I am new to R. I wanted to define a R function, find_element, that takes as its inputs a list and a value of any type, and returns the value of the matched element in the input list that matches the value. thanks for your help
find_element <- function(arr, val){
count = 0
for(i in arr){
if (i == val){
print(count)
} else
count = count + 1
print ("No Match")
}
}
e.g.
arr <- 1:10
find_element(arr, 10)
# 10
find_element(arr, 12)
# NULL
Just for educational purposes, please, try (although this is not recommended practice in R!):
find_element <- function(arr, val) {
count = 1
for (i in arr) {
if (i == val) {
return(count)
} else
count = count + 1
}
return("No Match")
}
This will yield
arr <- 1:10
find_element(arr, 10)
#[1] 10
find_element(arr, 12)
#[1] "No Match"
Please, note
In R, elements of vectors, etc are numbered starting with 1
You have to use return instead of print to indicate the return value of a function (well, I know there's a short cut - but it's for the purpose of education, here)
The final return must come after the for loop.
Built-in function
Also for educational purposes, please, note that Sotos already has shown the R way in his comment:
which(arr == 10)
#[1] 10
which(arr == 12)
#integer(0)
In R, it's almost always better to use the well-documented built-in functions or those from packages. And, yes, try to avoid for loops in R.
Learnig R online
As pointed out in the (now deleted) answer of engAnt there are several ressources to learn R. https://www.rstudio.com/online-learning/#R lists a number of resources.
Related
I am trying to learn how to implement control structures such as FOR and while loops.
I created a function the simulates betting in a famous Brazilian lotto.
In the lotto, a player bets on 6 unique integers from a 1:60 vector (called your_bet).
The function samples 6 values from the 1 to 60 universe ("result") and tests how many values in result match your_bet, printing out:
your_bet
result
total points (out of 6 possible)
one of three possible commentaries on the result of the bet.
Code as follows:
```
LOTTO<-function(your_bet=sample(1:60, size=6, replace=FALSE)){
result<-sample(1:60, size=6, replace=FALSE)
logical_vector<-(your_bet %in% result)
total_points<-sum(as.integer(logical_vector))
print(paste(c("Your bet:", as.character(your_bet))), collapse="")
print(paste(c("Result", as.character(result))), collapse="")
print(paste(c("Total points", as.character(total_points))), collapse="")
if (total_points==6)
print("You are a millonaire")
else if (total_points==5)
print("5 points, you are rich!")
else print("good luck next time")
}
```
I then tried to implement a loop that would make the function go over and over again in a loop until total points>= a given target (here as target_points), modifying the function as bellow.
```
LOTTO<-function(your_bet=sample(1:60, size=6, replace=FALSE), stubborn_until_x_points=FALSE,
target_points)#inserted stubborn_until_x_points and target_points arguments{
result<-sample(1:60, size=6, replace=FALSE)
logical_vector<-(your_bet %in% result)
total_points<-sum(as.integer(logical_vector))
print(paste(c("Your bet:", as.character(your_bet))), collapse="")
print(paste(c("Result", as.character(result))), collapse="")
print(paste(c("Total points", as.character(total_points))), collapse="")
if (total_points==6)
print("You are a millonaire")
else if (total_points==5)
print("5 points, you are rich!")
else print("good luck next time")
if (stubborn_until_x_points==TRUE)#Inserted WHILE loop here{
while(total_points < target_points){
LOTTO(your_bet, stubborn_until_x_points=TRUE, target_points)}
}
}
```
This did make the function repeat in a loop, but for some reason, it keeps looping even when the condition is met.
Observation - I realized that when the condition is met on the first run, it actually stops, but after entering a loop, it goes on forever.
I can´t find what is wrong. Any ideas?
Thanks in advance.
I haven't dug quite deep into the function, but the only reason that this should not break at any point is when target_points > n_samples (6 here).
