Results Transposed with R Apply

R - apply / mapply - Transpose is Required

Just have a look at the documentation (?apply)

If each call to FUN returns a vector of length n, then apply returns an array of dimension c(n, dim(X)[MARGIN]) if n > 1.

In other words: The number of rows of the resulting matrix is given by the length of the vectors in your function to be applied.

Why apply() returns a transposed xts matrix?

That's what apply is documented to do. From ?apply:

Value:

 If each call to ‘FUN’ returns a vector of length ‘n’, then ‘apply’
 returns an array of dimension ‘c(n, dim(X)[MARGIN])’ if ‘n > 1’.

In your case, 'n'=48 (because you're looping over rows), so apply will return an array of dimension c(48, 7429).

Also note that myxts.2 is not an xts object. It's a regular array. You have a couple options:

1. transpose the results of apply before re-creating your xts object:

   data(sample_matrix)
   myxts <- as.xts(sample_matrix)
   dim(myxts)    # [1] 180   4
   myxts.2 <- apply(myxts, 1 , identity)
   dim(myxts.2)  # [1]   4 180
   myxts.2 <- xts(t(apply(myxts, 1 , identity)), index(myxts))
   dim(myxts.2)  # [1] 180   4
   

2. Vectorize your function so it operates on all the rows of an xts
   object and returns an xts object. Then you don't have to worry
   about apply at all.

Finally, please start providing reproducible examples. It's not that hard and it makes it a lot easier for people to help. I've provided an example above and I hope you can use it in your following questions.

R apply result is inconsistent

With a little help from friends, I came up with this simpler, base-R solution:

df$id<-unlist(apply(df,1,function(x)
  ifelse(x["first"]=="none",0, which(as.integer(x["first"])==as.integer(x[2:10])))))

See the answers there for an explanation of why apply was problematic -- briefly, it transformed all your data to character, but then padded it in a way that made the comparisons fail.

On a related note, when you read.csv, you might want to add stringsAsFactors=FALSE to avoid making the first column a factor.

Transpose result of combn in R

I would simply use t() to transpose it

c2 <- t(combn(mylist, 2))

How to transpose a list of data frames?

Here is an approach using lapply().

# generate some data frames
set.seed(102134)
id <- 1:5
aList <- lapply(id,function(x){
     data.frame(matrix(runif(50),nrow=10,ncol=5))
})
# transpose and remove first 2 rows
transposeList <- lapply(aList,function(x){
     t(x)[-c(1,2),]
})
# print first transposed data frame
transposeList[[1]]

..and the output, noting that rows representing variables X1 and X2 in the original data frame have been omitted from the data frame we print:

> # print first transposed data frame
> transposeList[[1]]
        [,1]      [,2]      [,3]      [,4]      [,5]       [,6]       [,7]
X3 0.1128006 0.5884873 0.8532827 0.5957727 0.6995990 0.09765447 0.69149804
X4 0.1239681 0.7624771 0.9756067 0.1251610 0.4954070 0.92652298 0.04800376
X5 0.3698154 0.6789413 0.9660355 0.6613972 0.5099627 0.97766102 0.97139575
        [,8]       [,9]     [,10]
X3 0.7731442 0.03916568 0.8787288
X4 0.3829343 0.41939016 0.9668663
X5 0.8865407 0.63437436 0.6774895
> 

An important subtlety in this answer is the line of code t(x)[-c(1,2),] in the anonymous function within lapply(). Since the result of t(x) is an object, we can immediately use the [ form of the extract operator to remove the first two rows.

Diff function transposes matrix when used with apply over rows

To elaborate on the "transpose effect" of apply:

According to ?apply, apply applies a function to the row vectors (MARGIN = 1) or column vectors (MARGIN = 2) of an array (e.g. a matrix) and returns

an array of dimension ‘c(n, dim(X)[MARGIN])’ if ‘n > 1’

where n is the length of the vector returned by an individual call of the function to either the row or column vector.

So in your case dim(m) is 2 10 (i.e. a 2x10 matrix) and MARGIN = 1, so the array dimension of the return object is 9 2, which means a 9x2 matrix (as diff returns a vector of length n=9).

You can see the same "transpose effect" when you do

apply(m, 1, c)
#      [,1] [,2]
# [1,]    1    2
# [2,]    3    4
# [3,]    5    6
# [4,]    7    8
# [5,]    9   10
# [6,]   11   12
# [7,]   13   14
# [8,]   15   16
# [9,]   17   18
#[10,]   19   20

R transposing numeric data.frame results in character variables

After transposing it, convert the columns to numeric with type.convert

out <- as.data.frame(t(sampleDF), stringsAsFactors = FALSE)
out[] <- lapply(out, type.convert, as.is = TRUE)
row.names(out) <- NULL
out
#   V1 V2 V3     V4
#1  1  2  3 String
#2  1  2  3      4
#3  5  6  7      8

str(out)
#'data.frame':  3 obs. of  4 variables: 
#  $ V1: int  1 1 5
#  $ V2: int  2 2 6
#  $ V3: int  3 3 7    
#  $ V4: chr  "String" "4" "8"

---

Or rbind the first column converted to respective 'types' with the transposed other columns

rbind(lapply(sampleDF[,1], type.convert, as.is = TRUE), 
               as.data.frame(t(sampleDF[2:3])))

NOTE: The first method would be more efficient

---

Or another approach would be to paste the values together in each column and then read it again

read.table(text=paste(sapply(sampleDF, paste, collapse=" "), 
     collapse="\n"), header = FALSE, stringsAsFactors = FALSE)
#  V1 V2 V3     V4
#1  1  2  3 String
#2  1  2  3      4
#3  5  6  7      8

---

Or we can convert the 'data.frame' to 'data.matrix' which changes the character elements to NA, use the is.na to find the index of elements that are NA for replacing with the original string values

m1 <- data.matrix(sampleDF)
out <- as.data.frame(t(m1))
out[is.na(out)] <- sampleDF[is.na(m1)]

---

Or another option is type_convert from readr

library(dplyr)
library(readr)
sampleDF %>% 
     t %>%
     as_data_frame %>%
     type_convert
# A tibble: 3 x 4
#     V1    V2    V3 V4    
#   <int> <int> <int> <chr> 
#1     1     2     3 String
#2     1     2     3 4     
#3     5     6     7 8     

---

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