I am trying to get better with functions in R and I was working on a function to pull out every odd value from 100 to 500 that was divisible by 3. I got close with the function below. It keeps returning all of the values correctly but it also includes the first number in the sequence (101) when it should not. Any help would be greatly appreciated. The code I wrote is as follows:
Test=function(n){
if(n>100){
s=seq(from=101,to=n,by=2)
p=c()
for(i in seq(from=101,to=n,by=2)){
if(any(s==i)){
p=c(p,i)
s=c(s[(s%%3)==0],i)
}}
return (p)}else{
stop
}}
Test(500)
Here is a function that gets all non even multiples of 3. It's fully vectorized, no loops at all.
Check if n is within the range [100, 500].
Create an integer vector N from 100 to n.
Create a logical index of the elements of N that are divisible by 3 but not by 2.
Extract the elements of N that match the index i.
The main work is done in 3 code lines.
Test <- function(n){
stopifnot(n >= 100)
stopifnot(n <= 500)
N <- seq_len(n)[-(1:99)]
i <- ((N %% 3) == 0) & ((N %% 2) != 0)
N[i]
}
Test(500)
Here is a vectorised one-liner which optionally allows you to change the lower bound from a default of 100 to anything you like. If the bounds are wrong, it returns an empty vector rather than throwing an error.
It works by creating a vector of 1:500 (or more generally, 1:n), then testing whether each element is greater than 100 (or whichever lower bound m you set), AND whether each element is odd AND whether each element is divisible by 3. It uses the which function to return the indices of the elements that pass all the tests.
Test <- function(n, m = 100) which(1:n > m & 1:n %% 2 != 0 & 1:n %% 3 == 0)
So you can use it as specified in your question:
Test(500)
# [1] 105 111 117 123 129 135 141 147 153 159 165 171 177 183 189 195 201 207 213 219
# [21] 225 231 237 243 249 255 261 267 273 279 285 291 297 303 309 315 321 327 333 339
# [41] 345 351 357 363 369 375 381 387 393 399 405 411 417 423 429 435 441 447 453 459
# [61] 465 471 477 483 489 495
Or play around with upper and lower bounds:
Test(100, 50)
# [1] 51 57 63 69 75 81 87 93 99
Here is a function example for your objective
Test <- function(n) {
if(n<100 | n> 500) stop("out of range")
v <- seq(101,n,by = 2)
na.omit(ifelse(v%%2==1 & v%%3==0,v,NA))
}
stop() is called when your n is out of range [100,500]
ifelse() outputs desired odd values + NA
na.omit filters out NA and produce the final results
Related
I've just started my adventure with programming in R. I need to create a program summing numbers divisible by 3 and 5 in the range of 1 to 1000, using the '%%' operator. I came up with an idea to create two matrices with the numbers from 1 to 1000 in one column and their remainders in the second one. However, I don't know how to sum the proper elements (kind of "sum if" function in Excel). I attach all I've done below. Thanks in advance for your help!
s1<-1:1000
in<-s1%%3
m1<-matrix(c(s1,in), 1000, 2, byrow=FALSE)
s2<-1:1000
in2<-s2%%5
m2<-matrix(c(s2,in2),1000,2,byrow=FALSE)
Mathematically, the best way is probably to find the least common multiple of the two numbers and check the remainder vs that:
# borrowed from Roland Rau
# http://r.789695.n4.nabble.com/Greatest-common-divisor-of-two-numbers-td823047.html
gcd <- function(a,b) if (b==0) a else gcd(b, a %% b)
lcm <- function(a,b) abs(a*b)/gcd(a,b)
s <- seq(1000)
s[ (s %% lcm(3,5)) == 0 ]
# [1] 15 30 45 60 75 90 105 120 135 150 165 180 195 210
# [15] 225 240 255 270 285 300 315 330 345 360 375 390 405 420
# [29] 435 450 465 480 495 510 525 540 555 570 585 600 615 630
# [43] 645 660 675 690 705 720 735 750 765 780 795 810 825 840
# [57] 855 870 885 900 915 930 945 960 975 990
Since your s is every number from 1 to 1000, you could instead do
seq(lcm(3,5), 1000, by=lcm(3,5))
Just use sum on either result if that's what you want to do.