The problem in this case is quite obvious and simple to fix. Reduce target_points to be less than 6 or add n_samples (6 atm.) and make it greater than target_points. Rather than this I suspect the main problem lies within the recursive function though. R is rather restrictive when it comes to recursion, if one tries to do a simple recursion for example
i <- 0
f <- function(){
i <<- i + 1
f()
}
f()
i
# 896 on my pc
one can see that we cannot use recursion for very deep recursive functions (in R). This throws the very unhelpful error
Error: C stack usage 7974196 is too close to the limit
To alleviate this, one simply has to remove the recursion (simply in italian because sometimes it is not simple). In this case we just move the while loop to contain the main body of the function, and use an if statement to break early if necessary.
Below is a slightly modified version of the function (note that sample_n and number_range has been added as arguments).
In this function the while loop has been moved to contain the main body, and the result is instead printed at the end (using the variable res to figure out the result). At the end of the loop I use a if(isFALSE(stubborn_until_x_points))break statement to exit early if necessary.
LOTTO <- function(your_bet,
sample_n = 6,
number_range = 1:60,
stubborn_until_x_points = FALSE,
target_points){
if(missing(target_points) || target_points > sample_n)
stop('missing target_points or target_points too large')
total_points <- -Inf # Always smaller than target_points
i <- 0
res <- 0
# If you want a new bet every iteration.
# Add this at the end of the loop, but remove the 'if'
if(missing(your_bet))
your_bet <- sample(number_range, size=sample_n, replace=FALSE)
while(total_points < target_points){
result <- sample(number_range, size=sample_n, replace=FALSE)
logical_vector <- your_bet %in% result
total_points <- sum(logical_vector)
if (total_points==6){
res <- 1
}else if (total_points==5){
res <- 2
}
i <- i + 1
if(isFALSE(stubborn_until_x_points))
break
}
if(res == 1)
cat(res <- 'You\'re a millionaire!\n', sep = '\n')
else if(res == 2)
cat(res <- '5 points, you are rich!\n', sep = '\n')
else
cat(res <- 'Better luck next time.', sep = '\n')
c(result = res, number_of_tries = i)
}
The function is called as before, but now also returns the number of attempts and the result obtained from trials as shown below.
LOTTO(target_points = 6, stubborn_until_x_points = TRUE)
You're a millionaire!
#Output:
result number_of_tries
"You're a millionaire!\n" "8297820"
Removing recursion, including the body of the function inside the while loop, assigning -Inf to the initial total_points and adding the break statement were much usefull.
Addapting the answer by #Oliver yielded exactly what I was looking for:
LOTTO<-function(your_bet=sample(1:60, size=6, replace=FALSE), stubborn_until_x_points=FALSE,
target_points=0){
total_points<--Inf
while(total_points < target_points){
result<-sample(1:60, size=6, replace=FALSE)
logical_vector<-(your_bet %in% result)
total_points<-sum(as.integer(logical_vector))
print(paste(c("Your bet:", as.character(your_bet))), collapse="")
print(paste(c("Result", as.character(result))), collapse="")
print(paste(c("Total points", as.character(total_points))), collapse="")
if (total_points==6)
print("You are a millonaire")
else if (total_points==5)
print("5 points, you are rich!")
else print("good luck next time")
if (isFALSE(stubborn_until_x_points==TRUE))
break
}
His answer, however, yields interesting additional results, and better control over inadequate argument inputs
I am trying to understand the for and if-statement in r, so I run a code where I am saying that if the sum of rows are bigger than 3 then return 1 else zero:
Here is the code
set.seed(2)
x = rnorm(20)
y = 2*x
a = cbind(x,y)
hold = c()
Now comes the if-statement
for (i in nrow(a)) {
if ([i,1]+ [i,2] > 3) hold[i,] == 1
else ([i,1]+ [i,2]) <- hold[i,] == 0
return (cbind(a,hold)
}
I know that maybe combining for and if may not be ideal, but I just want to understand what is going wrong. Please keep the explanation at a dummy level:) Thanks
You've got some issues. #mnel covered a better way to go about doing this, I'll focus on understanding what went wrong in this attempt (but don't do it this way at all, use a vectorized solution).
