R, about "rm" function - r

Let
x = 3
If I type rm(x) in the R console, there is no variable in the workspace.
Since x is a numeric type of data, I thought rm(3) would also work.
But R pops up an error message. What's wrong with the statement rm(3) ?

In R, just because x = 3 does not mean that 3 = x. You're defining x to be equal to the value of three, but 3 can also be the value of other things as well. It wouldn't make any sense to remove any usage of the number 3. x is a string; 3 i a number. Make sense?

Related

How to suppress warnings from stats:::regularize.values?

In newer versions of R (I have 3.6 and previously had 3.2), the stats::regularize.values function has been changed to have a default value of warn.collapsing as TRUE. This function is used in splinefun and several other interpolation functions in R. In a microsimulation model, I am using splinefun to smooth a large amount (n > 100,000) of data points of the form (x, f(x)). Here, x is a simulated vector of positive-valued scalers, and f(x) is some function of (x). With an n that large, there are often some replications of pseudo-randomly generated values (i.e., not all values of x are unique). My understanding is that splinefun gets rid of ties in the x values. That is not a problem for me, but, because of the new default, I get a warning message printed each time (below)
"In regularize.values(x, y, ties, missing(ties)) : collapsing to
unique 'x' values"
Is there a way to either change the default of the warn.collapsing argument of the stats::regularize.values function back to F? Or can I somehow suppress that particular warning? This matters because it's embedded in a long microsimulation code and when I update it I often run into bugs. So I can't just ignore warning messages.
I tried using the formalize function. I was able to get the default arguments of stats::regularize.values printed, but when I tried to assign new values using the alist function it said there is no object 'stats'.
I had this problem too, and fixed it by adding ties=min to the argument list of splinefun().
The value of missing(ties) is now passed as warn.collapsing to regularize.values().
https://svn.r-project.org/R/trunk/src/library/stats/R/splinefun.R
https://svn.r-project.org/R/trunk/src/library/stats/R/approx.R
Also see:
https://cran.r-project.org/doc/manuals/r-release/NEWS.html
and search for regularize.values().
Referencing this article
Wrap your call of regularize.values like this:
withCallingHandlers(regularize.values(x), warning = function(w){
if (grepl("collapsing to unique 'x' values", w$message))
invokeRestart("muffleWarning")
})
Working example (adapted from the above link to call a function):
f1 <- function(){
x <- 1:10
x + 1:3
}
f1()
# if we just call f1() we get a warning
Warning in x + 1:3 :
longer object length is not a multiple of shorter object length
[1] 2 4 6 5 7 9 8 10 12 11
withCallingHandlers(f1(), warning=function(w){invokeRestart("muffleWarning")})
[1] 2 4 6 5 7 9 8 10 12 11

R: Produce Index Values to Group Increasing Values in Vector

I have a list of increasing year values that occasionally has breaks in it and I want to create a grouping value for each unbroken sequence. Think of a vector like this one (missing 2005,2011):
x <- c(2001,2002,2003,2004,2006,2007,2008,2009,2010,2013,2014,2015,2016)
I would like to produce an equal length vector that numbers every value in a run with the same index to end up with something like this.
[1] 1 1 1 1 2 2 2 2 2 3 3 3 3
I would like to do this using best R practices so I am trying to avoid falling back to a for loop but I am not sure how to get from Vector A to Vector B. Does anyone have any suggestions?
Some things I know I can do:
I can flag the record before or after a gap as true with an ifelse
I can get the index of when the counter should change by wrapping that in a which statement
This is the code to do each
ifelse(!is.na(lag(x)) & x == lag(x)+1, FALSE, TRUE)
which(ifelse(!is.na(lag(x)) & x == lag(x)+1, FALSE, TRUE))
I think there a couple solutions to this problem. One as d.b posted in the comment above that will produce a sequence that increments every time there is a break in the sequence.
cummax(c(1, diff(x)))
There is a similar solution that I chose to use with ifelse() flagging breaks and cumsum(). I chose this solution because additional information,like other vectors, can be included in the decision and diff seems to have problems with very erratic up and down values.
cumsum(ifelse(!is.na(lag(x)) & x == lag(x) + 1, FALSE, TRUE))

