Get the Vector of Values from Different Columns of a Matrix

Get the vector of values from different columns of a matrix

This should do it for you:

M.mat[cbind(seq_along(Idx),Idx)]

Get different columns of each row from matrix

Since you want row-major order (and Matlab works in column major order), transpose A first. Then build a logical mask to be used as index into A transposed:

At = A.'; %'
mask = (bsxfun(@ge, (1:size(At,1)), B(:,1)) & bsxfun(@le, 1:size(At,1), B(:,2))).'; %'
result = At(mask);

How to select elements of a matrix given a vector with same number of columns?

We can construct a matrix of row/column indexes to extract. The column sequence can be seq_len(ncol

mat1[cbind(vec1, seq_len(ncol(mat1)))]
[1] 13  6 19  4

Or another option is to extract the rows and use diag

diag(mat1[vec1,])
[1] 13  6 19  4

Using row-wise column indices in a vector to extract values from data frame

Just use matrix indexing, like this:

dframe[cbind(seq_along(i), i)]
# [1] "g" "b" "f"

The cbind(seq_along(i), i) part creates a two column matrix of the relevant row and column that you want to extract.

Values from multiple dataframe columns into one vector

I've found that converting to a matrix first makes getting to levels a bit easier.

as.vector(as.matrix(df[,c("alpha", "gamma", "zeta")]))

Of course, you could have just done stringsAsFactors=FALSE when you read the data in initially.

Subset a matrix according to a columns vector

Try

H[cbind(seq_len(nrow(H)), P)]
## [1] 0.6733731 0.7396847 0.5953580

Here we are indexing by consecutive rows and columns indicated in P

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Regarding your question, so the reason H[, P] returns a matrix is because you are telling R:

select all rows in columns: 2, 1, 2 from matrix "H"

thus the result that you are getting is a matrix with identical first and third columns.

In R, split vector valued column in data frame into multiple columns

In this case, with a matrix, you can use cbind:

cbind(results[1], results[[2]])
#   y         1         2         3
# 1 0 0.4710224 0.4280053 0.3206661
# 2 1 0.5769064 0.6220120 0.2683387

Using [ on the first argument makes sure it is a data.frame, so cbind.data.frame is used and the result is not coerced to matrix. But using [[ on the second argument makes sure that we are binding to the 2x3 matrix in the second column rather than a data frame containing that matrix.

More generally, tidyr::unnest works well with vector or list columns, but the result is in long form - you would have to add a column id and spread it to get it back to wide form.

However, the simplest way may be to use dplyr and summarize_at with a custom .funs argument:

library(dplyr)
descrip = funs(m =  mean, med = median, sd = sd)
Df %>% group_by(y) %>% summarize_at("x", .funs = descrip)
# # A tibble: 2 × 4
#       y         m       med        sd
#   <dbl>     <dbl>     <dbl>     <dbl>
# 1     0 0.4710224 0.4280053 0.3206661
# 2     1 0.5769064 0.6220120 0.2683387

How to compare two columns of matrix in r and output the column name in a new vector

You need the vectorized version of if/else. The matrix version:

ifelse(data[,2] > data[,1], "second", "first")

Vectorizing a for loop that changes columns of a matrix

An alternative to the t(apply in @Jon Spring's answer is matrixStats::rowCumsums.

library(matrixStats)

n <- 1e4L
n10 <- n/10L
age.mat <- outer(sample(150, n, TRUE), seq(0, 20, 5), "+")
x.mat <- matrix(FALSE, n, 5) # create the empty harvest matrix
# sample harvests so that no tree is harvested twice
x.mat[matrix(c(sample(n, n/2L), sample(n10:(6L*n10 - 1L)) %/% n10), n/2L)] <- TRUE

f1 <- function(age, x) {
  age[x[,1],] <- 0
  for (i in 2:5){ # we don't need to calculate over the first year
    age[,i] <- age[,i - 1] + 5L # add 5 to previous year
    age[x[,i], i] <- 0L # reset age of harvested trees to zero
  }
  age
}

f2 <- function(age, x) {
  age - rowCumsums(x*age)
}

microbenchmark::microbenchmark(f1 = f1(age.mat, x.mat),
                               f2 = f2(age.mat, x.mat),
                               check = "equal")
#> Unit: microseconds
#>  expr   min    lq     mean median     uq     max neval
#>    f1 294.4 530.2 1023.450  566.6 629.35 33222.8   100
#>    f2 135.2 263.6  334.622  284.2 307.15  4343.6   100

Tidyverse approach to selecting values from different columns in a dataframe with row-column index pairs

You could do

rain %>%
  mutate(id = row_number()) %>%
  pivot_longer(-id) %>%
  inner_join(tibble(id = seq_along(v), name = v)) 

which returns

# A tibble: 5 x 3
     id name  value
  <int> <chr> <dbl>
1     1 els       1
2     2 ceres     1
3     3 els       1
4     4 mond      0
5     5 ceres     3

Adding

pull(value, name)

returns the named vector

#>  els ceres   els  mond ceres 
#>    1     1     1     0     3 

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