Line 1
for (i in nrow(a)) {
a has 20 rows. nrow(a) is 20. Thus your code is equivalent to for (i in 20), which means i will only ever be 20.
Fix:
for (i in 1:nrow(a)) {
Line 2
if ([i,1]+ [i,2] > 3) hold[i,] == 1
[i,1] isn't anything, it's the ith row and first column of... nothing. You need to reference your data: a[i,1]
You initialized hold as a vector, c(), so it only has one dimension, not rows and columns. So we want to assign to hold[i], not hold[i,].
== is used for equality testing. = or <- are for assignment. Right now, if the >3 condition is met, then you check if hold[i,] is equal to 1. (And do nothing with the result).
Fix:
if (a[i,1]+ a[i,2] > 3) hold[i] <- 1
Line 3
else ([i,1]+ [i,2]) <- hold[i,] == 0
As above for assignment vs equality testing. (Here you used an arrow assignment, but put it in the wrong place - as if you're trying to assign to the else)
else happens whenever the if condition isn't met, you don't need to try to repeat the condition
Fix:
else hold[i] <- 0
Fixed code together:
for (i in 1:nrow(a)) {
if (a[i,1] + a[i,2] > 3) hold[i] <- 1
else hold[i] <- 0
}
You aren't using curly braces for your if and else expressions. They are not required for single-line expressions (if something do this one line). They are are required for multi-line (if something do a bunch of stuff), but I think they're a good idea to use. Also, in R, it's good practice to put the else on the same line as a } from the preceding if (inside the for loop or a function it doesn't matter, but otherwise it would, so it's good to get in the habit of always doing it). I would recommend this reformatted code:
for (i in 1:nrow(a)) {
if (a[i, 1] + a[i, 2] > 3) {
hold[i] <- 1
} else {
hold[i] <- 0
}
}
Using ifelse
ifelse() is a vectorized if-else statement in R. It is appropriate when you want to test a vector of conditions and get a result out for each one. In this case you could use it like this:
hold <- ifelse(a[, 1] + a[, 2] > 3, 1, 0)
ifelse will take care of the looping for you. If you want it as a column in your data, assign it directly (no need to initialize first)
a$hold <- ifelse(a[, 1] + a[, 2] > 3, 1, 0)
Such operations in R are nicely vectorised.
You haven't included a reference to the dataset you wish to index with your call to [ (eg a[i,1])
using rowSums
h <- rowSums(a) > 3
I am going to assume that you are new to R and trying to learn about the basic function of the for loop itself. R has fancy functions called "apply" functions that are specifically for doing basic math on each row of a data frame. I am not going to talk about these.
You want to do the following on each row of the array.
Sum the elements of the row.
Test that the sum is greater than 3.
Return a value of 1 or 0 representing the result of 2.
For 1, luckily "sum" is a built in function. It pays off to check out the built in functions within every programming language because they save you time. To sum the elements of a row, just use sum(a[row_number,]).
For 2, you are evaluating a logical statement "is x >3?" where x is the result from 1. The ">3" statement returns a value of true or false. The logical expression is a fancy "if then" statement without the "if then".
> 4>3
[1] TRUE
> 2>3
[1] FALSE
For 3, a true or false value is a data structure called a "logical" value in R. A 1 or 0 value is a data structure called a "numeric" value in R. By converting the "logical" into a "numeric", you can change the TRUE to 1's and FALSE to 0's.