R commands for finding mode in R seem to be wrong

I watched video on YouTube re finding mode in R from list of numerics. When I enter commands they do not work. R does not even give an error message. The vector is
X <- c(1,2,2,2,3,4,5,6,7,8,9)
Then instructor says use
temp <- table(as.vector(x))
to basically sort all unique values in list. R should give me from this command 1,2,3,4,5,6,7,8,9 but nothing happens except when the instructor does it this list is given. Then he says to use command,
names(temp)[temp--max(temp)]
which basically should give me this: 1,3,1,1,1,1,1,1,1 where 3 shows that the mode is 2 because it is repeated 3 times in list. I would like to stay with these commands as far as is possible as the instructor explains them in detail. Am I doing a typo or something?
You're kind of confused.
X <- c(1,2,2,2,3,4,5,6,7,8,9) ## define vector
temp <- table(as.vector(X))
to basically sort all unique values in list.
That's not exactly what this command does (sort(unique(X)) would give a sorted vector of the unique values; note that in R, lists and vectors are different kinds of objects, it's best not to use the words interchangeably). What table() does is to count the number of instances of each unique value (in sorted order); also, as.vector() is redundant.
R should give me from this command 1,2,3,4,5,6,7,8,9 but nothing happens except when the instructor does it this list is given.
If you assign results to a variable, R doesn't print anything. If you want to see the value of a variable, type the variable's name by itself:
temp
you should see
1 2 3 4 5 6 7 8 9
1 3 1 1 1 1 1 1 1
the first row is the labels (unique values), the second is the counts.
Then he says to use command, names(temp)[temp--max(temp)] which basically should give me this: 1,3,1,1,1,1,1,1,1 where 3 shows that the mode is 2 because it is repeated 3 times in list.
No. You already have the sequence of counts stored in temp. You should have typed
names(temp)[temp==max(temp)]
(note =, not -) which should print
[1] "2"
i.e., this is the mode. The logic here is that temp==max(temp) gives you a logical vector (a vector of TRUE and FALSE values) that's only TRUE for the elements of temp that are equal to the maximum value; names(temp)[temp==max(temp)] selects the elements of the names vector (the first row shown in the printout of temp above) that correspond to TRUE values ...

R in simple terms - why do I have to feel like such an idiot? [closed]

