I have a dataframe like this:
V1 = paste0("AB", seq(1:48))
V2 = seq(1:48)
test = data.frame(name = V1, value = V2)
I want to calculate the means of the value-column and specific rows.
The pattern of the rows is pretty complicated:
Rows of MeanA1: 1, 5, 9
Rows of MeanA2: 2, 6, 10
Rows of MeanA3: 3, 7, 11
Rows of MeanA4: 4, 8, 12
Rows of MeanB1: 13, 17, 21
Rows of MeanB2: 14, 18, 22
Rows of MeanB3: 15, 19, 23
Rows of MeanB4: 16, 20, 24
Rows of MeanC1: 25, 29, 33
Rows of MeanC2: 26, 30, 34
Rows of MeanC3: 27, 31, 35
Rows of MeanC4: 28, 32, 36
Rows of MeanD1: 37, 41, 45
Rows of MeanD2: 38, 42, 46
Rows of MeanD3: 39, 43, 47
Rows of MeanD4: 40, 44, 48
As you see its starting at 4 different points (1, 13, 25, 37) then always +4 and for the following 4 means its just stepping 1 more row down.
I would like to have an output of all these means in one list.
Any ideas? NOTE: In this example the mean is of course always the middle number, but my real df is different.
Not quite sure about the output format you require, but the following codes can calculate what you want anyhow.
calc_mean1 <- function(x) mean(test$value[seq(x, by = 4, length.out = 3)])
calc_mean2 <- function(x){sapply(x:(x+3), calc_mean1)}
output <- lapply(seq(1, 37, 12), calc_mean2)
names(output) <- paste0('Mean', LETTERS[seq_along(output)]) # remove this line if more than 26 groups.
output
## $MeanA
## [1] 5 6 7 8
## $MeanB
## [1] 17 18 19 20
## $MeanC
## [1] 29 30 31 32
## $MeanD
## [1] 41 42 43 44
An idea via base R is to create a grouping variable for every 4 rows, split the data every 12 rows (nrow(test) / 4) and aggregate to find the mean, i.e.
test$new = rep(1:4, nrow(test)%/%4)
lapply(split(test, rep(1:4, each = nrow(test) %/% 4)), function(i)
aggregate(value ~ new, i, mean))
# $`1`
# new value
# 1 1 5
# 2 2 6
# 3 3 7
# 4 4 8
# $`2`
# new value
# 1 1 17
# 2 2 18
# 3 3 19
# 4 4 20
# $`3`
# new value
# 1 1 29
# 2 2 30
# 3 3 31
# 4 4 32
# $`4`
# new value
# 1 1 41
# 2 2 42
# 3 3 43
# 4 4 44
And yet another way.
fun <- function(DF, col, step = 4){
run <- nrow(DF)/step^2
res <- lapply(seq_len(step), function(inc){
inx <- seq_len(run*step) + (inc - 1)*run*step
dftmp <- DF[inx, ]
tapply(dftmp[[col]], rep(seq_len(step), run), mean, na.rm = TRUE)
})
names(res) <- sprintf("Mean%s", LETTERS[seq_len(step)])
res
}
fun(test, 2, 4)
#$MeanA
#1 2 3 4
#5 6 7 8
#
#$MeanB
# 1 2 3 4
#17 18 19 20
#
#$MeanC
# 1 2 3 4
#29 30 31 32
#
#$MeanD
# 1 2 3 4
#41 42 43 44
Since you said you wanted a long list of the means, I assumed it could also be a vector where you just have all these values. You would get that like this:
V1 = paste0("AB", seq(1:48))
V2 = seq(1:48)
test = data.frame(name = V1, value = V2)
meanVector <- NULL
for (i in 1:(nrow(test)-8)) {
x <- c(test$value[i], test$value[i+4], test$value[i+8])
m <- mean(x)
meanVector <- c(meanVector, m)
}
Related
For example : I have frame with 4 columns and I want divide columns A and B by C, but I want unchanged column ID
A B C ID
4 8 23 1
5 12 325 2
6 23 56 3
73 234 21 4
23 23 213 5
The result which i expect is
A B C ID
0,173913043 0,347826087 1 1
0,015384615 0,036923077 1 2
0,107142857 0,410714286 1 3
3,476190476 11,14285714 1 4
0,107981221 0,107981221 1 5
or without the column C, doesn't matter
So, I have the code which give me only columns A and B without the column 'ID'
columns_to_divide <- c(1,2)
results <- results[,columns_to_divide ]/results[,3]
We can use mutate, which creates or alters the values in a column. across says to alter columns A and B, and then we can define a function to divide both of these columns by C.
