I have data in a csv file in a single column with 6954 values. I want to split this column into multiple columns such that each column has 122 data and next column has the next 122 data and so on. I guess, I will have a final matrix of 122 rows and 57 columns. Any help will be appreciated.
Thanks
Like this ?
x <- rep(1:122, 5)
xx <- matrix(x, nrow=122)
xx[1:5, ]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 1 1
[2,] 2 2 2 2 2
[3,] 3 3 3 3 3
[4,] 4 4 4 4 4
[5,] 5 5 5 5 5
Or this will do the trick as well:
x = 1:6954
dim(x) <- c(122, 57)
A column can be split using the colsplit function which is part of the reshape package
http://r.ramganalytics.com/r/split-a-column-by-a-character-using-colsplit-function/
Related
I'm a complete R novice, and I'm really struggling on this problem. I need to take a vector, evens, and subtract it from the first column of a matrix, top_secret. I tried to call up only that column using top_secret[,1] and subtract the vector from that, but then it only returns the column. Is there a way to do this inside the matrix so that I can continue to manipulate the matrix without creating a bunch of separate columns?
Sure, you can. Here is an example:
m <- matrix(c(1,2,3,4),4,4, byrow = TRUE)
> m
[,1] [,2] [,3] [,4]
[1,] 1 2 3 4
[2,] 1 2 3 4
[3,] 1 2 3 4
[4,] 1 2 3 4
m[,4] <- m[,4] - c(5,5,5,5)
which gives:
> m
[,1] [,2] [,3] [,4]
[1,] 1 2 3 -1
[2,] 1 2 3 -1
[3,] 1 2 3 -1
[4,] 1 2 3 -1
Or another option is replace
replace(m, cbind(seq_len(nrow(m)), 4), m[,4] - 5)
data
m <- matrix(c(1,2,3,4),4,4, byrow = TRUE)
Given the sequence 1 2 3 4, I would like to generate a matrix of pairs
1 2
2 3
3 4
to use as indexes for another matrix. What would be the fastest way to achieve this?
You could use embed(), reversing the columns on the output.
embed(1:4, 2)[, 2:1]
# [,1] [,2]
# [1,] 1 2
# [2,] 2 3
# [3,] 3 4
I have this code and can't understand how rbind.fill.matrix is used.
dtmat is a matrix with the documents on rows and words on columns.
word <- do.call(rbind.fill.matrix,lapply(1:ncol(dtmat), function(i) {
t(rep(1:length(dtmat[,i]), dtmat[,i]))
}))
I read the description of the function and says that binds matrices but cannot understand which ones and fills with NA missing columns.
From what I understand, the function replaces columns that dont bind with NA.
Lets say I have 2 matrices A with two columns col1 and col2, B with three columns col1, col2 and colA. Since I want to bind all both these matrices, but rbind only binds matrices with equal number of columns and same column names, rbind.fill.matrix binds the columns but adds NA to all values that should be in both the matrices that are not. The code below will explain it more clearly.
a <- matrix(c(1,1,2,2), nrow = 2, byrow = T)
> a
[,1] [,2]
[1,] 1 1
[2,] 2 2
>
> b <- matrix(c(1,1,1,2,2,2,3,3,3), nrow = 3, byrow = T)
> b
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
>
> library(plyr)
> r <- rbind.fill.matrix(a,b)
> r
1 2 3
[1,] 1 1 NA
[2,] 2 2 NA
[3,] 1 1 1
[4,] 2 2 2
[5,] 3 3 3
>
>
The documentation also mentions about column names, which I think you can also understand from the example.
I have a matrix
[,1] [,2]
[1,] 2 3
[2,] 3 5
[3,] 7 9
[4,] 11 3
[5,] 11 8
and I want to merge row 1 2 4 5 by their common value.
the result should be output
2 3 5 11 8
Test case:
m <- matrix(c(2,3,7,11,11,3,5,9,3,8),ncol=2)
I'm not sure this is what you want, but it gives the right answer:
unique(c(t(m[c(1,2,4,5),])))
Only two tricky bits here:
need to use c() to collapse the matrix into a single vector
need to use t() to get the matrix collapsed row-wise rather than column-wise to get the ordering as you specified.
I am trying to cut one row
x = [1 2 3 4 5 6 7 8 9 10 11 12]
into multiple rows of equal length so that
y(row1) = [1 2 3 4
y(row2) = 5 6 7 8
y(row3) = 9 10 11 12]
I know I can achieve this using a combination of rbind and cbind, but the dataset I am trying to apply this to is much larger than the example, so I am looking for a way to do it more quickly and automatically. I tried cut and cut2 but those didnt work either
jelle
The function matrix() is your friend here:
> matrix(1:12, nrow = 3, byrow = TRUE)
[,1] [,2] [,3] [,4]
[1,] 1 2 3 4
[2,] 5 6 7 8
[3,] 9 10 11 12
Note the optional parameter, byrow. The default is FALSE and will fill the matrix by columns, setting it to true in this case gets the data arranged in the order that you described.Just something to be careful about, since R won't throw an error if you fill by column, but your data won't be in the right format!
Use matrix:
> y <- 1:12
> y
[1] 1 2 3 4 5 6 7 8 9 10 11 12
> matrix(y,3,4,byrow=1)
[,1] [,2] [,3] [,4]
[1,] 1 2 3 4
[2,] 5 6 7 8
[3,] 9 10 11 12
Edit: I included the byrow=TRUE argument to matrix (pointed out by Chase in the comments) which fills the matrix along the rows instead of down the columns.