I have this data set
obs <- data.frame(replicate(8,rnorm(10, 0, 1)))
and this coefficients
coeff <- data.frame(replicate(8,rnorm(2, 0, 1)))
For each column of obs, I need to multiply the first element of first column, and add the second element of the first column too. I need to do the same for the 8 columns. I read somewhere that if someone copy and paste code more than once you are doing something wrong... and that's exactly what I did.
obs.transformed.X1 <-(obs[1]*coeff[1,1])+coeff[2,1]
obs.transformed.X2 <-(obs[2]*coeff[1,2])+coeff[2,2]
.
.
.
.
.
obs.transformed.X8 <-(obs[8]*coeff[1,8])+coeff[2,8]
I know there is a smarter way to do this (loop?), but I just couldn't figure it out. Any help will be appreciated.
This is what I've tried but I am only getting the last column
for (i in 1:length(obs)) {
results=(obs[i]*coeff[1,i])+coeff[2,i]
}
If you coerce to matrix class you can use the sweep function in a sequential fashion first multiplying columns by the first row of coeff and then by adding hte second row, again column-wise:
obs <- data.frame(matrix(1:60, 10)) # I find checking with random numbers difficult
coeff <- data.frame(matrix(1:12,2))
sweep(
sweep(as.matrix(obs), 2, as.matrix(coeff)[1,], "*"), # first operation is "*"
2, as.matrix(coeff)[2,], "+" ) # arguments for the addition
#--------------------------------
X1 X2 X3 X4 X5 X6
[1,] 3 37 111 225 379 573
[2,] 4 40 116 232 388 584
[3,] 5 43 121 239 397 595
[4,] 6 46 126 246 406 606
[5,] 7 49 131 253 415 617
[6,] 8 52 136 260 424 628
[7,] 9 55 141 267 433 639
[8,] 10 58 146 274 442 650
[9,] 11 61 151 281 451 661
[10,] 12 64 156 288 460 672
Decreased number of columns because your original code was too wide for my Rstudio console. But this should be very general. I suspect there's an equivalent matrix operator method but It didn't come to me
I came up with this solution..
results = list()
for (i in 1:length(obs)) {
results[[i]]=(obs[i]*coeff[1,i])+coeff[2,i]
}
results <- as.data.frame(results)
Is there any efficient way to do this?
I used Map
results <- as.data.frame(Map(`+`, Map(`*`, obs, coeff[1,]), coeff[2,]))
This should also give what you are looking for.
Related
I have a matrix that consists of two columns and a number (n) of rows, while each row represents a point with the coordinates x and y (the two columns).
This is what it looks (LINK):
V1 V2
146 17
151 19
153 24
156 30
158 36
163 39
168 42
173 44
...
now, I would like to use a subset of three consecutive points starting from 1 to do some fitting, save the values from this fit in another list, an den go on to the next 3 points, and the next three, ... till the list is finished. Something like this:
Data_Fit_Kasa_1 <- CircleFitByKasa(Data[1:3,])
Data_Fit_Kasa_2 <- CircleFitByKasa(Data[3:6,])
....
Data_Fit_Kasa_n <- CircleFitByKasa(Data[i:i+2,])
I have tried to construct a loop, but I can't make it work. R either tells me that there's an "unexpected '}' in "}" " or that the "subscript is out of bonds". This is what I've tried:
minimal runnable code
install.packages("conicfit")
library(conicfit)
CFKasa <- NULL
Data.Fit <- NULL
for (i in 1:length(Data)) {
row <- Data[i:(i+2),]
CFKasa <- CircleFitByKasa(row)
Data.Fit[i] <- CFKasa[3]
}
RStudio Version 0.99.902 – © 2009-2016 RStudio, Inc.; Win10 Edu.
The third element of the fitted circle (CFKasa[3]) represents the radius, which is what I am really interested in. I am really stuck here, please help.
Many thanks in advance!
Best, David
Turn your data into a 3D array and use apply:
DF <- read.table(text = "V1 V2
146 17
151 19
153 24
156 30
158 36
163 39", header = TRUE)
a <- t(DF)
dim(a) <-c(nrow(a), 3, ncol(a) / 3)
a <- aperm(a, c(2, 1, 3))
# , , 1
#
# [,1] [,2]
# [1,] 146 17
# [2,] 151 19
# [3,] 153 24
#
# , , 2
#
# [,1] [,2]
# [1,] 156 30
# [2,] 158 36
# [3,] 163 39
center <- function(m) c(mean(m[,1]), mean(m[,2]))
t(apply(a, 3, center))
# [,1] [,2]
#[1,] 150 20
#[2,] 159 35
center(DF[1:3,])
#[1] 150 20
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 have over 5000 data sets that i want to find p values for using fishers exact test in R. They are saved in a csv file and look something like this;
100 5000 400 500
250 400 600 400
... ... ... ...
where each row represents a contingency table.
Right now, I'm having to do a contingency table at a time, which will take me forever.
I used this code so far
alltables<-read.table("untitled1.csv") ##to read my data
apply(alltables,1, function(x) fisher.test(matrix(x,nr=2))$p.value)
But then I get the error "Error in fisher.test(matrix(x, nr = 2)) : 'x' must have at least 2 rows and columns"
You can do something like the following. But since you didn't really give a reproducible example, I first create some toy-data:
set.seed(1)
print(dat <- matrix(rbinom(n = 40, size = 1000, prob = 0.5), ncol = 4))
# [,1] [,2] [,3] [,4]
# [1,] 500 526 494 505
# [2,] 497 500 512 493
# [3,] 480 488 500 512
# [4,] 464 513 498 497
# [5,] 527 503 518 508
# [6,] 504 517 511 483
# [7,] 519 493 522 471
# [8,] 486 490 497 507
# [9,] 492 499 475 509
#[10,] 530 486 488 501
# Function to be applied row-wise
rowFisher <- function(x, ...) {
return(fisher.test(matrix(x, nrow = 2, ...))$p.value)
}
# Apply the function row-wise
apply(dat, 1, rowFisher)
# [1] 0.7557946 0.6548804 0.9641424 0.2603181 0.7912598 0.3729036 0.5916955 0.9283668 0.5585135
#[10] 0.2111895
Edit I didn't see your commments. But this should do the trick. If not, probably you have some NAs or other non-numeric values somewhere in your data.
