Attempting to calculate differences between every two values in a row then sum the total differences for each dataframe in a list. I know for/while loops in R absolutely suck. I had this working before, but I've broken it. Can someone suggest how to optimize this using an alternative in the apply family? Current code:
for (i in 1:length(refdata)) { #for each dataframe in a list
refdif <- as.data.frame(matrix(0, ncol = 1, nrow = nrow(refdata[[i]])))
refdif1 <- c()
for (z in 1:ncol(refdata[[i]])) { #for each column in a dataframe
for(x in 1:nrow(refdata[[i]])) { #for each row in a dataframe
refdif <- (refdata[[i]][x,z] - refdata[[i]][x,z+1]) #difference of first value + the enxt
refdif1[x,1] <- (refdif1[x,1] + refidf) #sum of latest difference
}
}
print(refdif1) #where I can conduct tests on each individual dataframe with a column of sums of differences
}
example data:
list 1 refdata[[1]]
$`1`
var1 var2 var3 var4
1 1 2 3 4
2 5 6 7 8
$`2`
var1 var2 var3 var4
1 1 2 3 4
2 5 6 7 8
var 1 + 2 has the difference calculated, var 3 and 4 has the difference calculated, then each difference is summed together and placed in a new dataframe in a single column. (5-6) + (7-8), (1-2) + (3-4), etc etc:
$`1`
dif
1 -2
2 -2
$`2`
dif
1 -2
2 -2
One way to do it (per unlisted dataframe) could be by using logical vectors for indexing - their values are recycled - that way calculating the difference between every other column and finally summing the resulting df row-wise.
refdata1<-rowSums(refdata[c(T,F)]-refdata[c(F,T)])
Edit
Exact output can be obtained by
lapply(refdata, function(df){ data.frame(dif=rowSums(df[c(T,F)]-df[c(F,T)])) })
thx Heroka
# Create test data
x <- rbind(1:4, 5:8)
refdata <- list(x,x)
# Calculate results (all elements should have an even number of columns)
lapply(refdata, FUN = function(x) x %*% rep_len(c(1, -1), NCOL(x)))
Related
does anyone know how to have a row in R that is calculated from another row automatically? i.e.
lets say in excel, i want to make a row C, which is made up of (B2/B1)
e.g. C1 = B2/B1
C2 = B3/B2
...
Cn = Cn+1/Cn
but in excel, we only need to do one calculation then drag it down. how do we do it in R?
In R you work with columns as vectors so the operations are vectorized. The calculations as described could be implemented by the following commands, given a data.frame df (i.e. a table) and the respective column names as mentioned:
df["C1"] <- df["B2"]/df["B1"]
df["C2"] <- df["B3"]/df["B2"]
In R you usually would name the columns according to the content they hold. With that, you refer to the columns by their name, although you can also address the first column as df[, 1], the first row as df[1, ] and so on.
EDIT 1:
There are multiple ways - and certainly some more elegant ways to get it done - but for understanding I kept it in simple base R:
Example dataset for demonstration:
df <- data.frame("B1" = c(1, 2, 3),
"B2" = c(2, 4, 6),
"B3" = c(4, 8, 12))
Column calculation:
for (i in 1:ncol(df)-1) {
col_name <- paste0("C", i)
df[col_name] <- df[, i+1]/df[, i]
}
Output:
B1 B2 B3 C1 C2
1 1 2 4 2 2
2 2 4 8 2 2
3 3 6 12 2 2
So you iterate through the available columns B1/B2/B3. Dynamically create a column name in every iteration, based on the number of the current iteration, and then calculate the respective column contents.
EDIT 2:
Rowwise, as you actually meant it apparently, works similarly:
a <- c(10,15,20, 1)
df <- data.frame(a)
for (i in 1:nrow(df)) {
df$b[i] <- df$a[i+1]/df$a[i]
}
Output:
a b
1 10 1.500000
2 15 1.333333
3 20 0.050000
4 1 NA
You can do this just using vectors, without a for loop.
a <- c(10,15,20, 1)
df <- data.frame(a)
df$b <- c(df$a[-1], 0) / df$a
print(df)
a b
1 10 1.500000
2 15 1.333333
3 20 0.050000
4 1 0.000000
Explanation:
In the example data, df$a is the vector 10 15 20 1.
df$a[-1] is the same vector with its first element removed, 15 20 1.
And using c() to add a new element to the end so that the vector has the same lenght as before:
c(df$a[-1],0) which is 15 20 1 0
What we want for column b is this vector divided by the original df$a.
