Suppose a vector of integers like this:
vect<-c(2,3,4)
i want to change this vector to another vector in which all elements are multiplied by 2 and subtracted from 20 .and get result like this:
result<-c(16,14,12)
in Python, there is a simple method like this:
result=[20-(item*2) for item in vect]
i know it can be easily done with a loop but i was wondering if there is any similar short method as in python in R? its probably a very basic question but im new to R and could not find the answer anywhere else.
Simply
vect2 <- 20 - vect * 2
Related
Suppose I want to create a vector in Julia. The vector should be [0,0.5,1,1.5,2,....20].
What's the quick command to create a vector like this, from 0 to 20, increasing by 0.5?
This will create a range, which is a kind of vector:
0:0.5:20
This does the same thing
range(0, 20; step=0.5)
If you absolutely need to make a Vector, you can use collect on either of these, but in most cases you should not, and just use 0:0.5:20 directly.
There are many ways to do this. While I prefer the colon array syntax already mentioned, this could be used for more general construction:
[0.5*i for i in 1:20]
I have a basic question in regards to the R programming language.
I'm at a beginners level and I wish to understand the meaning behind two lines of code I found online in order to gain a better understanding. Here is the code:
as.data.frame(y[1:(n-k)])
as.data.frame(y[(k+1):n])
... where y and n are given. I do understand that the results are transformed into a data frame by the function as.data.frame() but what about the rest? I'm still at a beginners level so pardon me if this question is off-topic or irrelevant in this forum. Thank you in advance, I appreciate every answer :)
Looks like you understand the as.data.frame() function so let's look at what is happening inside of it. We're looking at y[1:(n-k)]. Here, y is a vector which is a collection of data points of the same type. For example:
> y <- c(1,2,3,4,5,6)
Try running that and then calling back y. What you get are those numbers listed out. Now, consider the case you want to just call out the number 1 in that vector. How would you do that? Well, this is where the brackets come into play. If you wanted to just call the number 1 in y:
> y[1]
[1] 1
Therefore, the brackets are a way of calling out or indexing specific items in the vector. Note that the indexing starts at the value 1 and goes up to the number of items in the vector, or length. One last thing before we go back to the example you gave. What if we want to index the numbers 1, 2, and 3 from the vector but not the rest?
> y[1:3]
[1] 1 2 3
This is where the colon comes into play. It allows us to reference a subset of the numbers. However, it will reference all the numbers between the index left of the colon and right of it. Try this out for yourself in R! Play around and see what happens.
Finally going back to your example:
y[1:(n-k)]
How would this work based on what we discussed? Well, the colon means that we are indexing all values in the vector y from two index values. What are those values? Well, they are the numbers to the left and right of the colon. Therefore, we are asking R to give us the values from the first position (index of 1) to the (n-k) position. Therefore, it's important to know what n and k are. If n is 4 and k is 1 then the command becomes:
y[1:3]
The same logic can apply to the second as.data.frame() command in your question. Essentially, R is picking out different numbers from a vector y and multiplying them together.
Hope this helps. The best way to learn R is to play around with a command, throw different numbers at it, guess what will happen, and then see what happens!
I would like to do in r something equivalent that I do in excel:
IF(B3-[#[lagged Date]]>2,1,0)
In fact, it is just the previous row. Writing pseudo language could be considered as follows:
If(x-x_i>y,1,0)
Any way to do it with an explanation, please?
I am struggling with the way to go back and iterate over a column and go through each row using r.
I have 75 matrices that I want to search through. The matrices are named a1r1, a1r2, a1r3, a1r4, a1r5, a2r1,...a15r5, and I have a list with all 75 of those names in it; each matrix has the same number of rows and columns. Inside some nested for loops, I also have a line of code that, for the first matrix looks like this:
total <- (a1r1[row,i]) + (a1r1[row,j]) + (a1r1[row,k])
(i, j, k, and row are all variables that I am looping over.) I would like to automate this line so that the for loops would fully execute using the first matrix in the list, then fully execute using the second matrix and so on. How can I do this?
(I'm an experienced programmer, but new to R, so I'm willing to be told I shouldn't use a list of the matrix names, etc. I realize too that there's probably a better way in R than for loops, but I was hoping for sort of quick and dirty at my current level of R expertise.)
Thanks in advance for the help.
Here The R way to do this :
lapply(ls(pattern='a[0-9]r[0-9]'),
function(nn) {
x <- get(nn)
sum(x[row,c(i,j,k)])
})
ls will give a list of variable having a certain pattern name
You loop through the resulted list using lapply
get will transform the name to a varaible
use multi indexing with the vectorized sum function
It's not bad practice to build automatically lists of names designating your objects. You can build such lists with paste, rep, and sequences as 0:10, etc. Once you have a list of object names (let's call it mylist), the get function applied on it gives the objects themselves.
I have been trying to produce a command in R that allows me to produce a new vector where each row is the sum of 25 rows from a previous vector.
I've tried making a function to do this, this allows me to produce a result for one data point.
I shall put where I haver got to; I realise this is probably a fairly basic question but it is one I have been struggling with... any help would be greatly appreciated;
example<-c(1;200)
fun.1<-function(x)
{sum(x[1:25])}
checklist<-sapply(check,FUN=fun.1)
This then supplies me with a vector of length 200 where all values are NA.
Can anybody help at all?
Your example is a bit noisy (e.g., c(1;200) has no meaning, probably you want 1:200 there, or, if you would like to have a list of lists then something like rep, there is no check variable, it should have been example, etc.).
Here's the code what I think you need probably (as far as I was able to understand it):
x <- rep(list(1:200), 5)
f <- function(y) {y[1:20]}
sapply(x, f)
Next time please be more specific, try out the code you post as an example before submitting a question.