This looks like a solid, reproducible bug.  It&#39;s been filed as [Bug #2791](https://r-forge.r-project.org/tracker/index.php?func=detail&amp;aid=2791&amp;group_id=240&amp;atid=975). 

It appears that repeating the column attempts to delete the subsequent columns.   
If no columns remain, then R crashes. 

---

**UPDATE** : Now fixed in v1.8.11. From [NEWS][1] :

&gt; Assigning to the same column twice in the same query is now an error rather than a crash in some circumstances; e.g., DT[,c(&quot;B&quot;,&quot;B&quot;):=NULL] (delete by reference the same column twice). Thanks to Ricardo ([#2751][2]) and matt_k ([#2791][3]) for reporting. Tests added.


  [1]: https://r-forge.r-project.org/scm/viewvc.php/pkg/NEWS?view=markup&amp;root=datatable
  [2]: https://r-forge.r-project.org/tracker/?func=detail&amp;group_id=240&amp;aid=2751&amp;atid=975
  [3]: https://r-forge.r-project.org/tracker/index.php?func=detail&amp;aid=2791&amp;group_id=240&amp;atid=975

This Q has been answered but regard this as a side note.

I prefer the following syntax to drop multiple columns

    DT[ ,`:=`(x = NULL, y = NULL)]

because it matches the one to add multiple columns (variables)

    DT[ ,`:=`(x = letters, y = &quot;Male&quot;)]

This also check for duplicated column names. So trying to drop `x` twice will throw an error message.

I&#39;m using the support library v4 and my questions are, How to know if a Fragment is Visible? and How can I change the propierties of the Layout inflated in the Fragment? Thanks in advance.

---Edit---

I&#39;m using fragments like in the android developers tutorial with a FragmentActivity

How to know if a Fragment is Visible?

From reading the docs, it looks like you have to (or should) assign a model to a route like so:

    App.PostRoute = Ember.Route.extend({
        model: function() {
            return App.Post.find();
        }
    });

What if I need to use several objects in a certain route? i.e. Posts, Comments and Users? How do I tell the route to load those?

How to use multiple models with a single route in EmberJS / Ember Data?

What&#39;s the correct way to remove multiple columns from a data.table?  I&#39;m currently using the code below, but was getting unexpected behavior when I accidentally repeated one of the column names.  I wasn&#39;t sure if this was a bug, or if I shouldn&#39;t be removing columns this way.
   
    library(data.table)
    DT &lt;- data.table(x = letters, y = letters, z = letters)
    DT[ ,c(&quot;x&quot;,&quot;y&quot;) := NULL]
    names(DT)
    [1] &quot;z&quot;

The above works fine, but

    DT &lt;- data.table(x = letters, y = letters, z = letters)
    DT[ ,c(&quot;x&quot;,&quot;x&quot;) := NULL]
    names(DT)
    [1] &quot;z&quot;

Remove multiple columns from data.table

What's the correct way to remove multiple columns from a data.table? I'm currently using the code below, but was getting unexpected behavior when I accidentally repeated one of the column names. I wasn't sure if this was a bug, or if I shouldn't be removing columns this way.
<pre><code class="hljs language-scss">library(data.table)
DT &#x3C;- data.table(x = letters, y = letters, z = letters)
DT[ ,c("x","y") := NULL]
names(DT)
[1] "z"
</code></pre>
The above works fine, but
<pre><code class="hljs language-css">DT &#x3C;- data.table(x = letters, y = letters, z = letters)
DT[ ,c("x","x") := NULL]
names(DT)
[1] "z"
</code></pre>

If I have a data frame

    set.seed(12345) 
    df=data.frame(a=rnorm(5),b=rnorm(5))

I can add a row by e.g.

`df[6,] =c(5,6)`

If I now do the equivalent in data.table

    library(data.table)
    dt=data.table(df)
    dt[6,]=c(5,6)

It fails with an error. What is the right way to insert a row into a data.table?


Insert a row in a data.table

If I call a function using `parSapply` then `print`, `message` or `cat` statements inside that function don&#39;t seem to output to the console.

My process takes a very long time, so I need some way of seeing the progress and getting the results output as they are done.  Are there any special commands that would allow me to print to the console from a parallel process?