> class(4>3)
[1] "logical"
> as.numeric(4>3)
[1] 1
> class(as.numeric(4>3))
[1] "numeric"
A for loop has a min, a max, a counter, and an executable. The counter starts at the min, and increments until it goes to the max. The executable will run for each run of the counter. You are starting at the first row and going to the last row. Putting all the elements together looks like this.
for (i in 1:nrow(a)){
hold[i] <- as.numeric(sum(a[i,])>3)
}
UPDATE
Thanks to the help and suggestions of #CarlWitthoft my code was simplified to this:
model <- unlist(sapply(1:length(model.list),
function(i) ifelse(length(model.list[[i]][model.lookup[[i]]] == "") == 0,
NA, model.list[[i]][model.lookup[[i]]])))
ORIGINAL POST
Recently I read an article on how vectorizing operations in R instead of using for loops are a good practice, I have a piece of code where I used a big for loop and I'm trying to make it a vector operation but I cannot find the answer, could someone help me? Is it possible or do I need to change my approach? My code works fine with the for loop but I want to try the other way.
model <- c(0)
price <- c(0)
size <- c(0)
reviews <- c(0)
for(i in 1:length(model.list)) {
if(length(model.list[[i]][model.lookup[[i]]] == "") == 0) {
model[i] <- NA
} else {
model[i] <- model.list[[i]][model.lookup[[i]]]
}
if(length(model.list[[i]][price.lookup[[i]]] == "") == 0) {
price[i] <- NA
} else {
price[i] <- model.list[[i]][price.lookup[[i]]]
}
if(length(model.list[[i]][reviews.lookup[[i]]] == "") == 0) {
reviews[i] <- NA
} else {
reviews[i] <- model.list[[i]][reviews.lookup[[i]]]
}
size[i] <- product.link[[i]][size.lookup[[i]]]
}
Basically the model.list variable is a list from which I want to extract a particular vector, the location from that vector is given by the variables model.lookup, price.lookup and reviews.lookup which contain logical vectors with just one TRUE value which is used to return the desired vector from model.list. Then every cycle of the for loop the extracted vectors are stored on variables model, price, size and reviews.
Could this be changed to a vector operation?
In general, try to avoid if when not needed. I think your desired output can be built as follows.
model <- unlist(sapply(1:length(model.list), function(i) model.list[[i]][model.lookup[[i]]]))
model[model=='']<-NA
And the same for your other variables. This assumes that all model.lookup[[i]] are of length one. If they aren't, you won't be able to write the output to a single element of model in the first place.
I would also note that you are grossly overcoding, e.g. x<-0 is better than x<-c(0), and don't bother with length evaluation on a single item.
I have a command that generates a variable every 10 loops in R (index1, index2, index3... and so on). The command I have is functional, but I am thinking of a smarter way to write this command. Here's what my command looks like:
for (counter in 1:10){
for (i in 1:100){
if (counter == 1){
index1 <- data1 ## some really long command here, I just changed it to this simple command to illustrate the idea
}
if (counter == 2){
index2 <- data2
}
.
.
.
# until I reach index10
} indexing closure
} ## counter closure
Is there a way to write this without having to write the conditional if commands? I would like to generate index1, index2.... I am sure there is some easy way to do this but I just cannot think of it.
Thanks.
What you need is the modulo operator %%. inside the inner loop. Ex: 100%%10 returns 0 101%%10 returns 1 92%%10 returns 2 - in other words if it is multiple of 10 then you get 0. And the assign function.
Note: You no longer need the outer loop used in your example.
So to create a variable at every 10 iteration do something like this
for(i in 1:100){
#check if i is multiple of 10
if(i%%10==0){
myVar<-log(i)
assign(paste("index",i/10,sep=""), myVar)
}
}
ls() #shows that index1, index2, ...index10 objects have been created.
index1 #returns 2.302585
update:
Alternatively, you can store results in a vector
index<-vector(length=10)
for(i in 1:100){
#check if i is multiple of 10
if(i%%10==0){
index[i/10]<-log(i)
}
}
index #returns a vector with 10 elements, each a result at end of an iteration that is a multiple of 10.
