I have a matrix of one column and 6 rows. I would like to replicate that column i times but change one value randomly each time, and after each iteration, calculate the mean and variance across all columns.
For example:
values = rnorm(6, 6, 1); matrix1 = matrix(values, 6)
After i=1, would look like:
values2 = values
values2[sample(1:6, 1)] = values2[sample(1:6, 1)]+runif(1, 0, 1)
matrix2 = matrix(c(values, values2), 6)
At the end, I would like to output a data frame that looks like so:
i mean var
1 1.23 2.31
2 1.24 2.33 etc...
For many i's. I imagine there is a way to do this with loops, but my skills are not such that I can figure it out. Thanks for all your help!
If you know how many times you're doing this, it would be best to construct your final matrix beforehand, especially if i is large. However, without that:
jitter.func <- function(x, vec) {
cell <- sample(1:length(vec), 1)
vec[cell] <- vec[cell] + runif(1, 0, 1)
return(c(mean=mean(vec), var=var(vec)))
}
i <- 10
sapply(1:i, jitter.func, vec=values)
j <- 20 # Number of columns
i <- 6 # Number of rows
vec <- matrix(rnorm(i,6,1),ncol=j,nrow=i)# vector replicated j times
idx <- sample(seq(i),j,replace=TRUE) # j random rows
vec[cbind(idx, seq(j))] <- vec[cbind(idx, seq(j))]+runif(j) # add random number to random row in each column
apply(vec,2,plyr::each(mean,var)) # summary statistics
Related
recently I am trying to mimic a game.
I am going to throw 2 dice at the same time. If the sum of 2 dice is greater than or equals to 10, I win 1 point.
If it is lower than 10, I lose 1 point. I will do this for 1000 times.
At the very beginning, I draw 2000 random samples with set.seed (1234)
set.seed(1234)
d = sample(c(1:6), size = 2000, replace = T)
d
And then, I turn it into a matrix, and sum each row
a = matrix(d, nrow=1000, ncol=2, byrow=T)
t = rowSums(a)
t
Now, I have 1000 elements (sum of two dice each time). I would like to create a vector X to calculate the point that I can get.
However, how can I apply if statement to create vector X in this time?
Thank you very much
Do you mean this?
X <- ifelse(t>=10,1,-1)
or
X <- 2*(t>=10)-1
Using case_when
library(dplyr)
case_when(t >= 10 ~ 1, TRUE ~ -1)
You could assign a temporary variable and assign points by comparing the values.
tmp <- t
t[tmp >= 10] <- 1
t[tmp < 10] <- -1
Or without a temporary variable.
t1 <- c(-1, 1)[(t >= 10) + 1]
I have a data frame with 30 rows and 4 columns (namely, x, y, z, u). It is given below.
mydata = data.frame(x = rnorm(30,4), y = rnorm(30,2,1), z = rnorm(30,3,1), u = rnorm(30,5))
Further, I have a sequence values, which represent row number in my data frame.
myseq = c(seq(1, 30, by = 5))
myseq
[1] 1 6 11 16 21 26
Now, I wanted to compute the prob values for each segment of 99 rows.
filt= subset(mydata[1:6,], mydata[1:6,]$x < mydata[1:6,]$y & mydata[1:6,]$z < mydata[1:6,]$u
filt
prob = length(filt$x)/30
prob
Then I need to compute the above prob for 1:6,.., 27:30 and so on . Here, I have only 6 prob values. So, I can do one by one. If I have 100 values it would be tedious. Are there any way to compute the prob values?.
Thank you in advance.
BTW: in subset(DF[1:99,], ...), use DF[1:99,] in the first argument, not again, ala
subset(DF[1:99,], cumsuml < inchivaluel & cumsumr < inchivaluer)
Think about how to do this in a list.
The first step is to break your data into the va starting points. I'll start with a list of the indices to break it into:
inds <- mapply(seq, va, c(va[-1], nrow(DF)), SIMPLIFY=FALSE)
this now is a list of sequences, starting with 1:99, then 100:198, etc. See str(inds) to verify.
Now we can subset a portion of the data based on each element's vector of indices:
filts <- lapply(inds, function(ind) subset(DF[ind,], cumsuml < inchivaluel & cumsumr < inchivaluer))
We now have a list of vectors, let's summarize it:
results <- sapply(filts, function(filt) length(filt$cumsuml)/length(alpha))
Bottom line, it helps to think about how to break this problem into lists, examples at http://stackoverflow.com/a/24376207/3358272.
