Let's say I have this simple data frame:
df <- data.frame(x=c(1,3,3,1,3,1), y = c(2,2,2,2,2,2),z = c('a','b','c','d','e','f'))
> df
x y z
1 1 2 a
2 3 2 b
3 3 2 c
4 1 2 d
5 3 2 e
6 1 2 f
I would like to subset where x= 3, return only column x and y and include a calculated colum x+y.
I can get the first 2 things done, but I can't get the caclulated column to also appear.
df[df$x==3,c("x","y")]
How can I do that, but using base R only.
Staying in base, just do a rowSums before your subset.
df$xy <- rowSums(df[, c("x", "y")])
df[df$x == 3, c("x", "y", "xy")]
# x y xy
# 2 3 2 5
# 3 3 2 5
# 5 3 2 5
Personally, I do prefer the dplyr approach, which #akrun commented on your question.
You can also do like this
df <- data.frame(x=c(1,3,3,1,3,1), y = c(2,2,2,2,2,2),z = c('a','b','c','d','e','f'))
df$z <- ifelse(df$x == 3, (df$x + df$y), df$y)
df
x y z
1 1 2 2
2 3 2 5
3 3 2 5
4 1 2 2
5 3 2 5
6 1 2 2
Related
Suppose I have a dataframe that looks like this:
> data <- data.frame(x = c(1,1,2,2,3,4,5,6), y = c(1,2,3,4,5,6,7,8))
> data
x y
1 1 1
2 1 2
3 2 3
4 2 4
5 3 5
6 4 6
7 5 7
8 6 8
I want to use mutate and case_when to create a new id variable that will identify rows using the variable x, and give rows missing x a unique id. In other words, I should have the same id for rows one and two, rows three and four, while rows 5-8 should have their own unique ids. Suppose I want to generate these id values with a function:
id_function <- function(x, n){
set.seed(x)
res <- character(n)
for(i in seq(n)){
res[i] <- paste0(sample(c(letters, LETTERS, 0:9), 32), collapse="")
}
res
}
id_function(1, 1)
[1] "4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf"
I am trying to use this function on the RHS of a case_when expression like this:
data %>%
mutate(my_id = id_function(1234, nrow(.)),
my_id = dplyr::case_when(!is.na(x) ~ id_function(x, 1),
TRUE ~ my_id))
But the RHS does not seem to be vectorized and I get the same value for all non-missing values of x:
x y my_id
1 1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
3 2 3 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
4 2 4 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
5 NA 5 0vnws5giVNIzp86BHKuOZ9ch4dtL3Fqy
6 NA 6 IbKU6DjvW9ypitl7qc25Lr4sOwEfghdk
7 NA 7 8oqQMPx6IrkGhXv4KlUtYfcJ5Z1RCaDy
8 NA 8 BRsjumlCEGS6v4ANrw1bxLynOKkF90ao
I'm sure there's a way to vectorize the RHS, what am I doing wrong? Is there an easier approach to solving this problem?
I guess rowwise() would do the trick:
data %>%
rowwise() %>%
mutate(my_id = id_function(x, 1))
x y my_id
1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 3 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
2 4 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
3 5 e5lMJNQEhtj4VY1KbCR9WUiPrpy7vfXo
4 6 3kYcgR7109DLbxatQIAKXFeovN8pnuUV
5 7 bQ4ok7OuDgscLUlpzKAivBj2T3m6wrWy
6 8 0jSn3Jcb2HDA5uhvG8g1ytsmRpl6CQWN
purrr map functions can be used for non-vectorized functions. The following will give you a similar result. map2 will take the two arguments expected by your id_function.
library(tidyverse)
data %>%
mutate(my_id = map2(x, 1, id_function))
Output
x y my_id
1 1 1 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
2 1 2 4dMaHwQnrYGu0PTjgioXKOyW75NRZtcf
3 2 3 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
4 2 4 uof7FhqC3lOXkacp54MGZJLUR6siSKDb
5 3 5 e5lMJNQEhtj4VY1KbCR9WUiPrpy7vfXo
6 4 6 3kYcgR7109DLbxatQIAKXFeovN8pnuUV
7 5 7 bQ4ok7OuDgscLUlpzKAivBj2T3m6wrWy
8 6 8 0jSn3Jcb2HDA5uhvG8g1ytsmRpl6CQWN
I have a list of dataframes. Each of these dataframes has the same number of columns and rows, and has a similar data structure:
df.list <- list(data.frame1, data.frame2, data.frame3)
I have a vector of characters:
charvec <- c("a","b","c")
I want to replace the column name of the second column in each data frame by iterating through the above character vector. For example, the first data frame's second column should be "a". The second data frame's second column should be "b".
