I have a matrix
mat<-matrix(c(272,237,266,272,225,265,117,223,262,241,210,216,252,203,170),nrow=5,ncol=3,byrow=T)
[,1] [,2] [,3]
[1,] 272 237 266
[2,] 272 225 265
[3,] 117 223 262
[4,] 241 210 216
[5,] 252 203 170
Is there a way so that if the first element in the Nth column doesn't equal to 272 that the rest of the rows be equal to that number in the Nth column? And if it is equal to 272, I will need to search skip to the row that doesn't equal 272 and then perform the same operation and let it propagate downwards?
So an example in the above data frame would turn to:
[,1] [,2] [,3]
[1,] 272 237 266
[2,] 272 237 266
[3,] 117 237 266
[4,] 117 237 266
[5,] 117 237 266
An easy approach would be to loop through the entire thing with double for loop, but that would take forever in R. Is there any better process then looping? I understand that in most cases, there will not be a ONE line answer.
I was thinking of using na.locf() but this works for only NAs. Is there a function for carrying value forward if it meets certain conditions?
Here's one way:
apply(mat, 2, function(col)
replace(col, seq(col) > which.max(col!=272), col[which.max(col!=272)]))
Related
I am trying to calculate my data's means by populating a new dataframe with data corrected by my experiment's blank.
So far, I have created my new data frame:
data_mean <- data.frame(matrix(ncol = 17, # As many columns as experimental conditions plus one for "Time(h)"
nrow = nrow(data)))
Copied the data corresponding to time:
data_mean[,1] <- data[,1]
And attempted to populate the dataframe by assigning the mean of every condition minus the mean of the blanks to each column:
data_mean[,2] <- rowMeans(data[,5:8])-rowMeans(data[,2:4])
data_mean[,3] <- rowMeans(data[,9:12])-rowMeans(data[,2:4])
data_mean[,4] <- rowMeans(data[,13:16])-rowMeans(data[,2:4])
data_mean[,5] <- rowMeans(data[,17:20])-rowMeans(data[,2:4])
and so on.
Is there an easier way to do this rather than typing the same code over and over?
res <- sapply(split.default(data[, -1], seq(ncol(data) - 1)%/%4), rowSums)
res[,-1] - res[,1] # Should give you all the differences above
example:
data <- data.frame(matrix(1:200, 10))
res <- sapply(split.default(data[, -1], seq(ncol(data) - 1)%/%4), rowSums)
res[,-1] - res[,1]
1 2 3 4
[1,] 161 321 481 641
[2,] 162 322 482 642
[3,] 163 323 483 643
[4,] 164 324 484 644
[5,] 165 325 485 645
[6,] 166 326 486 646
[7,] 167 327 487 647
[8,] 168 328 488 648
[9,] 169 329 489 649
[10,] 170 330 490 650
and you can check:
rowSums(data[, 5:8]) - rowSums(data[,2:4])
[1] 161 162 163 164 165 166 167 168 169 170 # first column
rowSums(data[, 9:12]) - rowSums(data[,2:4])
[1] 321 322 323 324 325 326 327 328 329 330 # second column
I would like to store a raster image into an empty matrix. I load my image and I create an empty matrix of same dimension.
setwd("C:/Users/Desktop/image/")
img_path <- "image.jpeg"
raster <- brick(img_path, package="raster")
nrow <- dim(raster)[1]
ncol <- dim(raster)[2]
img_matrix <- matrix(, nrow = nrow, ncol = ncol)
dim(raster)
[1] 896 1408 3
dim(img_matrix)
[1] 896 1408
When I try to load the raster into the matrix I got this error:
img_matrix[1, 1] <- raster
Error in img_matrix[1, 1] <- raster :
number of items to replace is not a multiple of replacement length
I think the problem is there are 3 layers in the raster and the matrix is just 1. So I think I create 3 matrix and paste each R, G, B layer of the raster into the associated matrix. Finally, assemble the 3 matrix to create the image.
How to do it?
There is an as.array method for objects of class raster.
library(raster)
raster <- brick("3.jpg", package="raster")
dim(raster)
#[1] 665 800 3
array <- as.array(raster)
dim(array)
#[1] 665 800 3
array[201:205,401:405,1]
# [,1] [,2] [,3] [,4] [,5]
#[1,] 244 244 244 244 244
#[2,] 244 244 244 244 244
#[3,] 244 244 244 244 244
#[4,] 244 244 244 244 244
#[5,] 244 244 244 244 244
array[201:205,401:405,2]
# [,1] [,2] [,3] [,4] [,5]
#[1,] 202 202 202 202 202
#[2,] 202 202 202 202 202
#[3,] 202 202 202 202 202
#[4,] 202 202 202 202 202
#[5,] 202 202 202 202 202
As you can see, this will coerce the raster object into a three dimensional array.