So:
df$b <- c(df$a[-1], 0) / df$a
Let's say i have the following list of df's (in reality i have many more dfs).
seq <- c("12345","67890")
li <- list()
for (i in 1:length(seq)){
li[[i]] <- list()
names(li)[i] <- seq[i]
li[[i]] <- data.frame(A = c(1,2,3),
B = c(2,4,6))
}
What i would like to do is calculate the mean within the same cell position between the lists, keeping the same amount of rows and columns as the original lists. How could i do this? I believe I can use the apply() function, but i am unsure how to do this.
The expected output (not surprising):
A B
1 1 2
2 2 4
3 3 6
In reality, the values within each list are not necessarily the same.
If there are no NAs, then we can Reduce to get the sum of observations for each element and divide by the length of the list
Reduce(`+`, li)/length(li)
# A B
#1 1 2
#2 2 4
#3 3 6
If there are NA values, then it may be better to use mean (which has na.rm argument). For this, we can convert it to array and then use apply
apply(array(unlist(li), dim = c(dim(li[[1]]), length(li))), c(1, 2), mean)
An equivalent option in tidyverse would be
library(tidyverse)
reduce(li, `+`)/length(li)
I would like to use the vector:
time.int<-c(1,2,3,4,5) #vector to be use as a "guide"
and the database:
time<-c(1,1,1,1,5,5,5)
value<-c("s","s","s","t","d","d","d")
dat1<- as.data.frame(cbind(time,value))
to create the following vector, which I can then add to the first vector "time.int" into a second database.
freq<-c(4,0,0,0,3) #wished result
This vector is the sum of the events that belong to each time interval, there are four 1 in "time" so the first value gets a four and so on.
Potentially I would like to generalize it so that I can decide the interval, for example saying sum in a new vector the events in "times" each 3 numbers of time.int.
EDIT for generalization
time.int<-c(1,2,3,4,5,6)
time<-c(1,1,1,2,5,5,5,6)
value<-c("s","s","s","t", "t","d","d","d")
dat1<- data.frame(time,value)
let's say I want it every 2 seconds (every 2 time.int)
freq<-c(4,0,4) #wished result
or every 3
freq<-c(4,4) #wished result
I know how to do that in excel, with a pivot table.
sorry if a duplicate I could not find a fitting question on this website, I do not even know how to ask this and where to start.
The following will produce vector freq.
freq <- sapply(time.int, function(x) sum(x == time))
freq
[1] 4 0 0 0 3
BTW, don't use the construct as.data.frame(cbind(.)). Use instead
dat1 <- data.frame(time,value))
In order to generalize the code above to segments of time.int of any length, I believe the following function will do it. Note that since you've changed the data the output for n == 1 is not the same as above.
fun <- function(x, y, n){
inx <- lapply(seq_len(length(x) %/% n), function(m) seq_len(n) + n*(m - 1))
sapply(inx, function(i) sum(y %in% x[i]))
}
freq1 <- fun(time.int, time, 1)
freq1
[1] 3 1 0 0 3 1
freq2 <- fun(time.int, time, 2)
freq2
[1] 4 0 4
freq3 <- fun(time.int, time, 3)
freq3
[1] 4 4
We can use the table function to count the event number and use merge to create a data frame summarizing the information. event_dat is the final output.
# Create example data
time.int <- c(1,2,3,4,5)
time <- c(1,1,1,1,5,5,5)
# Count the event using table and convert to a data frame
event <- as.data.frame(table(time))
# Convert the time.int to a data frame
time_dat <- data.frame(time = time.int)
# Merge the data
event_dat <- merge(time_dat, event, by = "time", all = TRUE)
# Replace NA with 0
event_dat[is.na(event_dat)] <- 0
# See the result
event_dat
time Freq
1 1 4
2 2 0
3 3 0
4 4 0
5 5 3
I have Valence Category for word stimuli in my psychology experiment.
1 = Negative, 2 = Neutral, 3 = Positive
I need to sort the thousands of stimuli with a pseudo-randomised condition.
Val_Category cannot have more than 2 of the same valence stimuli in a row i.e. no more than 2x negative stimuli in a row.
for example - 2, 2, 2 = not acceptable
2, 2, 1 = ok
I can't sequence the data i.e. decide the whole experiment will be 1,3,2,3,1,3,2,3,2,2,1 because I'm not allowed to have a pattern.
I tried various packages like dylpr, sample, order, sort and nothing so far solves the problem.
I think there's a thousand ways to do this, none of which are probably very pretty. I wrote a small function that takes care of the ordering. It's a bit hacky, but it appeared to work for what I tried.
To explain what I did, the function works as follows:
Take the vector of valences and samples from it.