I have a data which has two parameters, they are data/time and flow. The flow data is intermittent flow. Lets say at times there is zero flow and suddenly the flow starts and there will be non-zero values for sometime and then the flow will be zero again. I want to understand when the non-zero values occur and how long does each non-zero flow last. I have attached the sample dataset at this location https://www.dropbox.com/s/ef1411dq4gyg0cm/sampledataflow.csv
The data is 1 minute data.
I was able to import the data into R as follows:
flow <- read.csv("sampledataflow.csv")
summary(flow)
names(flow) <- c("Date","discharge")
flow$Date <- strptime(flow$Date, format="%m/%d/%Y %H:%M")
sapply(flow,class)
plot(flow$Date, flow$discharge,type="l")
I made plot to see the distribution but couldn't get a clue where to start to get the frequency of each non zero values. I would like to see a output table as follows:
Date Duration in Minutes
Please let me know if I am not clear here. Thanks.
Additional Info:
I think we need to check the non-zero value first and then find how many non zero values are there continuously before it reaches zero value again. What I want to understand is the flow release durations. For eg. in one day there might be multiple releases and I want to note at what time did the release start and how long did it continue before coming to value zero. I hope this explain the problem little better.
The first point is that you have too many NA in your data. In case you want to look into it.
If I understand correctly, you require the count of continuous 0's followed by continuous non-zeros, zeros, non-zeros etc.. for each date.
This can be achieved with rle of course, as also mentioned by #mnel under comments. But there are quite a few catches.
First, I'll set up the data with non-NA entries:
flow <- read.csv("~/Downloads/sampledataflow.csv")
names(flow) <- c("Date","discharge")
flow <- flow[1:33119, ] # remove NA entries
# format Date to POSIXct to play nice with data.table
flow$Date <- as.POSIXct(flow$Date, format="%m/%d/%Y %H:%M")
Next, I'll create a Date column:
flow$g1 <- as.Date(flow$Date)
Finally, I prefer using data.table. So here's a solution using it.
# load package, get data as data.table and set key
require(data.table)
flow.dt <- data.table(flow)
# set key to both "Date" and "g1" (even though, just we'll use just g1)
# to make sure that the order of rows are not changed (during sort)
setkey(flow.dt, "Date", "g1")
# group by g1 and set data to TRUE/FALSE by equating to 0 and get rle lengths
out <- flow.dt[, list(duration = rle(discharge == 0)$lengths,
val = rle(discharge == 0)$values + 1), by=g1][val == 2, val := 0]
> out # just to show a few first and last entries
# g1 duration val
# 1: 2010-05-31 120 0
# 2: 2010-06-01 722 0
# 3: 2010-06-01 138 1
# 4: 2010-06-01 32 0
# 5: 2010-06-01 79 1
# ---
# 98: 2010-06-22 291 1
# 99: 2010-06-22 423 0
# 100: 2010-06-23 664 0
# 101: 2010-06-23 278 1
# 102: 2010-06-23 379 0
So, for example, for 2010-06-01, there are 722 0's followed by 138 non-zeros, followed by 32 0's followed by 79 non-zeros and so on...
I looked a a small sample of the first two days
> do.call( cbind, tapply(flow$discharge, as.Date(flow$Date), function(x) table(x > 0) ) )
2010-06-01 2010-06-02
FALSE 1223 911
TRUE 217 529 # these are the cumulative daily durations of positive flow.
You may want this transposed in which case the t() function should succeed. Or you could use rbind.
If you jsut wante the number of flow-postive minutes, this would also work:
tapply(flow$discharge, as.Date(flow$Date), function(x) sum(x > 0, na.rm=TRUE) )
#--------
2010-06-01 2010-06-02 2010-06-03 2010-06-04 2010-06-05 2010-06-06 2010-06-07 2010-06-08
217 529 417 463 0 0 263 220
2010-06-09 2010-06-10 2010-06-11 2010-06-12 2010-06-13 2010-06-14 2010-06-15 2010-06-16
244 219 287 234 31 245 311 324
2010-06-17 2010-06-18 2010-06-19 2010-06-20 2010-06-21 2010-06-22 2010-06-23 2010-06-24
299 305 124 129 295 296 278 0
To get the lengths of intervals with discharge values greater than zero:
tapply(flow$discharge, as.Date(flow$Date), function(x) rle(x>0)$lengths[rle(x>0)$values] )
#--------
$`2010-06-01`
[1] 138 79
$`2010-06-02`
[1] 95 195 239
$`2010-06-03`
[1] 57 360
$`2010-06-04`
[1] 6 457
$`2010-06-05`
integer(0)
$`2010-06-06`
integer(0)
... Snipped output
If you want to look at the distribution of these durations you will need to unlist that result. (And remember that the durations which were split at midnight may have influenced the counts and durations.) If you just wanted durations without dates, then use this:
flowrle <- rle(flow$discharge>0)
flowrle$lengths[!is.na(flowrle$values) & flowrle$values]
#----------
[1] 138 79 95 195 296 360 6 457 263 17 203 79 80 85 30 189 17 270 127 107 31 1
[23] 2 1 241 311 229 13 82 299 305 3 121 129 295 3 2 291 278
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().