_along(x)] )
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)
I have a problem figuering this out:
suppose this is how my data looks like:
Num condition y
1 a 1
2 a 2
3 a 3
4 b 4
5 b 5
6 b 6
7 c 7
8 c 8
9 c 9
10 b 10
11 b 11
12 b 12
I now want to make calculation (e.g., mean) on b, depending on whether value was in the row before b, in this example a or c?
Thanks for any help!!!
Angelika
Is this what you want?
# in order to separate between different runs of condition 'b',
# get length and value of runs of equal values of 'condition'
rl <- rle(x = df$condition)
df$run <- rep(x = seq_len(length(rl$lengths)), times = rl$lengths)
# calculate sum of y, on data grouped by condition and run, and where condition is 'b'
aggregate(y ~ condition + run, data = df, subset = condition == "b", sum)
You can add a "lagged" condition column to your dataframe (assuming DF) using
> DF <- within(DF, lag_cond <- c(NA, head(as.character(condition), -1)))
Result:
Num condition y lag_cond
1 a 1 <NA>
2 a 2 a
3 a 3 a
4 b 4 a
5 b 5 b
6 b 6 b
7 c 7 b
8 c 8 c
9 c 9 c
10 b 10 c
11 b 11 b
12 b 12 b
Now you can identify rows you want like this:
> DF[with(DF, condition=="b" & lag_cond %in% c("a","c")),]
Num condition y lag_cond
4 b 4 a
10 b 10 c