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my question is simple... every reference I find in books and on the internet for learning R programming is presented in a very linear way with no context. When I try and learn things like functions, I see the code and my brain just freezes because it's looking for something to relate these R terms to and I have no frame of reference. I have a PhD and did a lot of statistics for my dissertation but that was years ago when we were using different programming languages and when it comes to R, I don't know why I can't get this into my head. Is there someone who can explain in plain english an example of this simple code? So for example:
above <- function(x, n){
use <- x > n
x[use]
}
x <- 1:20
above(x, 12)
## [1] 13 14 15 16 17 18 19 20
I'm trying to understand what's going on in this code but simply don't. As a result, I could never just write this code on my own because I don't have the language in my head that explains what is happening with this. I get stuck at the first line:
above <- function(x, n) {
Can someone just explain this code sample in plain English so I have some kind of context for understanding what I'm looking at and why I'm doing what I'm doing in this code? And what I mean by plain English is, walking through the code, step by step and not just repeating the official terms from R like vector and function and array and all these other things, but telling me, in a common sense way, what this means.
Since your background ( phd in statsitics) the best way to understand this
is in mathematics words.
Mathematically speaking , you are defining a parametric function named above that extracts all element from a vector x above a certain value n. You are just filtering the set or the vector x.
In sets notation you can write something like :
above:{x,n} --> {y in x ; y>n}
Now, Going through the code and paraphrasing it (in the left the Math side , in the right its equivalent in R):
Math R
---------------- ---------------------
above: (x,n) <---> above <- function(x, n)
{y in x ; y>n} <---> x[x > n]
So to wrap all the statments together within a function you should respect a syntax :
function_name <- function(arg1,arg2) { statements}
Applying the above to this example (we have one statement here) :
above <- function(x,n) { x[x>n]}
Finally calling this function is exactly the same thing as calling a mathematical function.
above(x,2)
ok I will try, if this is too detailed let me know, but I tried to go really slowly:
above <- function(x, n)
this defines a function, which is just some procedure which produces some output given some input, the <- means assign what is on the right hand side to what is on the left hand side, or in other words put everything on the right into the object on the left, so for example container <- 1 puts 1 into the container, in this case we put a function inside the object above,
function(x, n) everything in the paranthesis specifys what inputs the function takes, so this one takes two variables x and n,
now we come to the body of the function which defines what it does with the inputs x and n, the body of the function is everything inside the curley braces:
{
use <- x > n
x[use]
}
so let's explain that piece by piece:
use <- x > n
this part again puts whats on the right side into the object on the left, and what is happening on the right hand side? a comparison returning TRUE if x is bigger than n and FALSE if x is equal to or smaller then n, so if x is 5 and n is 3 the result will be TRUE, and this value will get stored inside use, so use contains TRUE now, now if we have more than one value inside x than every value inside x will get compared to n, so for example if x = [1, 2, 3] and n = 2
than we have
1 > 2 FALSE
2 > 2 FALSE
3 > 2 TRUE
, so use will contain FALSE, FALSE, TRUE
x[use]
now we are taking a part of x, the square brackets specify which parts of x we want, so in my example case x has 3 elements and use has 3 elements if we combine them we have:
x use
1 FALSE
2 FALSE
3 TRUE
so now we say I dont want 1,2 but i want 3 and the result is 3
so now we have defined the function, now we call it, or in normal words we use it:
x <- 1:20
above(x, 12)
first we assign the numbers 1 through 20 to x, and then we tell the function above to execute (do everything inside its curley braces with the inputs x = 1:20 and n = 12, so in other words we do the following:
above(x, 12)
execute the function above with the inputs x = 1:20 and n = 12
use <- 1:20 > 12
compare 12 to every number from 1:20 and return for each comparison TRUE if the number is in fact bigger than 12 and FALSE if otherwise, than store all the results inside use
x[use]
now give me the corresponding elements of x for which the vector use contains TRUE
so:
x use
1 FALSE
2 FALSE
3 FALSE
4 FALSE
5 FALSE
6 FALSE
7 FALSE
8 FALSE
9 FALSE
10 FALSE
11 FALSE
12 FALSE
13 TRUE
14 TRUE
15 TRUE
16 TRUE
17 TRUE
18 TRUE
19 TRUE
20 TRUE
so we get the numbers 13:20 back as a result
I'll give it a crack too. A few basic points that should get you going in the right direction.
1) The idea of a function. Basically, a function is reusable code. Say I know that in my analysis for some bizarre reason I will often want to add two numbers, multiply them by a third, and divide them by a fourth. (Just suspend disbelief here.) So one way I could do that would just be to write the operation over and over, as follows:
(75 + 93)*4/18
(847 + 3)*3.1415/2.7182
(999 + 380302)*-6901834529/2.5
But that's tedious and error-prone. (What happens if I forget a parenthesis?) Alternatively, I can just define a function that takes whatever numbers I feed into it and carries out the operation. In R:
stupidMath <- function(a, b, c, d){
result <- (a + b)*c/d
}
That code says "I'd like to store this series of commands and attach them to the name "stupidMath." That's called defining a function, and when you define a function, the series of commands is just stored in memory---it doesn't actually do anything until you "call" it. "Calling" it is just ordering it to run, and when you do so, you give it "arguments" ---the stuff in the parentheses in the first line are the arguments it expects, i.e., in my example, it wants four distinct pieces of data, which will be called 'a', 'b', 'c', and 'd'.
Then it'll do the things it's supposed to do with whatever you give it. "The things it's supposed to do" is the stuff in the curly brackets {} --- that's the "body" of the function, which describes what to do with the arguments you give it. So now, whenever you want to carry that mathematical operation you can just "call" the function. To do the first computation, for example, you'd just write stupidMath(75, 93, 4, 18) Then the function gets executed, treating 75 as 'a', 83 as 'b', and so forth.
In your example, the function is named "above" and it takes two arguments, denoted 'x' and 'n'.
2) The "assignment operator": R is unique among major programming languages in using <- -- that's equivalent to = in most other languages, i.e., it says "the name on the left has the value on the right." Conceptually, it's just like how a variable in algebra works.
3) so the "body" of the function (the stuff in the curly brackets) first assigns the name "use" to the expression x > n. What's going on there. Well, an expression is something that the computer evaluates to get data. So remember that when you call the function, you give it values for x and n. The first thing this function does is figures out whether x is greater than n or less than n. If it's greater than n, it evaluates the expression x > n as TRUE. Otherwise, FALSE.
So if you were to define the function in your example and then call it with above(10, 5), then the first line of the body would set the local variable (don't worry right now about what a 'local' variable is) 'use' to be 'TRUE'. This is a boolean value.
Then the next line of the function is a "filter." Filtering is a long topic in R, but basically, R things of everything as a "vector," that is, a bunch of pieces of data in a row. A vector in R can be like a vector in linear algebra, i.e., (1, 2, 3, 4, 5, 99) is a vector, but it can also be of stuff other than numbers. For now let's just focus on numbers.
The wacky thing about R (one of the many wacky things about R) is that it treats a single number (a "scalar" in linear algebra terms) just as a vector with only one item in it.
Ok, so why did I just go into that? Because in lots of places in R, a vector and a scalar are interchangable.
So in your example code, instead of giving a scalar for the first argument, when we call the function we've given 'above' a vector for its first argument. R likes vectors. R really likes vectors. (Just talk to R people for a while. They're all obsessed with doing every goddmamn thing in terms of a vector.) So it's no problem to pass a vector for the first argument. But what that means is that the variable 'use' is going to be a vector too. Specifically, 'use' is going to be a vector of booleans, i.e., of TRUE or FALSE for each individual value of X.
To take a simpler version: suppose you said:
mynums <- c(5, 10)
myresult <- above(mynums, 7)
when the code runs, the first thing it's going to do is define that 'use' variable. But x is a vector now, not a scalar (the c(5,10) code said "make a vector with two elements, and fill them with the numbers '5' and '10'), so R's going to go ahead and carry out the comparison for each element of x. Since 5 is less than 7 and 10 is greater than 7, use becomes the two item-vector of boolean values (FALSE, TRUE)
Ok, now we can talk about filtering. So a vector of boolean values is called a 'logical vector.' And the code x[use] says "filter x by the stuff in the variable use." When you tell R to filter something by a logical vector, it spits back out the elements of the thing being filtered which correspond to the values of 'TRUE'
So in the example just given:
mynums <- c(5, 10)
myresult <- above(mynums, 7)
the value of myresult will just be 10. Why? Because the function filtered 'x' by the logical vector 'use,' 'x' was (5, 10), and 'use' was (FALSE, TRUE); since the second element of the logical was the only true, you only got the second element of x.
And that gets assigned to the variable myresult because myresult <- above(mynums, 7) means "assign the name myresult to the value of above(mynums, 7)"
voila.