Example:

    library(parallel)

    oneloop = function(x) {
      for(i in 1:50) {
        a = rnorm(100000)
        a = sort(a)
      }
      print(x)
      message(x)
      cat(x)
    }
    
    cl &lt;- makeCluster(5)
    output = parSapply(cl, 1:10, oneloop)
    stopCluster(cl)

How can I `print` or `cat` when using parallel

I am new to R but have turned to it to solve a problem with a large data set I am trying to process. Currently I have a 4 columns of data (Y values) set against minute-interval timestamps (month/day/year hour:min) (X values) as below:

        timestamp          tr            tt         sr         st  
    1   9/1/01 0:00   1.018269e+02   -312.8622   -1959.393   4959.828  
    2   9/1/01 0:01   1.023567e+02   -313.0002   -1957.755   4958.935  
    3   9/1/01 0:02   1.018857e+02   -313.9406   -1956.799   4959.938  
    4   9/1/01 0:03   1.025463e+02   -310.9261   -1957.347   4961.095  
    5   9/1/01 0:04   1.010228e+02   -311.5469   -1957.786   4959.078

The problem I have is that some timestamp values are missing - e.g. there may be a gap between 9/1/01 0:13 and 9/1/01 0:27 and such gaps are irregular through the data set. I need to put several of these series into the same database and because the missing values are different for each series, the dates do not currently align on each row.

I would like to generate rows for these missing timestamps and fill the Y columns with blank values (no data, not zero), so that I have a continuous time series.

I&#39;m honestly not quite sure where to start (not really used R before so learning as I go along!) but any help would be much appreciated. I have thus far installed chron and zoo, since it seems they might be useful.

Thanks!

Insert rows for missing dates/times

I have a named character vector returned from xmlAttrs like this:

    testVect &lt;- structure(c(&quot;11.2.0.3.0&quot;, &quot;12.89&quot;, &quot;12.71&quot;), .Names = c(&quot;db_version&quot;, 
                 &quot;elapsed_time&quot;, &quot;cpu_time&quot;))

I would like to convert it to a data frame that looks like this:

    testDF &lt;- data.frame(&quot;db_version&quot;=&quot;11.2.0.3.0&quot;,&quot;elapsed_time&quot;=12.89,&quot;cpu_time&quot;=12.71)
    head(testDF)
      db_version elapsed_time cpu_time
    1 11.2.0.3.0        12.89    12.71


Convert Named Character Vector to data.frame

I don&#39;t understand why I got this warning message.

    &gt; fixed &lt;- data.frame(&quot;Type&quot; = character(3), &quot;Amount&quot; = numeric(3))
    &gt; fixed[1, ] &lt;- c(&quot;lunch&quot;, 100)
    Warning message:
    In `[&lt;-.factor`(`*tmp*`, iseq, value = &quot;lunch&quot;) :
      invalid factor level, NA generated
    &gt; fixed
      Type Amount
    1 &lt;NA&gt;    100
    2           0
    3           0

Warning message: In `...` : invalid factor level, NA generated

I have a data.table with which I&#39;d like to perform the same operation on certain columns.  The names of these columns are given in a character vector. In this particular example, I&#39;d like to multiply all of these columns by -1.

Some toy data and a vector specifying relevant columns: 

    library(data.table)
    dt &lt;- data.table(a = 1:3, b = 1:3, d = 1:3)
    cols &lt;- c(&quot;a&quot;, &quot;b&quot;)
  

Right now I&#39;m doing it this way, looping over the character vector:

    for (col in 1:length(cols)) {
       dt[ , eval(parse(text = paste0(cols[col], &quot;:=-1*&quot;, cols[col])))]
    }
    
Is there a way to do this directly without the for loop?


Apply a function to every specified column in a data.table and update by reference

Apparently in my last question I demonstrated confusion between `data.frame` and `data.table`.  Admittedly, I didn&#39;t realize there was a distinction.  

So I read the help for each but in practical, everyday terms, what is the difference, what are the implications and what are each used for that would help guide me to their appropriate usage?

What is the practical difference between data.frame and data.table in R

I have a script that reads in data from a CSV file into a `data.table` and then splits the text in one column into several new columns. I am currently using the `lapply` and `strsplit` functions to do this. Here&#39;s an example:

    library(&quot;data.table&quot;)
    df = data.table(PREFIX = c(&quot;A_B&quot;,&quot;A_C&quot;,&quot;A_D&quot;,&quot;B_A&quot;,&quot;B_C&quot;,&quot;B_D&quot;),
                    VALUE  = 1:6)
    dt = as.data.table(df)

    # split PREFIX into new columns
    dt$PX = as.character(lapply(strsplit(as.character(dt$PREFIX), split=&quot;_&quot;), &quot;[&quot;, 1))
    dt$PY = as.character(lapply(strsplit(as.character(dt$PREFIX), split=&quot;_&quot;), &quot;[&quot;, 2))

    dt 
    #    PREFIX VALUE PX PY
    # 1:    A_B     1  A  B
    # 2:    A_C     2  A  C
    # 3:    A_D     3  A  D
    # 4:    B_A     4  B  A
    # 5:    B_C     5  B  C
    # 6:    B_D     6  B  D 

In the example above the column `PREFIX` is split into two new columns `PX` and `PY` on the &quot;_&quot; character.