I am having a problem with summing the rows of my matrices. I have a list formed by 30 matrices
Matrix<-matrix(1:45, ncol=9)
List<-list(lapply(seq_len(30), function(X) Matrix))
The idea is to create 30 matrices size 5*3. Firstly, I need to sum some columns, 1:3 4:6 7:9, such that the result will be the following:
[,1] [,2] [,3]
[1,] 18 63 108
[2,] 21 66 111
[3,] 34 69 114
[4,] 47 72 117
[5,] 30 75 120
I am trying to get this matrix using this code:
Y<-lapply(List, function(x) rowSums(x[, 1:3]))
But, it only allows me to sum the 3 firsts columns.
After this, I need to sum the list and obtain only one matrix(5*3). I think that the command final<-reduce(Y,+) could help.
540 1890 3240
630 1980 3330
1020 2070 3420
1410 2160 3510
900 2250 3600
Thank you for your help
You need to find someway to group your columns by threes, for example:
grp = (1:ncol(Matrix) -1) %/% 3
or if you know the dimensions:
grp = rep(0:2,each=3)
To do rowSums in columns of threes, we can do this with a function:
SumCols = function(M,col_grp){
sapply(unique(col_grp),function(i)rowSums(M[,col_grp==i]))
}
SumCols(Matrix,grp)
[,1] [,2] [,3]
[1,] 18 63 108
[2,] 21 66 111
[3,] 24 69 114
[4,] 27 72 117
[5,] 30 75 120
So put this inside your List of matrices,
Reduce("+",lapply(List[[1]],SumCols,grp))
[,1] [,2] [,3]
[1,] 540 1890 3240
[2,] 630 1980 3330
[3,] 720 2070 3420
[4,] 810 2160 3510
[5,] 900 2250 3600
Here is another base R solution
out <- Reduce(`+`,Map(function(x) do.call(cbind,Map(rowSums, split.default(data.frame(x),ceiling(seq(ncol(x))/3)))),List[[1]]))
such that
> out
0 1 2
[1,] 540 1890 3240
[2,] 630 1980 3330
[3,] 720 2070 3420
[4,] 810 2160 3510
[5,] 900 2250 3600
I have a table similar to this one, and want to calculate the ratio between column A and B. For example:
A B C D E F
[1,] 187 174 183 115 101 104
[2,] 451 166 177 842 101 133
[3,] 727 171 187 12803 98 134
[4,] 1532 181 196 730 98 108
[5,] 4139 188 214 20358 105 159
[6,] 689 185 211 1633 110 162
[7,] 1625 184 195 2283 109 114
[8,] 771 181 190 904 105 110
[9,] 950 177 190 1033 106 112
[10,] 703 180 191 463 106 110
[11,] 2052 178 188 2585 100 105
[12,] 1161 178 187 2874 99 110
[13,] 214 175 184 173 98 110
[14,] 473 184 191 971 104 111
[15,] 756 185 193 14743 107 114
I want to create a new matrix that has all of those previous rows as new rows and columns (15 rows and 15 columns) like so (values in parentheses are placeholders for the calculated ratios):
[,1] [,2] [,3] [,4]
[1,] (A1:B1) (A1:B2) (A1:B3) (A1:B4) ...
[2,]
[3,]
[4,]
...
That is maybe not the best example, but I hope it is not too confusing.
To calculate the ratios A1:B1, A2:B2, A3:B3 I could do something like:
data.matrix(data["A"]/data["B"])
And to do it for all, I would do something like:
data.matrix(data[1,]/data[1,1])
data.matrix(data[1,]/data[1,2])
...
and so on.
This seems to be a lot of work and maybe someone knows a quicker and more efficient method.
EDIT
I thought the combn function would work, but then I figured out it doesn't. When I have a 2 column matrix, such as:
A B
[1,] 187 115
[2,] 451 842
[3,] 727 12803
[4,] 1532 730
[5,] 4139 20358
[6,] 689 1633
[7,] 1625 2283
[8,] 771 904
[9,] 950 1033
[10,] 703 463
[11,] 2052 2585
[12,] 1161 2874
[13,] 214 173
[14,] 473 971
[15,] 756 14743
And I use the combn function to calculate all possible ratios (A1:B1, A1:B2, ... A2:B1, A2:B2...) I get just the result for A1 vs all values of B.
> combn(ncol(data), 2, function(x) data[,x[1]]/data[,x[2]])
[,1]
[1,] 1.62608696
[2,] 0.53562945
[3,] 0.05678357
[4,] 2.09863014
[5,] 0.20331074
[6,] 0.42192284
[7,] 0.71178274
[8,] 0.85287611
[9,] 0.91965150
[10,] 1.51835853
[11,] 0.79381044
[12,] 0.40396660
[13,] 1.23699422
[14,] 0.48712667
[15,] 0.05127857
Or maybe I just don't understand the combn function and I am doing something wrong here.
