removing columns with NA values only [duplicate] - r

This question already has answers here:
Remove columns from dataframe where ALL values are NA
(13 answers)
Closed 1 year ago.
I am using this command to remove the columns where all the values are NA.
testing5 <- subset(testing4,
select = -c(kurtosis_picth_belt, skewness_roll_belt,
skewness_roll_belt.1, min_yaw_belt, amplitude_yaw_belt,
kurtosis_roll_arm, kurtosis_picth_arm, kurtosis_yaw_arm,
skewness_roll_arm, skewness_pitch_arm, kurtosis_picth_dumbbell,
skewness_roll_dumbbell, skewness_pitch_dumbbell, min_yaw_dumbbell,
kurtosis_roll_forearm, kurtosis_picth_forearm, skewness_roll_forearm,
skewness_pitch_forearm))
Is there a shorter (programmitic) method?
Thanks and Regards,
Partha

The tidyverse approach would look like this (also using #Rich Scriven data):
d %>% select_if(~any(!is.na(.)))
# x
# 1 NA
# 2 3
# 3 NA

You can remove the columns that contain all NA values with e.g.
d <- data.frame(x = c(NA, 3, NA), y = rep(NA, 3))
# x y
# 1 NA NA
# 2 3 NA
# 3 NA NA
d[!sapply(d, function(x) all(is.na(x)))]
# x
# 1 NA
# 2 3
# 3 NA
On your data, this would be
testing4[!sapply(testing4, function(x) all(is.na(x)))]

Yet another way (a bit more vectorized) using #Richards data
d[!is.nan(colMeans(d, na.rm = TRUE))]
# x
# 1 NA
# 2 3
# 3 NA

Related

Fast way to insert values in column of data frame in R

I am currently trying to find unique elements between two columns of a data frame and write these to a new final data frame.
This is my code, which works perfectly fine, and creates a result which matches my expectation.
set.seed(42)
df <- data.frame(a = sample(1:15, 10),
b=sample(1:15, 10))
unique_to_a <- df$a[!(df$a %in% df$b)]
unique_to_b <- df$b[!(df$b %in% df$a)]
n <- max(c(unique_to_a, unique_to_b))
out <- data.frame(A=rep(NA,n), B=rep(NA,n))
for (element in unique_to_a){
out[element, "A"] = element
}
for (element in unique_to_b){
out[element, "B"] = element
}
out
The problem is, that it is very slow, because the real data contains 100.000s of rows. I am quite sure it is because of the repeated indexing I am doing in the for loop, and I am sure there is a quicker, vectorized way, but I dont see it...
Any ideas on how to speed up the operation is much appreciated.
Cheers!
Didn't compare the speed but at least this is more concise:
elements <- with(df, list(setdiff(a, b), setdiff(b, a)))
data.frame(sapply(elements, \(x) replace(rep(NA, max(unlist(elements))), x, x)))
# X1 X2
# 1 NA NA
# 2 NA NA
# 3 NA 3
# 4 NA NA
# 5 NA NA
# 6 NA NA
# 7 NA NA
# 8 NA NA
# 9 NA NA
# 10 NA NA
# 11 11 NA

Perform calculations on row depending on individual cells [duplicate]

This question already has answers here:
Sum rows in data.frame or matrix
(7 answers)
Closed 2 years ago.
I have a data frame in R that looks like
1 3 NULL,
2 NULL 5,
NULL NULL 9
I want to iterate through each row and perform and add the two numbers that are present. If there aren't two numbers present I want to throw an error. How do I refer to specific rows and cells in R? To iterate through the rows I have a for loop. Sorry not sure how to format a matrix above.
for(i in 1:nrow(df))
Data:
df <- data.frame(
v1 = c(1, 2, NA),
v2 = c(3, NA, NA),
v3 = c(NA, 5, 9)
)
Use rowSums:
df$sum <- rowSums(df, na.rm = T)
Result:
df
v1 v2 v3 sum
1 1 3 NA 4
2 2 NA 5 7
3 NA NA 9 9
If you do need a for loop:
for(i in 1:nrow(df)){
df$sum[i] <- rowSums(df[i,], na.rm = T)
}
If you have something with NULL you can make it a data.frame, but that will make the columns with NULL a character vector. You have to convert those to numeric, which will then introduce NA for NULL.
rowSums will then create the sum you want.
df <- read.table(text=
"
a b c
1 3 NULL
2 NULL 5
NULL NULL 9
", header =T)
# make columns numeric, this will change the NULL to NA
df <- data.frame(lapply(df, as.numeric))
cbind(df, sum=rowSums(df, na.rm = T))
# a b c sum
# 1 1 3 NA 4
# 2 2 NA 5 7
# 3 NA NA 9 9

Is there a way to recode an SPSS function in R to create a single new variable?

