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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.
Related
Can you help me understand how R interprets square brackets with forms such as y[i:j - k]?
dummy data:
y <- c(1, 2, 3, 5, 7, 8)
Here's what I do understand:
y[i] is the ith element of vector y.
y[i:j] is the ith to jth element (inclusive) of vector y.
y[-i] is vector y without the first i elements. etc. etc.
However, what I don't understand is what happens when you start mixing these options, and I haven't found a good resource for explaining it.
For example:
y[1-1:4]
[1] 5 7 8
So y[1-1:4] returns the vector without the first three elements. But why?
and
y[1-4]
[1] 1 2 5 7 8
So y[1-4] returns the vector without the third element. Is that because 1-4 = -3 and it's interpretting it the same as y[-3]? If so, that doesn't seem consistent with my previous example where y[1-1:4] would presumably be interpretted as y[0:4], but that isn't the case.
and
y[1:1+2-1]
[1] 2
Why does this return the second element? I encountered this while I was trying to code something along the lines of: y[i:i + j - k] and it took me a while to figure out that I should write y[i:(i + j - k)] so the parenthesis captured the whole of the right-hand-side of the colon. But I still can't figure out what logic R was doing when I didn't have those brackets.
Thanks!
It's best to look closer at precedence and the integer sequences you use for subsetting. These are evaluated before subsetting with []. Note that - is a function with two arguments (1, 1:4) which are evaluated beforehand and so
> 1-1:4
[1] 0 -1 -2 -3
Negative indices in [] mean exclusion of the corresponding elements. There is no "0" element (and so subsetting at 0 returns an empty vector of the present type -- numeric(0)). We thus expect y[1-1:4] to drop the first three elements in y and return the remainder.
As you write correctly y[1-4] is y[-3], i.e. omission of the third element.
Similar as above, in 1:1+2-1, 1:1 evaluates to a one-element vector 1, the rest is simple arithmetic.
For more on operator precedence, see Hadley's excellent book.
I am trying to write a recursive function that will return true if second number is power of first number.
For example:
find_power 3 9 will return true
find_power 2 9 will return false because the power of 2 is 8 not 9
This is what I have tried but I need a recursive solution
let rec find_power first second =
if (second mod first = 0)
return true
else
false ;;
A recursive function has the following rough form
let rec myfun a b =
if answer is obvious then
obvious_answer
else
let (a', b') = smaller_example_of_same_problem a b in
myfun a' b'
In your case, I'd say the answer is obvious if the second number is not a multiple of the first or if it's 1. That is essentially all your code is doing now, it's testing the obvious part. (Except you're not handling the 0th power, i.e., 1.)
So, you need to figure out how to make a smaller example of the same problem. You know (by hypothesis) that the second number is a multiple of the first one. And you know that x * a is a power of a if and only if x is a power of a. Since x is smaller than x * a, this is a smaller example of the same problem.
This approach doesn't work particularly well in some edge cases, like when the first number is 1 (since x is not smaller than x * 1). You can probably handle them separately.
I usually have no problem with vectorization in r, but I am having a tough time in the example below where there are both iterative and non-iterative components in the for loop.
In the code below, I have a calculation that I have to perform based on a set of constants (Dini), a vector of values (Xs), where the ith value of the output vector (Ys) is also dependent on i-1 value:
Dini=128 #constant
Xs=c(6.015, 5.996, 5.989, 5.911, 5.851, 5.851, 5.858, 5.851)
Y0=125.73251 #starting Y value
Ys=c(Y0) #starting of output vector, first value is known
for (Vi in Xs[2:length(Xs)]){
ytm1=Ys[length(Ys)]
y=(955.74301-2*((Dini+ytm1-Vi)^2-ytm1^2)^0.5+2*ytm1*acos(ytm1/(Dini+ytm1-Vi)))/pi/2
Ys=c(Ys, y)
}
df=data.frame(Xs, Ys)
df
Xs Ys
1 6.015 125.7325
2 5.996 125.7273
3 5.989 125.7251
4 5.911 125.7036
5 5.851 125.6859
6 5.851 125.6849
7 5.858 125.6868
8 5.851 125.6850
For this case, where there is a mix of both iterative and non iterative components in the for loop, my mind has got twisted in a non-vectorized knot.
