names

Is there a reason why R won't allow me to have a number as the column name of my dataframe? Also noticed that if i do data.frame(XX) it adds an X to all the column headers that have numbers at the front. Shane Yes, because R won't allow names of objects to start with numbers. If you were to call attach() with the data.frame, this would cause some issues. data.frame (and read.table ) function has the check.names parameter (default is TRUE ) If TRUE then the names of the variables in the data frame are checked to ensure that they are syntactically valid variable names and are not duplicated. If

Related to this I want to know what exactly is a nested name specifier? I looked up in the draft but I could understand the grammar as I haven't taken any Compiler Design classes yet. void S(){} struct S{ S(){cout << 1;} void f(){} static const int x = 0; }; int main(){ struct S *p = new struct ::S; p->::S::f(); S::x; ::S(); // Is ::S a nested name specifier? delete p; } ::S is a qualified-id . In the qualified-id ::S::f , S:: is a nested-name-specifier . In informal terms 1 , a nested-name-specifier is the part of the id that begins either at the very beginning of a qualified-id or after the

I need to access list names inside the lapply function. I've found some threads online where it's said I should iterate through the names of the list to be able to fetch each list element name in my function: > n = names(mylist) > mynewlist = lapply(n, function(nameindex, mylist) { return(mylist[[nameindex]]) }, mylist) > names(mynewlist) NULL > names(mynewlist) = n The problem is that mynewlist loses the original mylist indexes and I have to add that last names() assignment to restore them. Is there a way to give an explicit index name to each element returned by the lapply function? Or a

Let's say I have a data.frame, like so: x <- c(1:10,1:10,1:10,1:10,1:10,1:10,1:10,1:10,1:10,1:10) df <- data.frame("Label 1"=x,"Label 2"=rnorm(100)) head(df,3) returns: Label.1 Label.2 1 1 1.9825458 2 2 -0.4515584 3 3 0.6397516 How do I get R to stop automagically replacing the space with a period in the column name? ie, "Label 1" instead of "Label.1". You don't. With the space you desire the format would not satisfy the requirements for an identifier that come to play when you use df$column.1 -- that could not cope with a space. So see the make.names() function for details or an example: >

Now following my series of "python newbie questions" and based on another question . Prerogative Go to http://python.net/~goodger/projects/pycon/2007/idiomatic/handout.html#other-languages-have-variables and scroll down to "Default Parameter Values". There you can find the following: def bad_append(new_item, a_list=[]): a_list.append(new_item) return a_list def good_append(new_item, a_list=None): if a_list is None: a_list = [] a_list.append(new_item) return a_list There's even an "Important warning" on python.org with this very same example, tho not really saying it's "better". One way to put

There are the standard A-Z, a-z characters, but also there are hyphens, em dashes, quotes, etc. Plus, there are all of the international characters, like umlauts, etc. So, for an English-based system, what's the complete set? What about sets for other languages? What about UTF8, UTF16, etc? Bonus question: How many name fields are needed, and what are their maximum lengths? EDIT: There are definitely two different types of characters involved in people's names, those that are there as part of the context, and those that are there for structural reasons. I don't want to limit or interfere with