Can somebody please help me with a recode from SPSS into R?
SPSS code:
RECODE variable1
(1,2=1)
(3 THRU 8 =2)
(9, 10 =3)
(ELSE = SYSMIS)
INTO variable2
I can create new variables with the different values. However, I'd like it to be in the same variable, as SPSS does.
Many thanks.
x <- y<- 1:20
x
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
y[x %in% (1:2)] <- 1
y[x %in% (3:8)] <- 2
y[x %in% (9:10)] <- 3
y[!(x %in% (1:10))] <- NA
y
[1] 1 1 2 2 2 2 2 2 3 3 NA NA NA NA NA NA NA NA NA NA
I wrote a function that has very similiar coding to the spss code recode. See here
variable1 <- -1:11
recodeR(variable1, c(1, 2, 1), c(3:8, 2), c(9, 10, 3), else_do= "missing")
NA NA 1 1 2 2 2 2 2 2 3 3 NA
This function now works also for other examples. This is how the function is defined
recodeR <- function(vec_in, ..., else_do){
l <- list(...)
# extract the "from" values
from_vec <- unlist(lapply(l, function(x) x[1:(length(x)-1)]))
# extract the "to" values
to_vec <- unlist(lapply(l, function(x) rep(x[length(x)], length(x)-1)))
# plyr is required for mapvalues
require(plyr)
# recode the variable
vec_out <- mapvalues(vec_in, from_vec, to_vec)
# if "missing" is written then all outside the defined range will be missings.
# Otherwise values outside the defined range stay the same
if(else_do == "missing"){
vec_out <- ifelse(vec_in < min(from_vec, na.rm=T) | vec_in > max(from_vec, na.rm=T), NA, vec_out)
}
# return resulting vector
return(vec_out)}

"summarize" multiple incomplete columns to 1 summary column [duplicate]

I have some columns in R and for each row there will only ever be a value in one of them, the rest will be NA's. I want to combine these into one column with the non-NA value. Does anyone know of an easy way of doing this. For example I could have as follows:
data <- data.frame('a' = c('A','B','C','D','E'),
'x' = c(1,2,NA,NA,NA),
'y' = c(NA,NA,3,NA,NA),
'z' = c(NA,NA,NA,4,5))
So I would have
'a' 'x' 'y' 'z'
A 1 NA NA
B 2 NA NA
C NA 3 NA
D NA NA 4
E NA NA 5
And I would to get
'a' 'mycol'
A 1
B 2
C 3
D 4
E 5
The names of the columns containing NA changes depending on code earlier in the query so I won't be able to call the column names explicitly, but I have the column names of the columns which contains NA's stored as a vector e.g. in this example cols <- c('x','y','z'), so could call the columns using data[, cols].
Any help would be appreciated.
Thanks
A dplyr::coalesce based solution could be as:
data %>% mutate(mycol = coalesce(x,y,z)) %>%
select(a, mycol)
# a mycol
# 1 A 1
# 2 B 2
# 3 C 3
# 4 D 4
# 5 E 5
Data
data <- data.frame('a' = c('A','B','C','D','E'),
'x' = c(1,2,NA,NA,NA),
'y' = c(NA,NA,3,NA,NA),
'z' = c(NA,NA,NA,4,5))
You can use unlist to turn the columns into one vector. Afterwards, na.omit can be used to remove the NAs.
cbind(data[1], mycol = na.omit(unlist(data[-1])))
a mycol
x1 A 1
x2 B 2
y3 C 3
z4 D 4
z5 E 5
Here's a more general (but even simpler) solution which extends to all column types (factors, characters etc.) with non-ordered NA's. The strategy is simply to merge the non-NA values of other columns into your merged column using is.na for indexing:
data$mycol = data$x # your new merged column. Start with x
data$mycol[!is.na(data$y)] = data$y[!is.na(data$y)] # merge with y
data$mycol[!is.na(data$z)] = data$z[!is.na(data$z)] # merge with z
> data
a x y z mycol
1 A 1 NA NA 1
2 B 2 NA NA 2
3 C NA 3 NA 3
4 D NA NA 4 4
5 E NA NA 5 5
Note that this will overwrite existing values in mycol if there are several non-NA values in the same row. If you have a lot of columns you could automate this by looping over colnames(data).
I would use rowSums() with the na.rm = TRUE argument:
cbind.data.frame(a=data$a, mycol = rowSums(data[, -1], na.rm = TRUE))
which gives:
> cbind.data.frame(a=data$a, mycol = rowSums(data[, -1], na.rm = TRUE))
a mycol
1 A 1
2 B 2
3 C 3
4 D 4
5 E 5
You have to call the method directly (cbind.data.frame) as the first argument above is not a data frame.
Something like this ?
data.frame(a=data$a, mycol=apply(data[,-1],1,sum,na.rm=TRUE))
gives :
a mycol
1 A 1
2 B 2
3 C 3
4 D 4
5 E 5
max works too. Also works on strings vectors.
cbind(data[1], mycol=apply(data[-1], 1, max, na.rm=T))
One possibility using dplyr and tidyr could be:
data %>%
gather(variables, mycol, -1, na.rm = TRUE) %>%
select(-variables)
a mycol
1 A 1
2 B 2
8 C 3
14 D 4
15 E 5
Here it transforms the data from wide to long format, excluding the first column from this operation and removing the NAs.
In a related link (suppress NAs in paste()) I present a version of paste with a na.rm option (with the unfortunate name of paste5).
With this the code becomes
cols <- c("x", "y", "z")
cbind.data.frame(a = data$a, mycol = paste2(data[, cols], na.rm = TRUE))
The output of paste5 is a character, which works if you have character data otherwise you'll need to coerce to the type you want.
Though this is not the OP case, it seems some people like the approach based on sums, how about thinking in mean and mode, to make the answer more universal. This answer matches the title, which is what many people will find.
data <- data.frame('a' = c('A','B','C','D','E'),
'x' = c(1,2,NA,NA,9),
'y' = c(NA,6,3,NA,5),
'z' = c(NA,NA,NA,4,5))
splitdf<-split(data[,c(2:4)], seq(nrow(data[,c(2:4)])))
data$mean<-unlist(lapply(splitdf, function(x) mean(unlist(x), na.rm=T) ) )
data$mode<-unlist(lapply(splitdf, function(x) {
tab <- tabulate(match(x, na.omit(unique(unlist(x) ))));
paste(na.omit(unique(unlist(x) ))[tab == max(tab) ], collapse = ", " )}) )
data
a x y z mean mode
1 A 1 NA NA 1.000000 1
2 B 2 6 NA 4.000000 2, 6
3 C NA 3 NA 3.000000 3
4 D NA NA 4 4.000000 4
5 E 9 5 5 6.333333 5
If you want to stick with base,
data <- data.frame('a' = c('A','B','C','D','E'),'x' = c(1,2,NA,NA,NA),'y' = c(NA,NA,3,NA,NA),'z' = c(NA,NA,NA,4,5))
data[is.na(data)]<-","
data$mycol<-paste0(data$x,data$y,data$z)
data$mycol <- gsub(',','',data$mycol)