Any suggestions?
You might want to look into use Reduce in this case. For example
Ys<-Reduce(function(prev, cur) {
(955.74301-2*((Dini+prev-cur)^2-prev^2)^0.5 + 2*prev*acos(prev/(Dini+prev-cur)))/pi/2
}, Xs, init=Y0, accumulate=T)[-1]
From the ?Reduce help page: "Reduce uses a binary function to successively combine the elements of a given vector and a possibly given initial value." This makes it easier to create vectors where a given value depends on a previous value.
I have a for loop to do a matrix manipulation in R. For some checks are true i need to come to the same row again., means i need to be reduced by 1.
for(i in 1:10)
{
if(some chk)
{
i=i-1
}
}
Actually i is not reduced for me. For an example in 5th row i'm reducing the i to 4, so again it should come as 5, but it is coming as 6.
Please advice.
My intention is:
Checking the first column values of a matrix, if I find any duplicate value, I take the second column value and append with the first row's second column and remove the duplicate row. So, when I'm removing a row I do not need increase the i in while loop. (This is just a map reduce method, append values of same key)
Variables in R for loops are read-only, you cannot modify them. What you have written would be solved completely differently in normal R code – the exact solution depending on the actual problem, there isn’t a generic, direct replacement (except by replacing the whole thing with a while loop but this is both ugly and probably unnecessary).
To illustrate this, consider these two typical examples.
Assume you want to filter all duplicated elements from a list. Instead of looping over the list and copying all duplicated elements, you can use the duplicated function which tells you, for each element, whether it’s a duplicate.
Secondly, you use standard R subsetting syntax to select just those elements which are not a duplicate:
x = x[! duplicated(x)]
(This example works on a one-dimensional vector or list, but it can be generalised to more dimensions.)
For a more complex case, let’s say that you have a vector of numbers and, for every even number in the vector, you want to double the preceding number (this is highly artificial but in signal processing you might face similar problems). In other words:
input = c(1, 3, 2, 5, 6, 7, 1, 8)
output = ???
output
# [1] 1 6 2 10 6 7 2 8
… we want to fill in ???. In the first step, we check which numbers are even:
even = input %% 2 == 0
# [1] FALSE FALSE TRUE FALSE TRUE FALSE FALSE TRUE
Next, we shift the result down – because we want to know whether the next number is even – by removing the first element, and appending a dummy element (FALSE) at the end.
even = c(even[-1], FALSE)
# [1] FALSE TRUE FALSE TRUE FALSE FALSE TRUE FALSE
And now we can multiply just these inputs by two:
output = input
output[even] = output[even] * 2
There, done.
This question already has an answer here:
Closed 10 years ago.
Possible Duplicate:
How to generate a vector containing a numeric sequence?
In R, how can I get the list of numbers from 1 to 100? Other languages have a function 'range' to do this. R's range does something else entirely.
> range(1,100)
[1] 1 100
Your mistake is looking for range, which gives you the range of a vector, for example:
range(c(10, -5, 100))
gives
-5 100
Instead, look at the : operator to give sequences (with a step size of one):
1:100
or you can use the seq function to have a bit more control. For example,
##Step size of 2
seq(1, 100, by=2)
or
##length.out: desired length of the sequence
seq(1, 100, length.out=5)
If you need the construct for a quick example to play with, use the : operator.
But if you are creating a vector/range of numbers dynamically, then use seq() instead.
Let's say you are creating the vector/range of numbers from a to b with a:b, and you expect it to be an increasing series. Then, if b is evaluated to be less than a, you will get a decreasing sequence but you will never be notified about it, and your program will continue to execute with the wrong kind of input.
In this case, if you use seq(), you can set the sign of the by argument to match the direction of your sequence, and an error will be raised if they do not match. For example,
seq(a, b, -1)
will raise an error for a=2, b=6, because the coder expected a decreasing sequence.