I am backtesting some investment strategy using R, I have a piece of script below:
set.seed(1)
output.df <- data.frame(action=sample(c("initial_buy","sell","buy"),
10000,replace=TRUE),stringsAsFactors=FALSE)
output.df[,"uid"] <- 1:nrow(output.df)
cutrow.fx <- function(output.df) {
loop.del <- 2
while (loop.del <= nrow(output.df)) {
if ((output.df[loop.del,"action"]=="initial_buy" &
output.df[loop.del-1,"action"]=="initial_buy")|
(output.df[loop.del,"action"]=="sell" &
output.df[loop.del-1,"action"]=="sell")|
(output.df[loop.del,"action"]=="buy" &
output.df[loop.del-1,"action"]=="sell")|
(output.df[loop.del,"action"]=="initial_buy" &
output.df[loop.del-1,"action"]=="buy")){
output.df <- output.df[-loop.del,]
} else {
loop.del <- loop.del + 1
}
}
output.df<<-output.df
}
print(system.time(cutrow.fx(output.df=output.df)))
The strategy will determine: 1) when to start buying a stock; 2) when to add additional contribution to the stock; and 3) when to sell all the stock. I have a dataframe with price of a stock for the past 10 years. I wrote 3 scripts to indicate which date should I buy/sell the stock, combine the 3 results and order them.
I need to remove some of the "impossible action", e.g. I cannot sell the same stock twice without buying new units beforehand, so I used the script above to delete those impossible action. But the for loop is kind of slow.
Any suggestion for speeding it up?
Update 01
I have updated the cutrow.fx into the following but fail:
cutrow.fx <- function(output.df) {
output.df[,"action_pre"] <- "NIL"
output.df[2:nrow(output.df),"action_pre"] <- output.df[1:(nrow(output.df)-1),"action"]
while (any(output.df[,"action_pre"]=="initial_buy" & output.df[,"action"]=="initial_buy")|
any(output.df[,"action_pre"]=="sell" & output.df[,"action"]=="sell")|
any(output.df[,"action_pre"]=="sell" & output.df[,"action"]=="buy")|
any(output.df[,"action_pre"]=="buy" & output.df[,"action"]=="initial_buy")) {
output.df <- output.df[!(output.df[,"action_pre"]=="initial_buy" & output.df[,"action"]=="initial_buy"),]
output.df <- output.df[!(output.df[,"action_pre"]=="sell" & output.df[,"action"]=="sell"),]
output.df <- output.df[!(output.df[,"action_pre"]=="sell" & output.df[,"action"]=="buy"),]
output.df <- output.df[!(output.df[,"action_pre"]=="buy" & output.df[,"action"]=="initial_buy"),]
output.df[,"action_pre"] <- "NIL"
output.df[2:nrow(output.df),"action_pre"] <- output.df[1:(nrow(output.df)-1),"action"]
}
output.df[,"action_pre"] <- NULL
output.df<<-output.df
}
I used the vector comparison as somehow inspired (I used somehow as I'm not sure if I get exact what he means in the answer) by John, use a while-loop to repeat. But the output is not the same.
Is the for-loop here inevitable?
It looks like all you're doing is checking the last action. This doesn't require a loop at all. All you have to do is shift the vector and do straight vector comparisons. Here's an artificial example.
x <- sample(1:11)
buysell <- sample(c('buy', 'sell'), 11, replace = TRUE)
So, I have 11 samples, x, and whether I've bought or sold them. I want to make a boolean that shows whether I bought or sold the last sample.
bought <- c(NA, buysell[1:10])
which( bought == 'buy' )
Examine the x and buysell variables and you'll see the results here are the index of the x items where a buy was made on the prior item.
Also, you might want to check out he function %in%.
I tried to do something clever with vectorization, but failed because previous iterations of the loop can change the data relationships for later iterations through. So I couldn't lag the data by a set amount and compare lagged to real results.
What I can do is minimize the copying operation involved. R is assign-by-copy, so when you write a statement like output.df <- output.df[-loop.del,], you are copying the entire data structure for each row that is deleted. Instead of changing (and copying) the data frame, I made changes to a logical vector. Some other attempts at speed-up include using logical and (&&) instead of bitwise and (&), using %in% to make fewer comparisons, and minimizing accesses on output.df.