library(dplyr)
dat %>% mutate(across(c(A, B), function(x) x/C))
A B C ID
1: 0.17391304 0.34782609 23 1
2: 0.01538462 0.03692308 325 2
3: 0.10714286 0.41071429 56 3
4: 3.47619048 11.14285714 21 4
5: 0.10798122 0.10798122 213 5
div = c("A", "B")
div_by = "C"
DF[div] <- DF[div] / DF[[div_by]]
# A B C
# 1 0.17391304 0.34782609 23
# 2 0.01538462 0.03692308 325
# 3 0.10714286 0.41071429 56
# 4 3.47619048 11.14285714 21
# 5 0.10798122 0.10798122 213
Data
DF data.frame(
A = c(4, 5, 6, 73, 23), B = c(8, 12, 23, 234, 23), C = c(23, 325, 56, 21, 213)
)
Create Columns
A <- c(4, 5, 6, 73, 23)
B <- c(8, 12, 23, 234, 23)
C <- c(23, 325, 56, 21, 213)
ID <- c(1, 2, 3, 4, 5)
Add to data frame
df = data.frame(A, B, C, ID)
divide by and print
df$A <- df$A / df$C
df$B <- df$B / df$C
df$C <- df$C / df$C
print(df)
I have the following R codes running in RStudio:
foo <- list(
c(1, 2, 3, 5, 6, 7, 8, 9),
c(11, 13, 15, 16, 17, 19),
c(21, 25, 28, 29),
c(31, 33, 34, 35, 37, 38, 40)
)
names(foo) <- c(2, 1, 2, 1)
is.odd <- function(x) x %% 2 != 0
countOdds <- function(dt) {
sum(sapply(dt, is.odd))
}
while(TRUE) {
list2 <- unlist(sapply(seq_along(foo), function(x)
sample(foo[[x]], names(foo[x]))))
if(countOdds(list2) == 2)
break
}
sort(list2)
What the actual codes do:
The output is a random output (1 single outcome) from all possible outcomes.
Example: 2 6 16 25 28 37
In this output, we have 2 ODD numbers and 4 EVEN numbers as specified in the last part of the codes.
What I want to achieve:
I need to modify the codes so that it gives me ALL the possible outcomes from what has been specified instead of a single random output.
Ideally, I would like to write this new output into a CSV file.
Say, something like:
write.csv(result1, file="result1.csv")
How can I do this?
This was interesting problem to solve and maybe mine is not the most efficient approach but one way to solve this is
#First get all possible combinations based on the names
#For example, from list 1 we want to select 2 element, from list 2 only 1 and so on
#simplify is kept FALSE for the same purpose so that we get a list as outcome
#and those elements are treated as a group and not individual elements
all_combn <- sapply(seq_along(foo) ,function(x) combn(foo[[x]],
as.numeric(names(foo[x])), simplify = FALSE))
#Now create a dataframe where each row is one possible outcome
expanded <- expand.grid(all_combn)
#however, the problem is where there are more than 1 items to select it has
#columns as list, so we need to expand those list elements to separate columns
all_values <- t(apply(expanded, 1, unlist))
#Now select rows where our condition is satisfied
final <- all_values[apply(all_values, 1, countOdds) == 2, ]
final
# Var11 Var12 Var2 Var31 Var32 Var4
# [1,] 2 6 16 21 28 31
# [2,] 2 8 16 21 28 31
# [3,] 6 8 16 21 28 31
# [4,] 2 6 16 25 28 31
# [5,] 2 8 16 25 28 31
# [6,] 6 8 16 25 28 31
# [7,] 2 6 16 28 29 31
# [8,] 2 8 16 28 29 31
# [9,] 6 8 16 28 29 31
# [10,] 2 6 16 21 28 33
#.....