Props to #HoneyDippedBadger for figuring out what the OP was after.
See if this helps
x =1:1000 ## Store no. 1 to 1000 in variable x
x ## print x
Div = x[x%%3==0 & x%%5==0] ## Extract Nos. divisible by 3 & 5 both b/w 1 to 1000
Div ## Nos. Stored in DIv which are divisible by 3 & 5 both
length(Div)
table(x%%3==0 & x%%5==0) ## To see how many are TRUE for given condition
sum(Div) ## Sums up no.s divisible by both 3 and 5 b/w 1 to 1000
Sorry for the confusing title, but i wasn't sure how to title what i am trying to do. My objective is to create a dataset of 1000 obs each would be the length of the run. I have created a phase1 dataset, from which a set of control limits are produced. What i am trying to do now is create a phase2 dataset most likely using rnorm. what im trying to do is create a repeat loop that will continuously create values in the phase2 dataset until one of those values is outside of the control limits produced from the phase1 dataset. for example if i had 3.0 and -3.0 as control limits the phase2 dataset would create a bunch of observations until obs 398 when the value here happens to be 3.45, thus stopping the creation of data. my objective is then to record the number 398. Furthermore, I am then trying to loop the code back to the phase1 dataset/ control limits portion and create a new set of control limits and then run another phase2, until i have 1000 run lengths recorded. the code i have for the phase1/ control limits works fine and looks like this:
nphase1=50
nphase2=1000
varcount=1
meanshift= 0
sigmashift= 1
##### phase1 dataset/ control limits #####
phase1 <- matrix(rnorm(nphase1*varcount, 0, 1), nrow = nphase1, ncol=varcount)
mean_var <- apply(phase1, 2, mean)
std_var <- apply(phase1, 2, sd)
df_var <- data.frame(mean_var, std_var)
Upper_SPC_Limit_Method1 <- with(df_var, mean_var + 3 * std_var)
Lower_SPC_Limit_Method1 <- with(df_var, mean_var - 3 * std_var)
df_control_limits<- data.frame(Upper_SPC_Limit_Method1, Lower_SPC_Limit_Method1)
I have previously created this code in SAS and it looks like this. might be a better reference for what i am trying to achieve then me trying to explain it.
%macro phase2_dataset (n=,varcount=, meanshift=, sigmashift=, nphase1=,simID=,);
%do z=1 %to &n;
%phase1_dataset (n=&nphase1, varcount=&varcount);
data phase2; set control_limits n=lastobs;
call streaminit(0);
do until (phase2_var1<Lower_SPC_limit_method1_var1 or
phase2_var1>Upper_SPC_limit_method1_var1);
phase2_var1 = rand("normal", &meanshift, &sigmashift);
output;
end;
run;
ods exclude all;
proc means data=phase2;
var phase2_var1;
ods output summary=x;
run;
ods select all;
data run_length; set x;
keep Phase2_var1_n;
run;
proc append base= QA.Phase2_dataset&simID data=Run_length force; run;
%end;
%mend;
Also been doing research about using a while loop in replace of the repeat loop.
Im new to R so Any ideas you are able to throw my way are greatly appreciated. Thanks!
