Apply Over Matrix by Column - Any Way to Get Column Name

apply over matrix by column - any way to get column name?

One common way to deal with this is to loop over the column names themselves:

m <- matrix(runif(16),4,4)
colnames(m) <- letters[1:4]
sapply(colnames(m),function(x){mean(m[,x])})
        a         b         c         d 
0.4720319 0.4992337 0.5026318 0.3514267

Extract matrix column values by matrix column name

Yes. But place your "test" after the comma if you want the column...

> A <- matrix(sample(1:12,12,T),ncol=4)

> rownames(A) <- letters[1:3]

> colnames(A) <- letters[11:14]
> A[,"l"]
 a  b  c 
 6 10  1 

see also help(Extract)

Extract column names for each value in column in matrix in R

We may use paste in a vectorized way by replicating the column names with col index and then get the unique after pasteing with the values of the n.mat (it is a data.frame after as.data.frame - so we used unlist)

unique(paste(colnames(n.mat)[col(n.mat)], unlist(n.mat), sep="_"))

-output

[1] "V1_M" "V1_F" "V2_A" "V2_B" "V3_T" "V3_G" "V3_H" "V4_D"
[9] "V4_G" "V5_T" "V5_X" "V6_Y" "V6_J"

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Or if need a loop, use Map

unlist(Map(function(x, y) unique(paste(y, x, sep = "_")), 
     unname(n.mat), names(n.mat)))

The apply wouldn't work here as the names for each column will be the row.names attribute (even if change the OP's code to names)

Get column names from a matrix using a condition

If it is really a matrix and not a data.frame then:

 colnames(your_matrix)

Gives the column names - a character vector.

Since price is in the first row

your_matrix[1, ] > 20000

Gives a logical vector indicating the columns where the price is greater than 20000.

To get "column names from the cars that their price is bigger than 20000", just put those pieces together, using the logical test you want to subset the column names vector.

colnames(your_matrix)[your_matrix[1, ] > 20000]

Colnames passed through an apply function R

Thanks to the commentors I arrived at the below which give me my desired result.

trialData <- data.frame('a' = rnorm(100),
                        'b' = rnorm(100),
                        'c' = rnorm(100))

someData <- function(dataInput){
  # lots of code here
  return(
    dataName = colnames(dataInput)
  )
}

dataOutput <- lapply(colnames(trialData), function(x){someData(trialData[x])})

print(dataOutput)

How do I retrieve a matrix column and row name by a matrix index value?

First you need to get the row and column of that index using arrayInd.

k <- arrayInd(4, dim(mdat))

You can then get the right name by getting that element of the row and column names

rownames(mdat)[k[,1]]
colnames(mdat)[k[,2]]

Or both at once using mapply:

mapply(`[[`, dimnames(mdat), k)

Returning Column Name from Matrix based on value in R

1

Run sapply over columns and find min and then check if the min meets your condition.

colnames(DF)[sapply(DF, min) <= 2]
#[1] "V3"

2

You can also run apply on columns (MARGIN = 2) to see if any value in each column meets the required condition

colnames(DF)[apply(DF, MARGIN = 2, function(a) any(a<=2))]
#[1] "V3"

3

Use arr.ind = TRUE with which. This will give the indices of the values that meet the condition of which. From that, extract the column information [,2].

unique(colnames(DF)[which(DF<=2, arr.ind = TRUE)[,2]])
#[1] "V3"

DATA

set.seed(42)
DF <- matrix(sample(1:9,9),ncol=3,nrow=3)
DF <- as.data.frame.matrix(DF)
DF
#  V1 V2 V3
#1  9  5  6
#2  8  4  1
#3  3  7  2

Keep column name when select one column from a data frame/matrix in R

Use the drop argument:

> x <- matrix(1,3,3)
> colnames(x) <- c("test1","test2","test3")
> x[,1, drop = FALSE]
     test1
[1,]     1
[2,]     1
[3,]     1

Print column name in an r apply and save as new column on a dataframe

I think a lot of people will look for this same issue, so I'm answering my own question (having eventually found the answers). As below, there are other answers to both parts (thanks!) but non-combining these issues (and some of the examples are more complex).

First, it seems the "colnames" element really isn't something you can get around (seems weird to me!), so you 'loop' over the column names, and within the function call the actual vectors by name [c(x)].

Then the key thing is that to assign, so create your new columns, within an apply, you use '<<'

apply(colnames(df[c("a","b","c")]),function(x) {
z <- (ChISEQCIS[c(paste0(x))]/ChISEQCIS[c("V1")])
ChISEQCIS[c(paste0(x,"ind"))] <<- z
}
)

The << is discussed e.g. https://stackoverflow.com/questions/2628621/how-do-you-use-scoping-assignment-in-r

I got confused because I only vaguely thought about wanting to save the outputs initially and I figured I needed both the column (I incorrectly assumed apply worked like a loop so I could use a counter as an index or something) and that there should be same way to get the name separately (e.g. colname(x)).

There are a couple of related stack questions:

• https://stackoverflow.com/questions/9624866/access-to-column-name-of-dataframe-with-apply-function
• https://stackoverflow.com/questions/21512041/printing-a-column-name-inside-lapply-function
• https://stackoverflow.com/questions/10956873/how-to-print-the-name-of-current-row-when-using-apply-in-r
• https://stackoverflow.com/questions/7681013/apply-over-matrix-by-column-any-way-to-get-column-name (easiest to understand)

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