RowSums NA + NA gives 0 [duplicate]

This question already has answers here:
There is pmin and pmax each taking na.rm, why no psum?
(3 answers)
Closed 6 years ago.
I'll just understand a (for me) weird behavior of the function rowSums. Imagine I have this super simple dataframe:
a = c(NA, NA,3)
b = c(2,NA,2)
df = data.frame(a,b)
df
a b
1 NA 2
2 NA NA
3 3 2
and now I want a third column that is the sum of the other two. I cannot use simply + because of the NA:
df$c <- df$a + df$b
df
a b c
1 NA 2 NA
2 NA NA NA
3 3 2 5
but if I use rowSums the rows that have NA are calculated as 0, while if there is only one NA everything works fine:
df$d <- rowSums(df, na.rm=T)
df
a b c d
1 NA 2 NA 2
2 NA NA NA 0
3 3 2 5 10
am I missing something?
Thanks to all
One option with rowSums would be to get the rowSums with na.rm=TRUE and multiply with the negated (!) rowSums of negated (!) logical matrix based on the NA values after converting the rows that have all NAs into NA (NA^)
rowSums(df, na.rm=TRUE) *NA^!rowSums(!is.na(df))
#[1] 2 NA 10
Because
sum(numeric(0))
# 0
Once you used na.rm = TRUE in rowSums, the second row is numeric(0). After taking sum, it is 0.
If you want to retain NA for all NA cases, it would be a two-stage work. I recommend writing a small function for this purpose:
my_rowSums <- function(x) {
if (is.data.frame(x)) x <- as.matrix(x)
z <- base::rowSums(x, na.rm = TRUE)
z[!base::rowSums(!is.na(x))] <- NA
z
}
my_rowSums(df)
# [1] 2 NA 10
This can be particularly useful, if the input x is a data frame (as in your case). base::rowSums would first check whether input is matrix or not. If it gets a data frame, it would convert it into a matrix first. Type conversion is in fact more costly than actual row sum computation. Note that we call base::rowSums two times. To reduce type conversion overhead, we should make sure x is a matrix beforehand.
For #akrun's "hacking" answer, I suggest:
akrun_rowSums <- function (x) {
if (is.data.frame(x)) x <- as.matrix(x)
rowSums(x, na.rm=TRUE) *NA^!rowSums(!is.na(x))
}
akrun_rowSums(df)
# [1] 2 NA 10

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