cv.uk.df$new.d[2:nrow(cv.uk.df)] <- tail(cv.uk.df$deaths, -1) - head(cv.uk.df$deaths, -1) # this line of code works
I wanted to know why do we -1 in the tail and -1 in head to create this new column.
I made an effort to understand by removing the -1 and "R"(The code is in R studio) throws me this error.
Could anyone shed some light on this? I can't explain how much I would appreciate it.
Look at what is being done. On the left-hand side of the assignment operator, we have:
cv.uk.df$new.d[2:nrow(cv.uk.df)] <-
Let's pick this apart.
cv.uk.df # This is the data.frame
$new.d # a new column to assign or a column to reassign
[2:nrow(cv.uk.df)] # the rows which we are going to assign
Specifically, this line of code will assign a new value all rows of this column except the first. Why would we want to do that? We don't have your data, but from your example, it looks like you want to calculate the change from one line to the next. That calculation is invalid for the first row (no previous row).
Now let's look at the right-hand side.
<- tail(cv.uk.df$deaths, -1) - head(cv.uk.df$deaths, -1)
The cv.uk.df$deaths column has the same number of rows as the data.frame. R gets grouchy when the numbers of elements don't follow sum rules. For data.frames, the right-hand side needs to have the same number of elements, or a number that can be recycled a whole-number of times. For example, if you have 10 rows, you need to have a replacement of 10 values. Or you can have 5 values that R will recycle.
If your data.frame has 100 rows, only 99 are being replaced in this operation. You cannot feed 100 values into an operation that expects 99. We need to trim the data. Let's look at what is happening. The tail() function has the usage tail(x, n), where it returns the last n values of x. If n is a negative integer, tail() returns all values but the first n. The head() function works similarly.
tail(cv.uk.df$deaths, -1) # This returns all values but the first
head(cv.uk.df$deaths, -1) # This returns all values but the last
This makes sense for your calculation. You cannot subtract the number of deaths in the row before the first row from the number in the first row, nor can you subtract the number of deaths in the last row from the number in the row after the last row. There are more intuitive ways to do this thing using functions from other packages, but this gets the job done.
I am trying to count the length of occurrances of a value in a vector such as
q <- c(1,1,1,1,1,1,4,4,4,4,4,4,4,4,4,4,4,4,6,6,6,6,6,6,6,6,6,6,1,1,4,4,4)
Actual vectors are longer than this, and are time based. What I would like would be an output for 4 that tells me it occurred for 12 time steps (before the vector changes to 6) and then 3 time steps. (Not that it occurred 15 times total).
Currently my ideas to do this are pretty inefficient (a loop that looks element by element that I can have stop when it doesn't equal the value I specified). Can anyone recommend a more efficient method?
x <- with(rle(q), data.frame(values, lengths)) will pull the information that you want (courtesy of d.b. in the comments).
From the R Documentation: rle is used to "Compute the lengths and values of runs of equal values in a vector – or the reverse operation."
y <- x[x$values == 4, ] will subset the data frame to include only the value of interest (4). You can then see clearly that 4 ran for 12 times and then later for 3.
Modifying the code will let you check whatever value you want.
I'm having trouble with a for loop in a simple piece of code in R...
So I have a data frame with a bunch of precipitation data, with a bunch of "stations" (1-75) in rows and across in columns is values per month. I created a vector (which I called "Jan")of precipitation values for just the Jan column, so it looks something like this (I've put in random values for the purpose of this post) :
V1
1 10
2 5
3 15
...
75 5
I want to use a "for" loop with "if" to return the highest value of this vector. However, the code I have below seems to just go through each value in the vector and always just returns the last value of then entire thing (i.e. in row 75, would return "5"). I know it's something in my "if" loop that isn't letting me actually text the values of the vector, rather it's testing the "row number" value. Any advice??
highest_ppt<- function(v) {
i=0
output<-v[i] #c(0,length(v))
for (i in 2:length(v)){
if (v[i] > (v[i-1])){
output <- (v[i])
}
}
return(output)
}
max_ppt <- highest_ppt(Jan)
max_ppt
Thank you!
A for loop is the wrong approach here, but if you insist:
Max <- -Inf
for (i in seq_along(v)){
if (v[i] > Max)
Max <- v[i]
}
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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.
This is quite literally the first problem in Project Euler. I created these two algorithms to solve it, but they each yield different answers. Basically, the job is to write a program that sums all the products of 3 and 5 that are under 1000.
Here is the correct one:
divisors<-0
for (i in 1:999){
if ((i %% 3 == 0) || (i %% 5 == 0)){
divisors <- divisors+i
}
}
The answer it yields is 233168
Here is the wrong one:
divisors<-0
for (i in 1:999){
if (i %% 3 == 0){
divisors <- divisors + i
}
if (i %% 5 == 0){
divisors <- divisors + i
}
}
This gives the answer 266333
Can anyone tell me why these two give different answers? The first is correct, and obviously the simpler solution. But I want to know why the second one isn't correct.
EDIT: fudged the second answer on accident.
Because multiples of 15 will add i once in the first code sample and twice in the second code sample. Multiples of 15 are multiples of both 3 and 5.
To make them functionally identical, the second would have to be something like:
divisors<-0
for (i in 1:999) {
if (i %% 3 == 0) {
divisors <- divisors + i
} else {
if (i %% 5 == 0) {
divisors <- divisors + i
}
}
}
But, to be honest, your first sample seems far more logical to me.
As an aside (and moot now that you've edited it), I'm also guessing that your second output value of 26633 is a typo. Unless R wraps integers around at some point, I'd expect it to be more than the first example (such as the value 266333 which I get from a similar C program, so I'm assuming you accidentally left of a 3).
I don't know R very well, but right off the bat, I see a potential problem.
In your first code block, the if statement is true if either of the conditions are true. Your second block runs the if statement twice if both conditions are met.
Consider the number 15. In your first code block, the if statement will trigger once, but in the second, both if statements will trigger, which is probably not what you want.
I can tell you exactly why that's incorrect, conceptually.
Take the summation of all integers to 333 and multiply is by 3, you'll get x
Take the summation of all integers to 200 and multiply it by 5, you'll get y
Take the summation of all integers to 66 and multiply it by 15, you'll get z
x + y = 266333
x + y - z = 233168
15 is divisible by both 3 and 5. You've counted all multiples of 15 twice.