Consider the following data frame:
df <- setNames(data.frame(1:5,rep(1,5)), c("id", "value"))
I want to change the names for multiple cells in the column "id". Let's say I want to change the following:
df$id[df$id %In% 2:3] <- 1
df$id[df$id == 4] <- 3
However, instead of using the code above, I want to create a function, where I can do the transformation more "smooth" (because I have a lot of data frames, where I need to change the names for the cells). I want to create a function:
mapping <- function(...) {
...
}
where I afterward can create a simple and smooth mapping function for my df, where I only have to specific the "old" and the "new" names for the cells. Something like this:
df_mapping <- function(...) {
2.1
3.1
4.3
}
And then I can apply the function on my data and specific which column it should do it for, and it will work in the same way as the code with gsub:
df <- df_mapping(df,id)
Is it possible to create that mapping function?
if we need a function, then can have a 'data' argument, column name, values to replace and replacer value, then create the logical condition, subset the column, assign with replacer_val and return the dataset after the assignment
f1 <- function(dat, colnm, values_to_replace, replacer_val) {
dat[[colnm]][dat[[colnm]] %in% values_to_replace] <- replacer_val
return(dat)
}
f1(df, "id", c(2, 3), 1)
-output
# id value
#1 1 1
#2 1 1
#3 1 1
#4 4 1
#5 5 1
To replace values with corresponding sets of replacers,
f2 <- function(dat, colnm, values_to_replace, replacer_vals) {
nm1 <- setNames(replacer_vals, values_to_replace)
v1 <- nm1[as.character(dat[[colnm]])]
i1 <- !is.na(v1)
dat[[colnm]][i1] <- v1[i1]
return(dat)
}
f2(df, "id", c(2, 3), c(5, 6))
# id value
#1 1 1
#2 5 1
#3 6 1
#4 4 1
#5 5 1
Or another option is to create a key/value dataset and use merge or join
library(data.table)
f3 <- function(dat, colnm, values_to_replace, replacer_vals) {
keydat <- data.frame(key = values_to_replace, val = replacer_vals)
names(keydat)[1] <- colnm
dt <- as.data.table(dat)
dt[keydat, (colnm) := val, on = colnm][]
return(dt)
}
f3(df, "id", c(2, 5), c(3, 6))
Maybe a mapping like below could help
mapping <- function(df, id, to_replace, obj_value) {
transform(df, id = replace(id, id %in% to_replace, obj_value))
}
e.g.,
> mapping(df, id, c(2, 3), 1)
id value
1 1 1
2 1 1
3 1 1
4 4 1
5 5 1
You can use dplyr's recode function
mapping <- function(data, col, old, new) {
data[[col]] <- dplyr::recode(data[[col]], !!!setNames(new, old))
data
}
mapping(df, "id", c(2, 3), c(7L, 8L))
# id value
#1 1 1
#2 7 1
#3 8 1
#4 4 1
#5 5 1
Related
I'd like to assign a dynamic name, i.e. the name of the variable I'm passing to the function, as a column name in the dataframe that is being created by the function.
I've tried
- deparse(substitute(x))
- toString(x)
but no success...
Code
a <- (1:3)
b <- (5:7)
df <- data.frame(a,b)
fun <- function(x){
x %>% mutate(c=a+b)
colnames(x)[3] <- deparse(substitute(x))
}
Expected behaviour
after running fun(df):
a b df
1 1 5 6
2 2 6 8
3 3 7 10
instead:
> fun(df)
Error in names(x) <- value :
'names' attribute [3] must be the same length as the vector [2]
We can use := with evaluation (!!)
fun <- function(x){
nm1 <- deparse(substitute(x))
x %>%
mutate(!! nm1 := a+b)
}
fun(df)
# a b df
#1 1 5 6
#2 2 6 8
#3 3 7 10
In the OP's function, the the output of x %>% mutate is not assigned back, therefore, the original dataset have only two columns and not three i.e. if we do
fun <- function(x){
nm1 <- deparse(substitute(x))
x <- x %>% # assign the output after mutate
mutate(c=a+b)
colnames(x)[3] <- nm1
x # return the dataset
}
fun(df)
# a b df
#1 1 5 6
#2 2 6 8
#3 3 7 10
To rename a specific variable I can do for instance
names(df1)[which(names(df1) == "C")] <- "X"
> df1
A B X
1 1 2 3
I wonder if this is also possible with setNames(), but without repeating the names I don't want to rename as in
df1 <- setNames(df1, c("A", "B", "X"))`
I've tried setNames(df1, c(rep(NA, 2), "X")) and setNames(df1[3], "X") but this won't work. The advantage I see in setNames() is that I can set names while doing other stuff in one step.
