Asked  6 Months ago    Answers:  5   Viewed   28 times

There is a similar question for PHP, but I'm working with R and am unable to translate the solution to my problem.

I have this data frame with 10 rows and 50 columns, where some of the rows are absolutely identical. If I use unique on it, I get one row per - let's say - "type", but what I actually want is to get only those rows which only appear once. Does anyone know how I can achieve this?

I can have a look at clusters and heatmaps to sort it out manually, but I have bigger data frames than the one mentioned above (with up to 100 rows) where this gets a bit tricky.



This will extract the rows which appear only once (assuming your data frame is named df):

df[!(duplicated(df) | duplicated(df, fromLast = TRUE)), ]

How it works: The function duplicated tests whether a line appears at least for the second time starting at line one. If the argument fromLast = TRUE is used, the function starts at the last line.

Boths boolean results are combined with | (logical 'or') into a new vector which indicates all lines appearing more than once. The result of this is negated using ! thereby creating a boolean vector indicating lines appearing only once.

Tuesday, June 1, 2021
answered 6 Months ago

Try this code instead:

$(document).ready(function () { 
    var oTable = $('#example').dataTable({
        stateSave: true

    var allPages = oTable.fnGetNodes();

    $('body').on('click', '#selectAll', function () {
        if ($(this).hasClass('allChecked')) {
            $('input[type="checkbox"]', allPages).prop('checked', false);
        } else {
            $('input[type="checkbox"]', allPages).prop('checked', true);

The magic should happen in fnGetNodes():

fnGetNodes(): Get an array of the TR nodes that are used in the table's body


This alternative solution is mostly for debugging (to see if it works). Hardly optimal code:

$(document).ready(function () { 
    var oTable = $('#example').dataTable({
        stateSave: true

    var allPages = oTable.cells( ).nodes( );

    $('#selectAll').click(function () {
        if ($(this).hasClass('allChecked')) {
            $(allPages).find('input[type="checkbox"]').prop('checked', false);
        } else {
            $(allPages).find('input[type="checkbox"]').prop('checked', true);
Sunday, June 13, 2021
answered 6 Months ago

One option is to run the combn function in a lapply loop. You can define it as your own function

allCombs <- function(x) c(x, lapply(seq_along(x)[-1L], 
                             function(y) combn(x, y, paste0, collapse = "")),
                             recursive = TRUE)

## [1] "Yellow" "Blue" "Red" "YellowBlue" "YellowRed" "BlueRed" "YellowBlueRed"

Explanation (per request)

Basically OP wants a vector of all possible combinations (not permutations) depending on the length of the input vector. Thus, we should run the combn function over k <- 1:length(x) and generate all combinations for every k. So when k == 1, it's just the original vector, so we can skip that part and just concatenate the original vector using c. The rest of the generated combinations return a list of vectors with different lengths (in a descending order for obvious reasons), thus we need to use recursive = TRUE within the c function in order to mimic the behaviour of unlist and combine the results into a single vector.

Wednesday, July 7, 2021
answered 5 Months ago

First, find the count of each element in df$data_values:

 x <- sapply(df$data_values, function(x) sum(as.numeric(df$data_values == x)))

> x
 [1] 1 2 2 2 1 2 2 2 1 1

Now extract the rows:

> df[x==1,]
   data_subsets data_values
1             A           1
5             A           5
9             B           6
10            B           7

Note that you missed "A 5" above. There is no "B 5".

Saturday, August 28, 2021
answered 3 Months ago

You can use textConnection() to pass the character vector to read.table(). An example:

x  <- "first,secondnthird,fourthn"
x1 <- read.table(textConnection(x), sep = ",")
# x1
     V1     V2
1 first second
2 third fourth

Answer found in the R mailing list.

2017 EDIT

Seven years later, I'd probably do it like this:

read.table(text = x, sep = ",")
Thursday, October 7, 2021
answered 2 Months ago
Only authorized users can answer the question. Please sign in first, or register a free account.
Not the answer you're looking for? Browse other questions tagged :