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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$C
print(df)
I have a dataframe like this:
V1 = paste0("AB", seq(1:48))
V2 = seq(1:48)
test = data.frame(name = V1, value = V2)
I want to calculate the means of the value-column and specific rows.
The pattern of the rows is pretty complicated:
Rows of MeanA1: 1, 5, 9
Rows of MeanA2: 2, 6, 10
Rows of MeanA3: 3, 7, 11
Rows of MeanA4: 4, 8, 12
Rows of MeanB1: 13, 17, 21
Rows of MeanB2: 14, 18, 22
Rows of MeanB3: 15, 19, 23
Rows of MeanB4: 16, 20, 24
Rows of MeanC1: 25, 29, 33
Rows of MeanC2: 26, 30, 34
Rows of MeanC3: 27, 31, 35
Rows of MeanC4: 28, 32, 36
Rows of MeanD1: 37, 41, 45
Rows of MeanD2: 38, 42, 46
Rows of MeanD3: 39, 43, 47
Rows of MeanD4: 40, 44, 48
As you see its starting at 4 different points (1, 13, 25, 37) then always +4 and for the following 4 means its just stepping 1 more row down.
I would like to have an output of all these means in one list.
Any ideas? NOTE: In this example the mean is of course always the middle number, but my real df is different.
Not quite sure about the output format you require, but the following codes can calculate what you want anyhow.
calc_mean1 <- function(x) mean(test$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)
}
Let's say I have data in wide format (samples in row and species in columns).
species <- data.frame(
Sample = 1:10,
Lobvar = c(21, 15, 12, 11, 32, 42, 54, 10, 1, 2),
Limtru = c(2, 5, 1, 0, 2, 22, 3, 0, 1, 2),
Pocele = c(3, 52, 11, 30, 22, 22, 23, 10, 21, 32),
Genmes = c(1, 0, 22, 1, 2,32, 2, 0, 1, 2)
)
And I want to automatically change the species names, based on a reference of functional groups that I have for all of the species (so it works even if I have more references than actual species in the dataset), for example:
reference <- data.frame(
Species_name = c("Lobvar", "Ampmis", "Pocele", "Genmes", "Limtru", "Secgio", "Nasval", "Letgos", "Salnes", "Verbes"),
Functional_group = c("Crustose", "Geniculate", "Erect", "CCA", "CCA", "CCA", "Geniculate", "Turf","Turf", "Crustose"),
stringsAsFactors = FALSE
)
EDIT
Thanks to #Dan Y suggestions, I can now changes the species names to their functional group names:
names(species)[2:ncol(species)] <- reference$Functional_group[match(names(species), reference$Species_name)][-1]
However, in my actual data.frame I have more species, and this creates many functional groups with the same name in different columns. I now would like to sum the columns that have the same names. I updated the example to give a results in which there is more than one functional group with the same name.
So i get this:
Sample Crustose CCA Erect CCA Crustose
1 21 2 3 1 2
2 15 5 52 0 3
3 12 1 11 22 4
4 11 0 30 1 1
5 32 2 22 2 0
6 42 22 22 32 0
and the final result I am looking for is this:
Sample Crustose CCA Erect
1 23 3 3
2 18 5 52
3 16 22 11
4 12 1 30
5 32 4 22
6 42 54 22
How do you advise on approaching this? Thanks for your help and the amazing suggestions I already received.
Re Q1) We can use match to do the name lookup:
names(species)[2:ncol(species)] <- reference$Functional_group[match(names(species), reference$Species_name)][-1]
Re Q2) Then we can mapply the rowSums function after some regular expression work on the colnames:
namevec <- gsub("\\.[[:digit:]]", "", names(df))
mapply(function(x) rowSums(df[which(namevec == x)]), unique(namevec))
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 would like to filter a data.frame based on the number of NA's in each row.
If I start with the following,
> d
A B C E
1 2 2 6 7
2 4 9 NA 10
3 6 NA NA 4
4 9 7 1 8
I would like to filter d to remove rows with 2 or more NA's in columns A, B, and C to yield:
A B C E
1 2 2 6 7
2 4 9 NA 10
4 9 7 1 8
We could use rowSums with is.na on the subset of columns of dataset to subset the rows
d[rowSums(is.na(d[1:3]))<2,]
# A B C E
#1 2 2 6 7
#2 4 9 NA 10
#4 9 7 1 8
The d[1:3] select the dataset with only 'A, B, C' columns. Applying is.na converts it to a logical matrix of TRUE/FALSE, do the sum of the TRUE values in each row with rowSums, and finally check if that number is less than 2 to get a logical vector which we use for subsetting the rows.
An alternative would be to use Reduce with +
d[Reduce(`+`,lapply(d[1:3], is.na)) <2,]
For reproducibility, define a data.frame as below with various numbers of NAs in each row.
df <- data.frame(
A = c(1, 2, 3, NA),
B = c(1, 2, NA, NA),
C = c(1, NA, NA, NA),
E = c(5, 6, 7, 8)
)
Define a function that counts the number if NA's in a given row:
countNA <- function(df) apply(df, MARGIN = 1, FUN = function(x) length(x[is.na(x)]))
Based on the wording of the question, exclude column E from this calculation:
df_noE <- subset(df, select=-E)
Now count NAs in each row using the function above:
na_count <- countNA(df_noE)
Now filter the original data.frame with this count:
df[na_count < 2,]
All together in a single line:
df[countNA(subset(df, select=-E)) < 2,]