BTW: instead of initially making a list of indices, we could just break up the data in that first step, ala
DF2 <- mapply(function(a,b) DF[a:b,], va, c(va[-1], nrow(DF)), SIMPLIFY=FALSE)
filts <- lapply(DF2, function(x) subset(x, cumsuml < inchivaluel & cumsumr < inchivaluer))
results <- sapply(filts, function(filt) length(filt$cumsuml)/length(alpha))
I have the following problem.
I have multiple subarrays (say 2) that I have populated with character labels (1, 2, 3, 4, 5). My algorithm selects labels at random based on occurrence probabilities.
How can I get R to instead select labels 1:3 for subarray 1 and 4:5 for subarray 2, say, without using subsetting (i.e., []). That is, I want a random subset of labels to be selected for each subarray, instead of all labels assigned to each subarray manually using [].
I know sample() should help.
Using subsetting (which I don't want) one would do
x <- 1:5
sample(x[1:3], size, prob = probs[1:3])
but this assigns labels 1:3 to ALL subarrays.
Would
sample(sample(x), size, replace = TRUE, prob = probs)
work?
Any ideas? Please let me know if this is unclear.
Here is a small example, which selects labels from 1:5 for each of 10 subarrays.
set.seed(1)
N <- 10
K <- 2
Hstar <- 5
probs <- rep(1/Hstar, Hstar)
perms <- 5
## Set up container(s) to hold the identity of each individual from each permutation ##
num.specs <- ceiling(N / K)
## Create an ID for each haplotype ##
haps <- 1:Hstar
## Assign individuals (N) to each subpopulation (K) ##
specs <- 1:num.specs
## Generate permutations, assume each permutation has N individuals, and sample those individuals' haplotypes from the probabilities ##
gen.perms <- function() {
sample(haps, size = num.specs, replace = TRUE, prob = probs) # I would like each subarray to contain a random subset of 1:5.
}
pop <- array(dim = c(perms, num.specs, K))
for (i in 1:K) {
pop[,, i] <- replicate(perms, gen.perms())
}
pop
Hopefully this helps.
I think what you actually want is something like that
num.specs <- 3
haps[sample(seq(haps),size = num.specs,replace = F)]
[1] 3 5 4
That is a random subset of your vector haps ?
Not quite what you want (returns list of matrices instead of 3D array) but this might help
lapply(split(1:5, cut(1:5, breaks=c(0, 2, 5))), function(i) matrix(sample(i, 25, replace=TRUE), ncol=5))
Use cut and split to partition your vector of character labels before sampling them. Here I split your character labels at the value 2. Also, rather than sampling 5 numbers 5 times, you can sample 25 numbers once, and convert to matrix.
I am currently working my way through the book 'R for Data Science'.
I am trying to solve this exercise question (21.2.1 Q1.4) but have not been able to determine the correct output before starting the for loop.
Write a for loop to:
Generate 10 random normals for each of μ= −10, 0, 10 and 100.
Like the previous questions in the book I have been trying to insert into a vector output but for this example, it appears I need the output to be a data frame?
This is my code so far:
values <- c(-10,0,10,100)
output <- vector("double", 10)
for (i in seq_along(values)) {
output[[i]] <- rnorm(10, mean = values[[i]])
}
I know the output is wrong but am unsure how to create the format I need here. Any help much appreciated. Thanks!
There are many ways of doing this. Here is one. See inline comments.
set.seed(357) # to make things reproducible, set random seed
N <- 10 # number of loops
xy <- vector("list", N) # create an empty list into which values are to be filled
# run the loop N times and on each loop...
for (i in 1:N) {
# generate a data.frame with 4 columns, and add a random number into each one
# random number depends on the mean specified
xy[[i]] <- data.frame(um10 = rnorm(1, mean = -10),
u0 = rnorm(1, mean = 0),
u10 = rnorm(1, mean = 10),
u100 = rnorm(1, mean = 100))
}
# result is a list of data.frames with 1 row and 4 columns
# you can bind them together into one data.frame using do.call
# rbind means they will be merged row-wise
xy <- do.call(rbind, xy)
um10 u0 u10 u100
1 -11.241117 -0.5832050 10.394747 101.50421
2 -9.233200 0.3174604 9.900024 100.22703
3 -10.469015 0.4765213 9.088352 99.65822
4 -9.453259 -0.3272080 10.041090 99.72397
5 -10.593497 0.1764618 10.505760 101.00852
6 -10.935463 0.3845648 9.981747 100.05564
7 -11.447720 0.8477938 9.726617 99.12918
8 -11.373889 -0.3550321 9.806823 99.52711
9 -7.950092 0.5711058 10.162878 101.38218
10 -9.408727 0.5885065 9.471274 100.69328
Another way would be to pre-allocate a matrix, add in values and coerce it to a data.frame.
xy <- matrix(NA, nrow = N, ncol = 4)
for (i in 1:N) {
xy[i, ] <- rnorm(4, mean = c(-10, 0, 10, 100))
}
# notice that i name the column names post festum
colnames(xy) <- c("um10", "u0", "u10", "u100")
xy <- as.data.frame(xy)
As this is a learning question I will not provide the solution directly.