Now you can write this final to csv using write.csv or any other method.
I apologize for the poor phrasing of this question, I am still a beginner in R and I am still getting used to the proper terminology. I have provided sample data below:
mydata <- data.frame(x = c(1, 2, 7, 19, 45), y=c(10, 12, 15, 19, 24))
View(mydata)
My intention is to find the x speed, and for this I would need to find the difference between 1 and 2, 2 and 7, 7 and 19, and so on. How would I do this?
You can use the diff function.
> diffs <- as.data.frame(diff(as.matrix(mydata)))
> diffs
x y
1 1 2
2 5 3
3 12 4
4 26 5
> mean(diffs$x)
[1] 11
You can use dplyr::lead() and dplyr::lag() depending on how you want the calculations to line up
library(dplyr)
mydata <- data.frame(x = c(1, 2, 7, 19, 45), y=c(10, 12, 15, 19, 24))
View(mydata)
mydata %>%
mutate(x_speed_diff_lead = lead(x) - x
, x_speed_diff_lag = x - lag(x))
# x y x_speed_diff_lead x_speed_diff_lag
# 1 1 10 1 NA
# 2 2 12 5 1
# 3 7 15 12 5
# 4 19 19 26 12
# 5 45 24 NA 26
Let's say I want to create a column in a data.table, in which the value in each row is equal to the standard deviation of the values in three other cells in the same row. E.g., if I make
DT <- data.table(a = 1:4, b = c(5, 7, 9, 11), c = c(13, 16, 19, 22), d = c(25, 29, 33, 37))
DT
a b c d
1: 1 5 13 25
2: 2 7 16 29
3: 3 9 19 33
4: 4 11 22 37
and I'd like to add a column that contains the standard deviation of a, b, and d for each row, like this:
a b c d abdSD
1: 1 5 13 23 12.86
2: 2 7 16 27 14.36
3: 3 9 19 31 15.87
4: 4 11 22 35 17.39
I could of course write a for-loop or use an apply function to calculate this. Unfortunately, what I actually want to do needs to be applied to millions of rows, isn't as simple a function as calculating a standard deviation, and needs to finish within a fraction of a second, so I really need a vectorized solution. I want to write something like
DT[, abdSD := sd(c(a, b, d))]
but unfortunately that doesn't give the right answer. Is there any data.table syntax that can create a vector out of different values within the same row, and make that vector accessible to a function populating a new cell within that row? Any help would be greatly appreciated. #Arun
Depending on the size of your data, you might want to convert the data into a long format, then calculate the result as follows:
complexFunc <- function(x) sd(x)
cols <- c("a", "b", "d")
rowres <- melt(DT[, rn:=.I], id.vars="rn", variable.factor=FALSE)[,
list(abdRes=complexFunc(value[variable %chin% cols])), by=.(rn)]
DT[rowres, on=.(rn)]
or if your complex function has 3 arguments, you can do something like
DT[, abdSD := mapply(complexFunc, a, b, d)]
As #Frank mentioned, I could avoid adding a column by doing by=1:nrow(DT)
DT[, abdSD:=sd(c(a,b,d)),by=1:nrow(DT)]
output:
a b c d abdSD
1: 1 5 13 25 12.85820
2: 2 7 16 29 14.36431
3: 3 9 19 33 15.87451
4: 4 11 22 37 17.38774
if you add a row_name column, it would be ultra easy
DT$row_id<-row.names(DT)
Simply by=row_id, would get you the result you want
DT[, abdSD:=sd(c(a,b,d)),by=row_id]
Result would have:
a b c d row_id abdSD
1: 1 5 13 25 1 12.85820
2: 2 7 16 29 2 14.36431
3: 3 9 19 33 3 15.87451
4: 4 11 22 37 4 17.38774
If you want row_id removed, simply adding [,row_id:=NULL]
DT[, abdSD:=sd(c(a,b,d)),by=row_id][,row_id:=NULL]
This line would get everything you want
a b c d abdSD
1: 1 5 13 25 12.85820
2: 2 7 16 29 14.36431
3: 3 9 19 33 15.87451
4: 4 11 22 37 17.38774
You just gotta do it by row.