Using a while loop indeed seems to be the way to go. Here's what I think you're looking for:
set.seed(10) #Making results reproducible
replicate(100, { #100 is easier to display here
phase1 <- matrix(rnorm(nphase1*varcount, 0, 1), nrow = nphase1, ncol=varcount)
mean_var <- colMeans(phase1) #Slightly better than apply
std_var <- apply(phase1, 2, sd)
df_var <- data.frame(mean_var, std_var)
Upper_SPC_Limit_Method1 <- with(df_var, mean_var + 3 * std_var)
Lower_SPC_Limit_Method1 <- with(df_var, mean_var - 3 * std_var)
df_control_limits<- data.frame(Upper_SPC_Limit_Method1, Lower_SPC_Limit_Method1)
#Phase 2
x <- 0
count <- 0
while(x > Lower_SPC_Limit_Method1 && x < Upper_SPC_Limit_Method1) {
x <- rnorm(1)
count <- count + 1
}
count
})
The result is:
[1] 225 91 97 118 304 275 550 58 115 6 218 63 176 100 308 844 90 2758
[19] 161 311 1462 717 2446 74 175 91 331 210 118 1517 420 32 39 201 350 89
[37] 64 385 212 4 72 730 151 7 1159 65 36 333 97 306 531 1502 26 18
[55] 67 329 75 532 64 427 39 352 283 483 19 9 2 1018 137 160 223 98
[73] 15 182 98 41 25 1136 405 474 1025 1331 159 70 84 129 233 2 41 66
[91] 1 23 8 325 10 455 363 351 108 3
If performance becomes a problem, perhaps it would be interesting to explore some improvements, like creating more numbers with rnorm() at a time and then counting how many are necessary to exceed the limits and repeat if necessary.
I'm setting up a script to extract the thickness and voltages from a single column text file and perform a Weibull distribution on it. When I try to use fitdistr() I get an error stating "'x' must be a non-empty numeric vector". R is supposed to interpret numbers in text files as numeric but that doesn't seem to be happening. Any thoughts?
filename <- "SampleBreakdownSet.txt"
d <- read.table(filename, header = FALSE, sep = "")
#Extract thickness from the dataset; set to variable t
t = d[1,1]
#Extract the breakdown voltages and toss into dataset, BDV
BDV = tail(d,(nrow(d)-1))
#Calculates the breakdown field from the thickness and BDV
BDF = (BDV*10000) / t
#Calculates the Weibull parameters from the input breakdown voltages.
fitdistr(BDF, densfun ="weibull", lower = 0)
fitdistr(BDF, densfun ="weibull", lower = 0)
Error in fitdistr(BDF, densfun = "weibull", lower = 0) :
'x' must be a non-empty numeric vector
Sample data I'm using:
2
200
250
450
320
100
400
200
403
502
203
420
120
342
304
253
423
534
534
243
253
423
123
433
534
234
633
432
342
543
532
123
453
231
532
342
213
243
You are passing a data.frame to fitdistr, but you should be passing the vector itself.
Try this:
d <- read.table(text='200
250
450
320
100
400
200
403
502
203
420
120
342
304
253
423
534
534
243
253
423
123
433
534
234
633
432
342
543
532
123
453
231
532
342
213
243', header=FALSE)
t <- d[1,1]
#Extract the breakdown voltages and toss into dataset, BDV
BDV <- d[-1, 1]
BDF <- (BDV*10000) / t
library(MASS)
fitdistr(BDF, densfun ="weibull", lower = 0)
You could also refer to the relevant column when calling fitdistr, e.g.:
fitdistr(BDF$V1, densfun ="weibull", lower = 0)
# shape scale
# 2.745485e+00 1.997509e+04
# (3.716797e-01) (1.283667e+03)
I would like to do 10 fold cross validation and then using MSE for model selection in R . I can divide the data into 10 groups, but I got the following error, how can I fix it?
crossvalind <- function(N, kfold) {
len.seg <- ceiling(N/kfold)
incomplete <- kfold*len.seg - N
complete <- kfold - incomplete
ind <- matrix(c(sample(1:N), rep(NA, incomplete)), nrow = len.seg, byrow = TRUE)
cvi <- lapply(as.data.frame(ind), function(x) c(na.omit(x))) # a list
return(cvi)
}
I am using logspline package for estimation of a density function.