To compare the two functions I slightly modified OP solution such that the original data frame was not overwritten. It looks like this can improve speeds by a factor of 10, but it still takes a noticeable about of time (>0.5 sec). I'd love to see any faster solutions.
OP's solution (slightly modified in return value and without global assign)
cutrow.fx <- function(output.df) {
loop.del <- 2
while (loop.del <= nrow(output.df)) {
if ((output.df[loop.del,"action"]=="initial_buy" &
output.df[loop.del-1,"action"]=="initial_buy")|
(output.df[loop.del,"action"]=="sell" &
output.df[loop.del-1,"action"]=="sell")|
(output.df[loop.del,"action"]=="buy" &
output.df[loop.del-1,"action"]=="sell")|
(output.df[loop.del,"action"]=="initial_buy" &
output.df[loop.del-1,"action"]=="buy")){
output.df <- output.df[-loop.del,]
} else {
loop.del <- loop.del + 1
}
}
return(output.df)
}
ans1 <- cutrow.fx(output.df)
my solution
cutrow.fx2 <- function(output.df) {
##edge case if output.df has too few rows
if (nrow(output.df) < 2) return(output.df)
##logical vector of indices of rows to keep
idx <- c(TRUE,logical(nrow(output.df)-1))
##keeps track of the previous row
prev.row <- 1
prev.act <- output.df[prev.row,"action"]
for (current.row in seq_len(nrow(output.df))[-1]) {
##access output.df only once per iteration
current.act <- output.df[current.row,"action"]
##checks to see if current row is bad
##if so, continue to next row and leave previous row as is
if ( (prev.act %in% c("initial_buy","buy")) &&
(current.act == "initial_buy") ) {
next
} else if ( (prev.act == "sell") &&
(current.act %in% c("buy","sell")) ) {
next
}
##if current row is good, mark it in idx and update previous row
idx[current.row] <- TRUE
prev.row <- current.row
prev.act <- current.act
}
return(output.df[idx,])
}
ans2 <- cutrow.fx2(output.df)
checks that answers are the same
identical(ans1,ans2)
## [1] TRUE
#benchmarking
require(microbenchmark)
mb <- microbenchmark(
ans1=cutrow.fx(output.df)
,ans2=cutrow.fx2(output.df),times=50)
print(mb)
# Unit: milliseconds
# expr min lq median uq max
# 1 ans1 9630.1671 9743.1102 9967.6442 10264.7000 12396.5822
# 2 ans2 481.8821 491.6699 500.6126 544.4222 645.9658
plot(mb)
require(ggplot2)
ggplot2::qplot(y=time, data=mb, colour=expr) + ggplot2::scale_y_log10()
Here is some code that is a bit simpler and much faster. It does not loop over all elements, but only loops between matches. It matches forward rather than backward.
First, modify your cutrow.fx function. Remove the <<-output.df on the last line, and simply return the result. Then you can run two functions and compare the results.
cutrow.fx1 <- function(d) {
len <- length(d[,1])
o <- logical(len)
f <- function(a) {
switch(a,
initial_buy=c('buy', 'sell'),
buy=c('buy', 'sell'),
sell='initial_buy'
)
}
cur <- 1
o[cur] <- TRUE
while (cur < len) {
nxt <- match(f(d[cur,1]), d[(cur+1):len,1])
if (all(is.na(nxt))) {
break
} else {
cur <- cur + min(nxt, na.rm=TRUE);
o[cur] <- TRUE
}
}
d[o,]
}
Show that the results are correct:
identical(cutrow.fx1(output.df), cutrow.fx(output.df))
## [1] TRUE
And it is quite a bit faster. This is due to the partial vectorization of the problem, using match to find the next row to keep, rather than iterating to discard rows.
print(system.time(cutrow.fx(output.df)))
## user system elapsed
## 5.688 0.000 5.720
print(system.time(cutrow.fx1(output.df)))
## user system elapsed
## 1.050 0.000 1.056