data.frame does it by row on default, data.table does it by column on default I think. It's a bit tricky
Hope this helps
I think you should try matrixStats package
library(matrixStats)
#sample data
dt <- data.table(a = 1:4, b = c(5, 7, 9, 11), c = c(13, 16, 19, 22), d = c(25, 29, 33, 37))
dt[, `:=`(abdSD = rowSds(as.matrix(.SD), na.rm=T)), .SDcols=c('a','b','d')]
dt
Output is:
a b c d abdSD
1: 1 5 13 25 12.85820
2: 2 7 16 29 14.36431
3: 3 9 19 33 15.87451
4: 4 11 22 37 17.38774
Not an answer, but just trying to show the difference between using apply and the solution provided by Prem above :
I have blown up the sample data to 40,000 rows to show solid time differences :
library(matrixStats)
#sample data
dt <- data.table(a = 1:40000, b = rep(c(5, 7, 9, 11),10000), c = rep(c(13, 16, 19, 22),10000), d = rep(c(25, 29, 33, 37),10000))
df <- data.frame(a = 1:40000, b = rep(c(5, 7, 9, 11),10000), c = rep(c(13, 16, 19, 22),10000), d = rep(c(25, 29, 33, 37),10000))
t0 = Sys.time()
dt[, `:=`(abdSD = rowSds(as.matrix(.SD), na.rm=T)), .SDcols=c('a','b','d')]
print(paste("Time taken for data table operation = ",Sys.time() - t0))
# [1] "Time taken for data table operation = 0.117115020751953"
t0 = Sys.time()
df$abdSD <- apply(df[,c("a","b","d")],1, function(x){sd(x)})
print(paste("Time taken for apply opertaion = ",Sys.time() - t0))
# [1] "Time taken for apply opertaion = 2.93488311767578"
Using DT and matrixStats clearly wins the race
It's not hard to vectorize the sd for this situation:
vecSD = function(x) {
n = ncol(x)
sqrt((n/(n-1)) * (Reduce(`+`, x*x)/n - (Reduce(`+`, x)/n)^2))
}
DT[, vecSD(.SD), .SDcols = c('a', 'b', 'd')]
#[1] 12.85820 14.36431 15.87451 17.38774
I input a vector vec<-c(2 3 4 8 10 12 15 19 20 23 27 28 39 47 52 60 64 75), and the size of intervals that I want to break the vector entries into.
In this example I want to break this into 9 different vectors based on the size of each entry.
In my case I want vector number 1 to be entries in the interval [1,9], then vector 2 to be entries in [10,18]...ect
In other words:
vec1: 2 3 4 8
vec2: 10 12 15
vec3: 19 20 23 27
ect...
I have tried using the split function but I do not know how to set a ratio that will work.
Maybe the following will do what you want.
f <- cut(vec, seq(0, max(vec), by = 9), include.lowest = TRUE)
sp <- split(vec, f)
sp <- sp[sapply(sp, function(x) length(x) != 0)]
sp
Use integer division %/% to return a vector of which group each value belongs in. Then split into separate vectors. Use (vec-1) to be "inclusive", i.e. 27 goes with group 3, not group 4.
split(vec,(vec-1) %/% 9)
Edit:
Another way using dplyr and cut which explicitly tags each interval
require(dplyr)
vec <- as.data.frame(vec)
df2 %>% mutate(interval = cut(vec,breaks=seq(0,((max(vec) %/% 9) +1) * 9,9),include.lowest=TRUE,right=TRUE))
vec interval
1 2 [0,9]
2 3 [0,9]
3 4 [0,9]
4 8 [0,9]
5 10 (9,18]
6 12 (9,18]
7 15 (9,18]
8 19 (18,27]
9 20 (18,27]
10 23 (18,27]
11 27 (18,27]
maybe this
library(purrr)
vec <- c(2, 3, 4, 8, 10 ,12, 15 ,19, 20, 23, 27, 28, 39, 47, 52, 60, 64, 75)
vec1 <- keep(vec, function(x) x >= 1 & (x) <= 9)
vec2 <- keep(vec, function(x) x >= 10 & (x) <= 18)