You can achieve what you want by using expand.grid, apply and matrix functions as below
I am assuming what you want is matrix like
A1/B1 A1/B2 A1/B3 ...
A2/B1 A2/B2 A2/B3 ...
... ... ... ...
... ... ... ...
Here is the code to do that. Explanation in comments
txt <- "A B C D E F\n187 174 183 115 101 104\n451 166 177 842 101 133\n727 171 187 12803 98 134\n1532 181 196 730 98 108\n4139 188 214 20358 105 159\n689 185 211 1633 110 162\n1625 184 195 2283 109 114\n771 181 190 904 105 110\n950 177 190 1033 106 112\n703 180 191 463 106 110\n2052 178 188 2585 100 105\n1161 178 187 2874 99 110\n214 175 184 173 98 110\n473 184 191 971 104 111\n756 185 193 14743 107 114"
data <- as.matrix(read.table(textConnection(txt), header = TRUE))
# expand.grid : creates every combination of one element each from column A and
# B with elements of B repeated first
# apply : calls function(x) { x[1]/x[2]) } for every combination outputted by
# expand.grid
# matrix : converts the result of apply into matrix. dimnames arguments sets
# rownames and colnames for easy verification for us
result <- matrix(apply(expand.grid(data[, "A"], data[, "B"]), 1, function(x) x[1]/x[2]),
nrow = nrow(data), dimnames = list(data[, "A"], data[, "B"]))
# note that we have set rownames for result to be values of A and colnames for
# result to be value of B
result
## 174 166 171 181 188 185 184
## 187 1.074713 1.126506 1.093567 1.033149 0.9946809 1.010811 1.016304
## 451 2.591954 2.716867 2.637427 2.491713 2.3989362 2.437838 2.451087
## 727 4.178161 4.379518 4.251462 4.016575 3.8670213 3.929730 3.951087
## 1532 8.804598 9.228916 8.959064 8.464088 8.1489362 8.281081 8.326087
## 4139 23.787356 24.933735 24.204678 22.867403 22.0159574 22.372973 22.494565
## 689 3.959770 4.150602 4.029240 3.806630 3.6648936 3.724324 3.744565
## 1625 9.339080 9.789157 9.502924 8.977901 8.6436170 8.783784 8.831522
## 771 4.431034 4.644578 4.508772 4.259669 4.1010638 4.167568 4.190217
## 950 5.459770 5.722892 5.555556 5.248619 5.0531915 5.135135 5.163043
## 703 4.040230 4.234940 4.111111 3.883978 3.7393617 3.800000 3.820652
## 2052 11.793103 12.361446 12.000000 11.337017 10.9148936 11.091892 11.152174
## 1161 6.672414 6.993976 6.789474 6.414365 6.1755319 6.275676 6.309783
## 214 1.229885 1.289157 1.251462 1.182320 1.1382979 1.156757 1.163043
## 473 2.718391 2.849398 2.766082 2.613260 2.5159574 2.556757 2.570652
## 756 4.344828 4.554217 4.421053 4.176796 4.0212766 4.086486 4.108696
## 181 177 180 178 178 175 184
## 187 1.033149 1.056497 1.038889 1.050562 1.050562 1.068571 1.016304
## 451 2.491713 2.548023 2.505556 2.533708 2.533708 2.577143 2.451087
## 727 4.016575 4.107345 4.038889 4.084270 4.084270 4.154286 3.951087
## 1532 8.464088 8.655367 8.511111 8.606742 8.606742 8.754286 8.326087
## 4139 22.867403 23.384181 22.994444 23.252809 23.252809 23.651429 22.494565
## 689 3.806630 3.892655 3.827778 3.870787 3.870787 3.937143 3.744565
## 1625 8.977901 9.180791 9.027778 9.129213 9.129213 9.285714 8.831522
## 771 4.259669 4.355932 4.283333 4.331461 4.331461 4.405714 4.190217
## 950 5.248619 5.367232 5.277778 5.337079 5.337079 5.428571 5.163043
## 703 3.883978 3.971751 3.905556 3.949438 3.949438 4.017143 3.820652
## 2052 11.337017 11.593220 11.400000 11.528090 11.528090 11.725714 11.152174
## 1161 6.414365 6.559322 6.450000 6.522472 6.522472 6.634286 6.309783
## 214 1.182320 1.209040 1.188889 1.202247 1.202247 1.222857 1.163043
## 473 2.613260 2.672316 2.627778 2.657303 2.657303 2.702857 2.570652
## 756 4.176796 4.271186 4.200000 4.247191 4.247191 4.320000 4.108696
## 185
## 187 1.010811
## 451 2.437838
## 727 3.929730
## 1532 8.281081
## 4139 22.372973
## 689 3.724324
## 1625 8.783784
## 771 4.167568
## 950 5.135135
## 703 3.800000
## 2052 11.091892
## 1161 6.275676
## 214 1.156757
## 473 2.556757
## 756 4.086486
Edit: I seem to have misunderstood the question. The answer is even more simpler using outer:
# gives the same 15*15 matrix as geektrader's
outer(mm[,1], mm[,2], '/')
Old answer (not correct):
You should use combn:
# combn(ncol(mm), 2) gives you all possible combinations
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15]
# [1,] 1 1 1 1 1 2 2 2 2 3 3 3 4 4 5
# [2,] 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6
# it also accepts a function argument. we can use it to divide
# respective columns
mm.div <- combn(ncol(mm), 2, function(x) mm[,x[1]]/mm[,x[2]])
# set column names the matrix
colnames(mm.div) <- combn(colnames(mm), 2, paste, collapse="")
I might be completely missing the point here, but why not just use a couple for loops? I wrote a quick function, then you could pass the pairs to.