Even though this works just fine, I was wondering if there is a better (more efficient) way to do this using `data.table`. My real datasets have  &gt;=10M+ rows, so time/memory efficiency becomes really important.

---

### UPDATE:

Following @Frank&#39;s suggestion I created a larger test case and used the suggested commands, but the `stringr::str_split_fixed` takes a lot longer than the original method.

    library(&quot;data.table&quot;)
    library(&quot;stringr&quot;)
    system.time ({
        df = data.table(PREFIX = rep(c(&quot;A_B&quot;,&quot;A_C&quot;,&quot;A_D&quot;,&quot;B_A&quot;,&quot;B_C&quot;,&quot;B_D&quot;), 1000000),
                        VALUE  = rep(1:6, 1000000))
        dt = data.table(df)
    })
    #   user  system elapsed 
    #  0.682   0.075   0.758 

    system.time({ dt[, c(&quot;PX&quot;,&quot;PY&quot;) := data.table(str_split_fixed(PREFIX,&quot;_&quot;,2))] })
    #    user  system elapsed 
    # 738.283   3.103 741.674 

    rm(dt)
    system.time ( {
        df = data.table(PREFIX = rep(c(&quot;A_B&quot;,&quot;A_C&quot;,&quot;A_D&quot;,&quot;B_A&quot;,&quot;B_C&quot;,&quot;B_D&quot;), 1000000),
                         VALUE = rep(1:6, 1000000) )
        dt = as.data.table(df)
    })
    #    user  system elapsed 
    #   0.123   0.000   0.123 

    # split PREFIX into new columns
    system.time ({
        dt$PX = as.character(lapply(strsplit(as.character(dt$PREFIX), split=&quot;_&quot;), &quot;[&quot;, 1))
        dt$PY = as.character(lapply(strsplit(as.character(dt$PREFIX), split=&quot;_&quot;), &quot;[&quot;, 2))
    })
    #    user  system elapsed 
    #  33.185   0.000  33.191 

So the `str_split_fixed` method takes about 20X times longer.



Split text string in a data.table columns

There are a lot of posts about replacing NA values. I am aware that one could replace NAs in the following table/frame with the following:

    x[is.na(x)]&lt;-0

But, what if I want to restrict it to only certain columns? Let&#39;s me show you an example.

First, let&#39;s start with a dataset.

    set.seed(1234)
    x &lt;- data.frame(a=sample(c(1,2,NA), 10, replace=T),
                    b=sample(c(1,2,NA), 10, replace=T), 
                    c=sample(c(1:5,NA), 10, replace=T))

Which gives:

        a  b  c
    1   1 NA  2
    2   2  2  2
    3   2  1  1
    4   2 NA  1
    5  NA  1  2
    6   2 NA  5
    7   1  1  4
    8   1  1 NA
    9   2  1  5
    10  2  1  1

Ok, so I only want to restrict the replacement to columns &#39;a&#39; and &#39;b&#39;. My attempt was:

    x[is.na(x), 1:2]&lt;-0

and:

    x[is.na(x[1:2])]&lt;-0

Which does not work.

My data.table attempt, where `y&lt;-data.table(x)`, was obviously never going to work:

    y[is.na(y[,list(a,b)]), ]

I want to pass columns inside the is.na argument but that obviously wouldn&#39;t work.

I would like to do this in a data.frame and a data.table. My end goal is to recode the 1:2 to 0:1 in &#39;a&#39; and &#39;b&#39; while keeping &#39;c&#39; the way it is, since it is not a logical variable. I have a bunch of columns so I don&#39;t want to do it one by one. And, I&#39;d just like to know how to do this.

Do you have any suggestions?

Content Type	Original Author	Original Content on Stackoverflow
Question	matt_k	View Question on Stackoverflow
Solution 1 - R	Ricardo Saporta	View Answer on Stackoverflow
Solution 2 - R	Pankil Shah	View Answer on Stackoverflow

Remove multiple columns from data.table

R Problem Overview

R Solutions

Solution 1 - R

Solution 2 - R

How to use multiple models with a single route in EmberJS / Ember Data?

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