I was prompted a question and am ever so close to solving what I need. The question is as follows-
"Write a while loop that computes and stores as a new object, the factorial of any non-negative integer mynum by decrementing mynum by 1 at each repetition of the braced code."
Another factor was that if 0 or 1 was entered, the output would be 1.
The code that I wrote as follows-
factorialcalc <- function(i){
factorial <- 1
if(i==0 | i==1){
factorial <- 1
} else{
while(i >= 1){
factorial <- factorial * i
i <- i-1
}
}
return (factorial)
}
with inputs-
mynum <- 5
factorialcalc(mynum)
and output-
[1] 120
You may be wondering, "your code works perfect, so what's the issue?"
My issue lies in the part of the question that says "computes AND stores."
How can I modify my code to put the answers of factorialcalc into a vector?
Example-
I input
mynum <- 5
factorialcalc(mynum)
and
mynum <- 3
factorialcalc(mynum)
and
mynum <- 4
factorialcalc(mynum)
When I call this new vector, I would like to see a vector with all three of their outputs
(so almost like I made a vector c(120,6,24))
I'm thinking there's a way to add this vector somewhere in my function or while loop, but I'm not sure where. Also, please note that the answer must contain a loop like in my code.
Option 1.
"Vectorize" your function
# simply wrap the whole thing in Vectorize()
Factorialcalc = Vectorize(function(i){
factorial <- 1
if(i==0 | i==1){
factorial <- 1
} else{
while(i >= 1){
factorial <- factorial * i
i <- i-1
}
}
return (factorial)
})
# Now when you supply it a vector, it runs on each element
> Factorialcalc(c(5, 3, 4))
[1] 120 6 24
Option 2.
Use functions that are designed to apply a single function to multiple elements of a supplied vector.
Using map_dbl from the purrr package, you can call:
map_dbl(c(5, 3, 4), factorialcalc)
Which supplies to your function factorialcalc each element in vector and concatenates each result before returning a vector.
Using base R you can simply use the apply-family functions:
sapply(c(5, 3, 4), factorialcalc)
and get the same result.
Example
> map_dbl(c(5, 3, 4), factorialcalc)
[1] 120 6 24
> sapply(c(5, 3, 4), factorialcalc)
[1] 120 6 24
My professor has assigned a question for programming in R and I am stuck. He wants us to make a function that will take the exponential (e^(x[i]) of all the numbers in a vector and then sum them. the equation is:
the summation of e^x(i), n, and i=1.
I have made a function that will give me the exponential of the first value in my vector. But I want to get the exponential of all the values and sum them. Here is my code
#Vector for summing
x=c(2,1,3,0.4)
#Code for function
mysum = 0
myfun=function(x){
for (i in 1:length(x)){
mysum = mysum + exp(x[i])
return(mysum)
}
}
myfun(x)
#returns 7.389056
I have also tried using i = 1:1 because the equation specifies i=1, even though I knew that would only go through 1 number, and it gave me the same answer.... obviously.
myfun=function(x){
for (i in 1:1)
Does anyone have any suggestions to get it to sum?
You need to set the initial value of mysum to the accumulation afterwards, and also move the line return(mysum) outsides your for loop to return the result, i.e.,
myfun=function(x){
mysum <- 0
for (i in 1:length(x)){
mysum = mysum + exp(x[i])
}
return(mysum)
}
or just
myfun=function(x){
mysum <- 0
for (i in x){
mysum = mysum + exp(x)
}
return(mysum)
}
Since exp operation is vectoroized, you can also define your function myfun like below
myfun <- function(x) sum(exp(x))
You could also use the fact that most base functions are already vectorized :
1) create a dummy vector
1:10
#> [1] 1 2 3 4 5 6 7 8 9 10
2) apply your function on that vector, you get vectorized result
exp(1:10)
#> [1] 2.718282 7.389056 20.085537 54.598150 148.413159
#> [6] 403.428793 1096.633158 2980.957987 8103.083928 22026.465795
3) Sum that vector
sum(exp(1:10))
#> [1] 34843.77
4) Write your function to gain (a little) time
my_fun <- function(x){sum(exp(x))}
my_fun(1:10)
#> [1] 34843.77
I have to create a function as: ans(x) which returns the value 2*abs(x), if x is
negative, and the value x otherwise. What command could i use?
Thanks
ans <- function(x){
ifelse(x < 0, 2*abs(x), x)
}
will do.
> ans(2)
[1] 2
> ans(-2)
[1] 4
Explanation:
We can use the built-in base R function ifelse(). The logic is pretty simple:
ifelse(condition, output if condition is TRUE, output if condition is FALSE)
Therefore, ifelse(x < 0, 2*abs(x), x) will do the following:
evaluate whether value x is negative (<0)
if TRUE, return 2*abs(x)
if FALSE, return x
The advantage of ifelse() over traditional if() is the vectorization. if() can only handle a single value, ifelse() will evaluate any vector given as input.