library(logspline)
x = rnorm(300, 0, 1)
kfold <- 10
cvi <- crossvalind(N = 300, kfold = 10)
for (i in 1:length(cvi)) {
xc <- x[cvi[-i]] # x in training set
xt <- x[cvi[i]] # x in test set
fit <- logspline(xc)
f.pred <- dlogspline(xt, fit)
f.true <- dnorm(xt, 0, 1)
mse[i] <- mean((f.true - f.pred)^2)
}
Error in x[cvi[-i]] : invalid subscript type 'list'
cvi is a list object, so cvi[-1] and cvi[1] are list objects, and then you try and get x[cvi[-1]] which is subscripting using a list object, which doesn't make sense because list objects can be complex objects containing numbers, characters, dates and other lists.
Subscripting a list with single square brackets always returns a list. Use double square brackets to get the constituents, which in this case are vectors.
> cvi[1] # this is a list with one element
$V1
[1] 101 78 231 82 211 239 20 201 294 276 181 168 207 240 61 72 267 75 218
[20] 177 127 228 29 159 185 118 296 67 41 187
> cvi[[1]] # a length 30 vector:
[1] 101 78 231 82 211 239 20 201 294 276 181 168 207 240 61 72 267 75 218
[20] 177 127 228 29 159 185 118 296 67 41 187
so you can then get those elements of x:
> x[cvi[[1]]]
[1] 0.32751014 -1.13362827 -0.13286966 0.47774044 -0.63942372 0.37453378
[7] -1.09954301 -0.52806368 -0.27923480 -0.43530831 1.09462984 0.38454106
[13] -0.68283862 -1.23407793 1.60511404 0.93178122 0.47314510 -0.68034783
[19] 2.13496564 1.20117869 -0.44558321 -0.94099782 -0.19366673 0.26640705
[25] -0.96841548 -1.03443796 1.24849113 0.09258465 -0.32922472 0.83169736
this doesn't work with negative indexes:
> cvi[[-1]]
Error in cvi[[-1]] : attempt to select more than one element
So instead of subscripting x with the list elements you don't want, subscript it with the negative of the indexes you do want (since you are partitioning here):
> x[-cvi[[1]]]
will return the other 270 elements. Note I've used 1 here for the first pass through your loop, replace with i and insert in your code.
I have a large data matrix (33183x1681), each row corresponding to one observation and each column corresponding to the variables.
I applied K-medoids clustering using PAM function in R, and I tried to visualize the clustering results using the built-in plots available with the PAM function. I got this error:
Error in princomp.default(x, scores = TRUE, cor = ncol(x) != 2) :
cannot use cor=TRUE with a constant variable
I think this problem is because of the high dimensionality of the data matrix I'm trying to cluster.
Any thoughts/ideas how to tackle this issue?
Check out the clara() function in package cluster which is shipped with all versions of R.
library("cluster")
## generate 500 objects, divided into 2 clusters.
x <- rbind(cbind(rnorm(200,0,8), rnorm(200,0,8)),
cbind(rnorm(300,50,8), rnorm(300,50,8)))
clarax <- clara(x, 2, samples=50)
clarax
> clarax
Call: clara(x = x, k = 2, samples = 50)
Medoids:
[,1] [,2]
[1,] -1.15913 0.5760027
[2,] 50.11584 50.3360426
Objective function: 10.23341
Clustering vector: int [1:500] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ...
Cluster sizes: 200 300
Best sample:
[1] 10 17 45 46 68 90 99 150 151 160 184 192 232 238 243 250 266 275 277
[20] 298 303 304 313 316 327 333 339 353 358 398 405 410 411 421 426 429 444 447
[39] 456 477 481 494 499 500
Available components:
[1] "sample" "medoids" "i.med" "clustering" "objective"
[6] "clusinfo" "diss" "call" "silinfo" "data"
Note that you should study the help for clara() (?clara) in some detail as well as the references cited in order to make the clustering performed by clara() as close to or identical to pam().