For example:
A <- rnorm(15)
B <- rnorm(15)
data <- data.frame(A,B)
ratio <- function(input1, input2){
out <- matrix(0, nrow=length(input1), ncol=length(input1))
k <- 1
for (i in 1:length(input1)){
for (j in 1:length(input1)){
out[k, j] <- input1[k] / input2[j]
}
k <- k + 1
}
return(out)
}
ratio(data$A, data$B)
EDIT
Another thought. To then use the function to do all possible pairs of ratios, you could simply add another for loop, like this:
combs <- combn(1:4, 2)
out <- list()
for (i in 1:(length(combs)/2)){
out[[i]] <- ratio(data[,combs[1,i]], data[,combs[2,i]])
}
Hope that helps!
I have two vectors, subject and target. I want to create a new vector based on comparisons between the two existing vectors, with elements being compared lagged. I've solved this okay using the loop below, but I'm essentially wondering whether there's a more elegant solution using apply?
subject <- c(200, 195, 190, 185, 185, 185, 188, 189, 195, 200, 210, 210)
target <- c(subject[1], subject[1]-cumsum(rep(perweek, length(subject)-1)))
adjtarget <- target
for (i in 1:(length(subject)-1)) {
if (subject[i] > adjtarget[i]) {
adjtarget[i+1] <- adjtarget[i]
} else {
adjtarget[i+1] <- adjtarget[i]-perweek }
}
}
This doesn't exactly solve your problem, but may point in a helpful direction. I'm disregarding the interplay between changing adjtarget and comparing to it, and show a similar problem, where we compare to the constant target. Then it's possible to change the if in the loop to a vector comparison:
lv <- but.last(subject) > but.last(target)
ind <- which(lv)
Prepare the result vector (I'll call it x, as it won't be the same result as your adjtarget) as a shifted copy of target and assign the changes to it:
x <- c(target[1], but.last(target)) # corresponds to the true branch of the `if`
x[ind+1] <- target[ind] - perweek # corresponds to the false branch
Alternatively,
x <- c(target[1], but.last(target) - (!lv)*perweek
As I said, this doesn't solve your problem, but perhaps we could start from here.
Just for clarification, if I understand your code, this is the kind of result you're looking for...
> (goal <- cbind(subject,target,adjtarget))
subject target adjtarget
[1,] 200 200 200
[2,] 195 198 198
[3,] 190 196 196
[4,] 185 194 194
[5,] 185 192 192
[6,] 185 190 190
[7,] 188 188 188
[8,] 189 186 186
[9,] 195 184 186
[10,] 200 182 186
[11,] 210 180 186
[12,] 210 178 186
If I'm right, then the challenge to vectorizing this is the repeated assignment of 186 in adjtarget. Vectorized code will evaluate the right hand side (RHS) before assigning it to the left hand side (LHS). So, the vectorized code won't see the updated value in adjtarget at row 9 until after the assignment is finished.
> y <- ifelse(subject > target, 1, 0) # matches TRUE case
> x <- target
> x[ind+1] <- target[ind]
> cbind(goal, x, y)
subject target adjtarget x y
[1,] 200 200 200 200 0
[2,] 195 198 198 198 0
[3,] 190 196 196 196 0
[4,] 185 194 194 194 0
[5,] 185 192 192 192 0
[6,] 185 190 190 190 0
[7,] 188 188 188 188 0
[8,] 189 186 186 186 1
[9,] 195 184 186 186 1 # assigned correctly (?)
[10,] 200 182 186 184 1 # incorrect x; should be 186
[11,] 210 180 186 182 1 # incorrect x; should be 186
[12,] 210 178 186 180 1 # incorrect x; should be 186