Comparison:
ans_if <- function(x){
if(x < 0){2*abs(x)}else{x}
}
This is the same function, using a traditional if() structure. Giving a single value as input will result in the same output for both functions:
> ans(-2)
[1] 4
> ans_if(-2)
[1] 4
But if you want to input multiple values, let's say
test <- c(-1, -2, 3, -4)
the ifelse() variant will evaluate every element of the vector and generate the correct output as a vector of the same length:
> ans(test)
[1] 2 4 3 8
whereas the if() variant will throw a warning
> ans_if(test)
[1] 2 4 6 8
Warning message:
In if (x < 0) { :
the condition has length > 1 and only the first element will be used
and return the wrong output, as only the first value was used for evaluation (-1) and the operation over the whole vector was based on this evaluation.
I am relatively new to R, and matrix-based scripting languages in general. I have written this function to return the index's of each row which has a content similar to any another row's content. It is a primitive form of spam reduction that I am developing.
if (!require("RecordLinkage")) install.packages("RecordLinkage")
library("RecordLinkage")
# Takes a column of strings, returns a list of index's
check_similarity <- function(x) {
threshold <- 0.8
values <- NULL
for(i in 1:length(x)) {
values <- c(values, which(jarowinkler(x[i], x[-i]) > threshold))
}
return(values)
}
is there a way that I could write this to avoid the for loop entirely?
We can simplify the code somewhat using sapply.
# some test data #
x = c('hello', 'hollow', 'cat', 'turtle', 'bottle', 'xxx')
# create an x by x matrix specifying which strings are alike
m = sapply(x, jarowinkler, x) > threshold
# set diagonal to FALSE: we're not interested in strings being identical to themselves
diag(m) = FALSE
# And find index positions of all strings that are similar to at least one other string
which(rowSums(m) > 0)
# [1] 1 2 4 5
I.e. this returns the index positions of 'hello', 'hollow', 'turtle', and 'bottle' as being similar to another string
If you prefer, you can use colSums instead of rowSums to get a named vector, but this could be messy if the strings are long:
which(colSums(m) > 0)
# hello hollow turtle bottle
# 1 2 4 5
function(q,b,Data1,Data2){
x<-sum(
ifelse(Data1[13+q,b]/Data1[12+q,b]>Data2[13+q,1]/Data2[12+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[11+q,b]>Data2[13+q,1]/Data2[11+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[10+q,b]>Data2[13+q,1]/Data2[10+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[9+q,b]>Data2[13+q,1]/Data2[9+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[8+q,b]>Data2[13+q,1]/Data2[8+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[7+q,b]>Data2[13+q,1]/Data2[7+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[6+q,b]>Data2[13+q,1]/Data2[6+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[5+q,b]>Data2[13+q,1]/Data2[5+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[4+q,b]>Data2[13+q,1]/Data2[4+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[3+q,b]>Data2[13+q,1]/Data2[3+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[2+q,b]>Data2[13+q,1]/Data2[2+q,1],1,0)+
ifelse(Data1[13+q,b]/Data1[1+q,b]>Data2[13+q,1]/Data2[1+q,1],1,0)
)/12
}
Is there a way to simplify this? (no characters, only numbers in the data sets)
Thank you
Two pieces of knowledge you can combine to improve your code:
Firstly, you can divide a single number by a vector and R will return a vector with elementwise divisions. For example:
5 / c(1,2,3,4,5,6)
# [1] 5.0000000 2.5000000 1.6666667 1.2500000 1.0000000 0.8333333
The numerator on both sides of the inequality are the same all the time, you can use the above. So instead of explicitly calling it for every inequality, you can just call it once.
Secondly, an expression with TRUE or FALSE will be coerced to 1 and 0 when you try to perform arithmetic operations (in your case division, or calculating a mean). Inequalities return TRUE or FALSE values. Explicitly telling R to convert them to 0 and 1 is wasted energy, because R will automatically do it in your last step.
Putting this together in a simplified function:
function(q, b, Data1, Data2){
qseq <- (1:12) + q # Replaces all "q+1", "q+2", ... , "q+12"
dat1 <- Data1[qseq, b] # Replaces all "Data1[q+1, b]", ... "Data1[q+12, b]"
dat2 <- Data2[qseq, 1] # Replaces all "Data2[q+1, 1]", ... "Data2[q+12, 1]"
mean( Data1[13+q, b]/dat1 > Data2[13+q, 1]/dat2 )
this simplify a bit:
function(q,b,Data1,Data2){
data1_num <- Data1[13+q,b]
data2_num <- Data2[13+q,1]
x <- 0
for (i in 1:12) {
x <- x + ((data1_num/Data1[i+q,b]) > (data2_num /Data2[i+q,1]))
}
x <- x /12
#return(x)
}
But If you provide data example, and the output your expecting, i'm sure there is way to simplify it better