> values <- c(-10,0,10,100)
> for (i in seq_along(values)) {print(i)} # Checking we iterate by position
[1] 1
[1] 2
[1] 3
[1] 4
> output <- vector("double", 10)
> output # Checking the place where the output will be
[1] 0 0 0 0 0 0 0 0 0 0
> for (i in seq_along(values)) { # Testing the full code
+ output[[i]] <- rnorm(10, mean = values[[i]])
+ }
Error in output[[i]] <- rnorm(10, mean = values[[i]]) :
more elements supplied than there are to replace
As you can see the error say there are more elements to put than space (each iteration generates 10 random numbers, (in total 40) and you only have 10 spaces. Consider using a data format that allows to store several values for each iteration.
So that:
> output <- ??
> for (i in seq_along(values)) { # Testing the full code
+ output[[i]] <- rnorm(10, mean = values[[i]])
+ }
> output # Should have length 4 and each element all the 10 values you created in the loop
# set the number of rows
rows <- 10
# vector with the values
means <- c(-10,0,10,100)
# generating output matrix
output <- matrix(nrow = rows,
ncol = 4)
# setting seed and looping through the number of rows
set.seed(222)
for (i in 1:rows){
output[i,] <- rnorm(length(means),
mean=means)
}
#printing the output
output
I have a data.frame, ordered by mean column that looks like this:
10SE191_2 10SE207 10SE208 mean
7995783 12.64874 13.06391 12.69378 12.73937
8115327 12.69979 12.52285 12.41582 12.50363
8108370 12.58685 12.87818 12.66021 12.45720
7945680 12.46392 12.26087 11.77040 12.36518
7923547 11.98463 11.96649 12.50666 12.33138
8016718 12.81610 12.71548 12.48164 12.32703
I would like to apply a t.test to each row, using as input the intensity values: df[i,1:3] and the mean values from the rows with lower intensities. For example, for the first row I want to compute a t.test for df[1,1:3] vs _mean values_ from row 2 to row 6. My code uses a for loop but my current data.frame has more than 20,000 rows and 24 columns and it takes a long time. Any ideas for improving the code?
Thanks
Code:
temp <- matrix(-9, nrow=dim(matrix.order)[1], ncol=2) #create a result matrix
l <- dim(matrix.order)[1]
for (i in 1:l){
j <- 1+i
if (i < l | j +2 == l) { #avoid not enough y observations
mean.val <- matrix.order[j:l,4]
p <- t.test(matrix.order[i, 1:3], mean.val)
temp[i,1] <- p$p.value
}
else {temp[i,1] <- 1}
}
dput for my df
structure(list(`10SE191_2` = c(12.6487418898415, 12.6997932097351,12.5868508174491, 12.4639169398277, 11.9846348627906, 12.8160978540904), `10SE207` = c(13.0639063105224, 12.522848114011, 12.8781769160682, 12.260865493177, 11.9664905651469, 12.7154788700468), `10SE208` = c(12.6937808736673, 12.4158248856386, 12.6602128982717, 11.7704045448312, 12.5066604109231, 12.4816357798965), mean = c(12.7393707471856, 12.5036313008127, 12.4572035036992, 12.3651842840775, 12.3313821056582, 12.3270331271091)), .Names = c("10SE191_2", "10SE207", "10SE208", "mean"), row.names = c("7995783", "8115327", "8108370", "7945680", "7923547", "8016718"), class = "data.frame")
You can obtain all p-values (if possible) with this command:
apply(df, 1, function(x) {
y <- df$mean[df$mean < x[4]]
if(length(y) > 1)
t.test(x[1:3], y)$p.value
else NA
})
The function will return NA if there are not enough values for y.
7995783 8115327 8108370 7945680 7923547 8016718
0.08199794 0.15627947 0.04993244 0.50885253 NA NA
Running 2E4 t.tests probably takes a lot of time no matter what. Try using Rprof to find the hot spots. You might also want to use mcapply or similar parallel processing tools, since your analysis of each row is independent of all other data (which means this is a task well-suited to multicore parallel processing).