If sequences are found that are larger than the desired length, then, (for each such sequence), take the last value of that sequence at places it "somewhere else".
Check if the problem is solved. If so, return the reordered vector. If not, then go back to 2.
# some vector of valences
val <- rep(1:3,each=50)
pseudoRandomize <- function(x, n){
# take an initial sample
out <- sample(val)
# check if the sample is "bad" (containing sequences longer than n)
bad.seq <- any(rle(out)$lengths > n)
# length of the whole sample
l0 <- length(out)
while(bad.seq){
# get lengths of all subsequences
l1 <- rle(out)$lengths
# find the bad ones
ind <- l1 > n
# take the last value of each bad sequence, and...
for(i in cumsum(l1)[ind]){
# take it out of the original sample
tmp <- out[-i]
# pick new position at random
pos <- sample(2:(l0-2),1)
# put the value back into the sample at the new position
out <- c(tmp[1:(pos-1)],out[i],tmp[pos:(l0-1)])
}
# check if bad sequences (still) exist
# if TRUE, then 'while' continues; if FALSE, then it doesn't
bad.seq <- any(rle(out)$lengths > n)
}
# return the reordered sequence
out
}
Example:
The function may be used on a vector with or without names. If the vector was named, then these names will still be present on the pseudo-randomized vector.
# simple unnamed vector
val <- rep(1:3,each=5)
pseudoRandomize(val, 2)
# gives:
# [1] 1 3 2 1 2 3 3 2 1 2 1 3 3 1 2
# when names assigned to the vector
names(val) <- 1:length(val)
pseudoRandomize(val, 2)
# gives (first row shows the names):
# 1 13 9 7 3 11 15 8 10 5 12 14 6 4 2
# 1 3 2 2 1 3 3 2 2 1 3 3 2 1 1
This property can be used for randomizing a whole data frame. To achieve that, the "valence" vector is taken out of the data frame, and names are assigned to it either by row index (1:nrow(dat)) or by row names (rownames(dat)).
# reorder a data.frame using a named vector
dat <- data.frame(val=rep(1:3,each=5), stim=rep(letters[1:5],3))
val <- dat$val
names(val) <- 1:nrow(dat)
new.val <- pseudoRandomize(val, 2)
new.dat <- dat[as.integer(names(new.val)),]
# gives:
# val stim
# 5 1 e
# 2 1 b
# 9 2 d
# 6 2 a
# 3 1 c
# 15 3 e
# ...
I believe this loop will set the Valence Category's appropriately. I've called the valence categories treat.
#Generate example data
s1 = data.frame(id=c(1:10),treat=NA)
#Setting the first two rows
s1[1,"treat"] <- sample(1:3,1)
s1[2,"treat"] <- sample(1:3,1)
#Looping through the remainder of the rows
for (i in 3:length(s1$id))
{
s1[i,"treat"] <- sample(1:3,1)
#Check if the treat value is equal to the previous two values.
if (s1[i,"treat"]==s1[i-1,"treat"] & s1[i-1,"treat"]==s1[i-2,"treat"])
#If so draw one of the values not equal to that value
{
a = 1:3
remove <- s1[i,"treat"]
a=a[!a==remove]
s1[i,"treat"] <- sample(a,1)
}
}
This solution is not particularly elegant. There may be a much faster way to accomplish this by sorting several columns or something.
I am working in R with a data frame d:
ID <- c("A","A","A","B","B")
eventcounter <- c(1,2,3,1,2)
numberofevents <- c(3,3,3,2,2)
d <- data.frame(ID, eventcounter, numberofevents)
> d
ID eventcounter numberofevents
1 A 1 3
2 A 2 3
3 A 3 3
4 B 1 2
5 B 2 2
where numberofevents is the highest value in the eventcounter for each ID.
Currently, I am trying to create an additional vector z <- c(6,6,6,3,3).
If the numberofevents == 3, it is supposed to calculate sum(1:3), equally to 3 + 2 + 1 = 6.
If the numberofevents == 2, it is supposed to calculate sum(1:2) equally to 2 + 1 = 3.
Working with a large set of data, I thought it might be convenient to create this additional vector
by using the sum function in R d$z<-sum(1:d$numberofevents), i.e.
sum(1:3) # for the rows 1-3
and
sum(1:2) # for the rows 4-5.
However, I always get this warning:
Numerical expression has x elements: only the first is used.
You can try ave
d$z <- with(d, ave(eventcounter, ID, FUN=sum))
Or using data.table
library(data.table)
setDT(d)[,z:=sum(eventcounter), ID][]
Try using apply sapply or lapply functions in R.
sapply(numberofevents, function(x) sum(1:x))
It works for me.