When I write the following:
tempx <- tempx (-1, c(-4, -2:-20))
I get the following error message:
Could not find function "tempx"
But when I write the following the code runs properly:
tempx <- tempx [-1, c(-4, -2:-20)]
Please let me know the diff between () & [].
() is used to call a function. [] is used for subsetting vectors, arrays and matrices (and other such objects).
I'd suggest if you haven't already, reading an introduction to R, also available by typing help.start() into R itself. In particular, you might like to look at sections 2.1 Vectors and assignment and 5.2 Array indexing. Subsections of an array.
Related
In sagemath, I would like to plot the following function foo (Coef is an array that is big enough) :
def foo(x):
x_approx = floor (x*4)
return Coef[x_approx]
I wanted to use the command plot(foo(x), (x,0,0.1)).
But I got the error unable to convert floor(4*x) to an integer.
Whereas when `foo is not using an array, it works:
def foo(x):
x_approx = floor (x*4)
return 4*x_approx
Use plot(foo, (x, 0, 0.1)) instead (that is, replace foo(x) with foo). If you use foo(x), then Sage tries to evaluate foo(x) first, in which case it treats x as a symbolic variable and can't turn it into a number to plot. If you use foo, then it knows to treat it as a plottable/callable function, and it does the right thing.
Edit: I think the issue is that for plotting, Sage requires a certain type of function, a symbolic function, and using a Python construct like Coef[...] doesn't fit into that framework.
So, I'm brushing up on how to work with data frames in R and I came across this little bit of code from https://cloud.r-project.org/web/packages/data.table/vignettes/datatable-intro.html:
input <- if (file.exists("flights14.csv")) {
"flights14.csv"
} else {
"https://raw.githubusercontent.com/Rdatatable/data.table/master/vignettes/flights14.csv"
}
Apparently, this assigns the strings (character vectors?) in the if and else statements to input based on the conditional. How is this working? It seems like magic. I am hoping to find somewhere in the official R documentation that explains this.
From other languages I would have just done:
if (file.exists("flights14.csv")) {
input <- "flights14.csv"
} else {
input <- "https://raw.githubusercontent.com/Rdatatable/data.table/master/vignettes/flights14.csv"
}
or in R there is ifelse which also seems designed to do exactly this, but somehow that first example also works. I can memorize that this works but I'm wondering if I'm missing the opportunity to understand the bigger picture about how R works.
From the documentation on the ?Control help page under "Value"
if returns the value of the expression evaluated, or NULL invisibly if none was (which may happen if there is no else).
So the if statement is kind of like a function that returns a value. The value that's returned is the result of either evaulating the if or the then block. When you have a block in R (code between {}), the brackets are also like a function that just return the value of the last expression evaluated in the block. And a string literal is a valid expression that returns itself
So these are the same
x <- "hello"
x <- {"hello"}
x <- {"dropped"; "hello"}
x <- if(TRUE) {"hello"}
x <- if(TRUE) {"dropped"; "hello"}
x <- if(TRUE) {"hello"} else {"dropped"}
And you only really need blocks {} with if/else statements when you have more than one expression to run or when spanning multiple lines. So you could also do
x <- if(TRUE) "hello" else "dropped"
x <- if(FALSE) "dropped" else "hello"
These all store "hello" in x
You are not really missing anything about the "big picture" in R. The R if function is atypical compared both to other languages as well as to R's typical behavior. Unlike most functions in R which do require assignment of their output to a "symbol", i.e a proper R name, if allows assignments that occur within its consequent or alternative code blocks to occur within the global environment. Most functions would return only the final evaluation, while anything else that occurred inside the function body would be garbage collected.
The other common atypical function is for. R for-loops only
retain these interior assignments and always return NULL. The R Language Definition calls these atypical R functions "control structures". See section 3.3. On my machine (and I suspect most Linux boxes) that document is installed at: http://127.0.0.1:10731/help/doc/manual/R-lang.html#Control-structures. If you are on another OS then there is probably a pulldown Help menu in your IDE that will have a pointer to it. Thew help document calls them "control flow constructs" and the help page is at ?Control. Note that it is necessary to quote these terms when you wnat to access that help page using one of those names since they are "reserved words". So you would need ?'if' rather than typing ?if. The other reserved words are described in the ?Reserved page.