[[1]]
col1 a
1 1 2
2 2 3
[[2]]
col1 b
1 1 2
2 2 3
A reproducible example:
charvec <- c("a","b","c")
df_list <- list(df1 = data.frame(x = seq_len(3), y = seq_len(3)), df2 = data.frame(x = seq_len(4), y = seq_len(4)), df3 = data.frame(x = seq_len(5), y = seq_len(5)))
for(i in seq_along(df_list)){
names(df_list[[i]])[2] <- charvec[i]
}
> df_list
$df1
x a
1 1 1
2 2 2
3 3 3
$df2
x b
1 1 1
2 2 2
3 3 3
4 4 4
$df3
x c
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
Also can use map2 from purrr. Thanks to #ismirsehregal for example data.
library(purrr)
map2(
df_list,
charvec,
\(x, y) {
names(x)[2] <- y
x
}
)
Output
$df1
x a
1 1 1
2 2 2
3 3 3
$df2
x b
1 1 1
2 2 2
3 3 3
4 4 4
$df3
x c
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
I'm trying to filter a very heterogeneous dataset.
I have numerous variables with several replicates each one. I have a factor with two levels (lets say X and Y), and I would like to subset the variables which present a fold change on its mean greater than 2 (X/Y >= 2 OR Y/X >= 2).
How can I achieve that in R? I can think of some ways but they seem too much of a hassle, I'm sure there is a better way. I would later run multivariate test on those filtered variables.
This would be an example dataset:
d <- read.table(text = "a b c d factor replicate
1 2 2 3 X 1
3 2 4 4 X 2
2 3 1 2 X 3
1 2 3 2 X 4
5 2 6 4 Y 1
7 4 5 5 Y 2
8 5 7 4 Y 3
6 4 3 3 Y 4", header = TRUE)
From this example, only variables a and c should be kept.
Using colMeans:
#subset
x <- d[ d$factor == "X", 1:4 ]
y <- d[ d$factor == "Y", 1:4 ]
# check colmeans, and get index
which(colMeans(x/y) >= 2 | colMeans(y/x) >= 2)
# a c
# 1 3
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
first the data, then my question:
df <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
df
Z Y A
1 2 1
2 2 2
3 2 3
1 3 4
2 3 5
3 3 6
I am using R and would like to generate a vector a with given vector Z and Y. Like if (z1==1 & y1==2), then a1=1, if (z3==3 & y1==2), then a1=3....If we only have one condition (Z or Y), I could generate A with switch function, however, how could I generate A with Z and Y?
If I understood the question correctly the answer is as follows. If I misunderstood then please correct me.
df <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
df$A <- NA # Empty out A
# Re-create A
df$A[df$Z == 1 & df$Y == 2] <- 1
df$A[df$Z == 2 & df$Y == 2] <- 2
df$A[df$Z == 3 & df$Y == 2] <- 3
df$A[df$Z == 1 & df$Y == 3] <- 4
df$A[df$Z == 2 & df$Y == 3] <- 5
df$A[df$Z == 3 & df$Y == 3] <- 6
df <- df[order(df$A),]
df
Z Y A
1 1 2 1
5 2 2 2
3 3 2 3
4 1 3 4
2 2 3 5
6 3 3 6
Another way to do it in one call is to use ifelse function as follows:
df <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
df$A <- ifelse(df$Z == 1 & df$Y == 2,1,
ifelse(df$Z == 3 & df$Y == 2,3,
ifelse(df$Z > 1,5,8)))
Z Y A
1 1 2 1
2 2 3 5
3 3 2 3
4 1 3 8
5 2 2 5
6 3 3 5
As the number of conditions increases, if you're just looking at exact matches, it becomes more useful to use a lookup table for the results and merge them. So we start with a lookup table as you've defined it
df.lookup <- data.frame(Z=c(1,2,3),Y=c(2,3),A=c(1,2,3,4,5,6))
And let's say that we have a dataframe we want to populate
set.seed(17) # To get repeatable random numbers
df <- data.frame(Z=sample(1:3, 6, replace=T), Y=sample(2:3, 6, replace=T))
So we have
> df
Z Y
1 1 2
2 3 2
3 2 3
4 3 2
5 2 2
6 2 2
> merge(df, df.lookup)
Z Y A
1 1 2 1
2 2 2 5
3 2 2 5
4 2 3 2
5 3 2 3
6 3 2 3
The merge function will automatically select same-named columns, but you can specify the names if they are different, see ?merge for more information.