Dynamic variables in base R

How to create please dependent variables in R ?
For example
a <- 1
b <- a*2
a <- 2
b
# [1] 2
But I expect the result 4. How can R maintained relations automatically ?
Thank you very much
Explanation - I'm trying to create something as excel spreeadsheet with the relationships (formula or functions) between cells. Input for R is for examle csv (same values, some function or formula) and output only values
It sounds like you're looking for makeActiveBinding
a <- 1
makeActiveBinding('b', function() a * 2, .GlobalEnv)
b
# [1] 2
a <- 2
b
# [1] 4
The syntax is simpler if you want to use Hadley's nifty pryr package:
library(pryr)
b %<a-% (a * 2)
Most people don't expect variables to behave like this, however. So if you're writing code that others will be reading, I don't recommend using this feature of R. Explicitly update b when a changes or make b a function of a.
Warning: This isn't a good idea and task callbacks really should only be used if you know what you're doing.
You can do something like this but it's tedious and there are better ways to achieve your goal. You can make a function that will be called after every top level evaluation that basically does the reassignment for you.
modified <- function(expr, value, ok, visible){
if(exists("a")){
assign("b", a*2, env = .GlobalEnv)
}
return(TRUE)
}
addTaskCallback(modified)
After running that you should be able to get this...
> a
Error: object 'a' not found
> b
Error: object 'b' not found
> a <- 2
> a
[1] 2
> b
[1] 4
> a <- 3
> a
[1] 3
> b
[1] 6
Note that if you want to emulate a spreadsheet it would probably just be better to define a function to take your input and do all the necessary calculations to get your desired output. R isn't Excel and it would be best if you don't treat it like Excel.
R doesn't work like that. Variables only change when assigned new values. This is a good thing, because it means things don't change magically. Suppose in 20 lines time you want to know the value of b? When did it change? What does it depend on?
R is not a spreadsheet.
Just to spell it out a bit more.
sales = 100
costs = 90
profit = sales - costs
now profit has the value 10.
sales = 120
Only sales has changed.
profit = sales - costs
That changes profits to 30.
If you have a complex calculation you would normally write a function:
computeProfit = function(sales, costs){return(sales - costs)}
and then do:
profit = computeProfit(sales, costs)
whenever you want to compute the profits from the sales and the costs.
Although what you want to do is not completely possible in R, with a simple modification of b into a function and thanks to lexical scoping, you actually can have a "dependent variable" (sort of).
Define a:
a <- 1
Define b like this:
b <- function() {
a*2
}
Then, instead of using b to get the value of b, use b()
b() ##gives 2
a <- 4
b() ##gives 8

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