?Control
?'if' ; ?'for'
?Reserved
# When you just type:
?if # and hit <return>
# you will see a "+"-sign which indicateds an incomplete expression.
# you nthen need to hit <escape> to get back to a regular R interaction.
In R, functions don't need explicit return. If not specified the last line of the function is automatically returned. Consider this example :
a <- 5
b <- 1
result <- if(a == 5) {
a <- a + 1
b <- b + 1
a
} else {b}
result
#[1] 6
The last line in if block was saved in result. Similarly, in your case the string values are "returned" implicitly.
I'm trying to do something like this:
-module(count).
-export([main/0]).
sum(X, Sum) -> X + Sum.
main() ->
lists:foldl(sum, 0, [1,2,3,4,5]).
but see a warning and code fails:
function sum/2 is unused
How to fix the code?
NB: this is just a sample which illustrates problem, so there is no reason to propose solution which uses fun-expression.
Erlang has slightly more explicit syntax for that:
-module(count).
-export([main/0]).
sum(X, Sum) -> X + Sum.
main() ->
lists:foldl(fun sum/2, 0, [1,2,3,4,5]).
See also "Learn you some Erlang":
If function names are written without a parameter list then those names are interpreted as atoms, and atoms can not be functions, so the call fails.
...
This is why a new notation has to be added to the language in order to let you pass functions from outside a module. This is what fun Module:Function/Arity is: it tells the VM to use that specific function, and then bind it to a variable.
We have a UI for users to put down LaTex responses. To evaluate those responses, we translate LaTex to R, then run some script similar to the following to compare the response to the correct answer.
x = #{random number}
y = #{random number}
abs(x*x*y - x^2*y) < 10^-6
It's working fine for those functions mapped well. For example:
\frac{x}{y} => x/y
\left|x\right| => abs(x)
{\cdot} => *
But I didn't find any match for the following functions/notations:
\hat{x} -- unit vector
\vec{x} -- vector
\sqrt[3]{x} -- nth root
Any suggestions?
PS:
There is a hat function in R, but I'm not sure if it's the same as the one in LaTex.
Also, \sqrt[3]{x} can be replaced by x^(1/3). Will this be always working? Any performance concern? Someone's saying x^(1/2) is slower than sqrt(x).
The manual states:
The operator ‘<-’ can be used anywhere,
whereas the operator ‘=’ is only allowed at the top level (e.g.,
in the complete expression typed at the command prompt) or as one
of the subexpressions in a braced list of expressions.
The question here mention the difference when used in the function call. But in the function definition, it seems to work normally:
a = function ()
{
b = 2
x <- 3
y <<- 4
}
a()
# (b and x are undefined here)
So why the manual mentions that the operator ‘=’ is only allowed at the top level??
There is nothing about it in the language definition (there is no = operator listed, what a shame!)
The text you quote says at the top level OR in a braced list of subexpressions. You are using it in a braced list of subexpressions. Which is allowed.
You have to go to great lengths to find an expression which is neither toplevel nor within braces. Here is one. You sometimes want to wrap an assignment inside a try block: try( x <- f() ) is fine, but try( x = f(x) ) is not -- you need to either change the assignment operator or add braces.
Expressions not at the top level include usage in control structures like if. For example, the following programming error is illegal.
> if(x = 0) 1 else x
Error: syntax error
As mentioned here: https://stackoverflow.com/a/4831793/210673
Also see http://developer.r-project.org/equalAssign.html
Other than some examples such as system.time as others have shown where <- and = have different results, the main difference is more philisophical. Larry Wall, the creater of Perl, said something along the lines of "similar things should look similar, different things should look different", I have found it interesting in different languages to see what things are considered "similar" and which are considered "different". Now for R assignment let's compare 2 commands:
myfun( a <- 1:10 )
myfun( a = 1:10 )
Some would argue that in both cases we are assigning 1:10 to a so what we are doing is similar.
The other argument is that in the first call we are assigning to a variable a that is in the same environment from which myfun is being called and in the second call we are assigning to a variable a that is in the environment created when the function is called and is local to the function and those two a variables are different.
So which to use depends on whether you consider the assignments "similar" or "different".
Personally, I prefer <-, but I don't think it is worth fighting a holy war over.