Data
df1 <- setNames(data.frame(matrix(1:3, 1)), LETTERS[1:3])
> df1
A B C
1 1 2 3
You can use replace,
setNames(df1, replace(names(df1), names(df1) == 'B', 'X'))
# A X C
#1 1 2 3
setNames(df1, replace(names(df1), names(df1) == 'A', 'X'))
# X B C
#1 1 2 3
setNames(df1, replace(names(df1), names(df1) == 'C', 'X'))
# A B X
#1 1 2 3
You can do it using setnames from library(data.table)
library(data.table)
setnames(DF, "oldName", "newName")
dplyr also has a special function for this:
dplyr::rename(df1, X = C)
# A B X
# 1 1 2 3
Because names of data is a vector, I try to use ifelse() to identify elements logically.
setNames(df1, ifelse(names(df1) == "A", "X", names(df1)))
X B C
1 1 2 3
setNames(df1, ifelse(names(df1) == "B", "X", names(df1)))
A X C
1 1 2 3
setNames(df1, ifelse(names(df1) == "C", "X", names(df1)))
A B X
1 1 2 3
Best I can do is this one, which doesn't seem any easier than using other methods. You could also write a function that would fit your needs..
df2 <- setNames(df1, c(colnames(df1)[1:2],"test"))
> df2
A B test
1 1 2 3
Edit: to change other names (for example column B), we can define a custom function:
dfrename <- function(mydf, mycolumns=1:ncol(mydf), mynewnames=c(letters[1:mycolumns])) {
if(!is.numeric(mycolumns)) {
toreplace <- colnames(mydf) %in% mycolumns
} else {
toreplace <- 1:ncol(mydf) %in% mycolumns
}
mycols <- colnames(mydf)
mycols[toreplace] <- mynewnames
res <- setNames(mydf, mycols)
return(res)
}
You can either use the indexes of the columns to replace or their names.
> dfrename(df1, 2, "test")
A test C
1 1 2 3
Another base R solution if you're ok to repeat the name of the old variable:
res <- transform(iris, a = Species, Species = NULL)
# [1] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width" "a"
Regarding efficiency I'm not sure if the data is copied or not.
I'm looking for the (1) name and (2) a (cleaner) method in R (base and data.table preferred) of the following.
Input
> d1
id x y
1 1 1 NA
2 2 NA 3
3 3 4 NA
> d2
id x y z
1 4 NA 30 a
2 3 20 2 b
3 2 14 NA c
4 1 15 97 d
(note that the actual data.frames have hundreds of columns)
Expected output:
> d1
id x y z
1 1 1 97 d
2 2 14 3 c
3 3 4 2 b
Data and current solution:
d1 <- data.frame(id = 1:3, x = c(1, NA, 4), y = c(NA, 3, NA))
d2 <- data.frame(id = 4:1, x = c(NA, 20, 14, 15), y = c(30, 2, NA, 97), z = letters[1:4])
for (col in setdiff(names(d1), "id")) {
# If missing look in d2
missing <- is.na(d1[[col]])
d1[missing, col] <- d2[match(d1$id[missing], d2$id), col]
}
for (col in setdiff(names(d2), names(d1))) {
# If column missing then add
d1[[col]] <- d2[match(d1$id, d2$id), col]
}
PS:
Likely this questions has been asked before but I'm lacking in vocabulary to search it.
Assuming you are working with 2 data.frames, here is a base solution
#expand d1 to have the same columns as d2
d <- merge(d1, d2[, c("id", setdiff(names(d2), names(d1))), drop=FALSE],
by="id", all.x=TRUE, all.y=FALSE)
#make sure that d2 also have same number of columns as d1
d2 <- merge(d2, d1[, c("id", setdiff(names(d1), names(d2))), drop=FALSE],
by="id", all.x=TRUE, all.y=FALSE)
#align rows and columns to match those in d1
mask <- d2[match(d1$id, d2$id), names(d)]
#replace NAs with those mask
replace(d, is.na(d), mask[is.na(d)])
If you dont mind, we can rewrite your question into a general matrix-coalesce question (i.e. any number of matrices, columns, rows) which seems like it has not been asked before.
edit:
Another base R solution is a hack of coalesce1a from How to implement coalesce efficiently in R
coalesce.mat <- function(...) {
ans <- ..1
for (elt in list(...)[-1]) {
rn <- match(ans$id, elt$id)
ans[is.na(ans)] <- elt[rn, names(ans)][is.na(ans)]
}
ans
}
allcols <- Reduce(union, lapply(list(d1, d2), names))
do.call(coalesce.mat,
lapply(list(d1, d2), function(x) {
x[, setdiff(allcols, names(x))] <- NA
x
}))
edit:
a possible data.table solution using coalesce1a from How to implement coalesce efficiently in R by Martin Morgan.
coalesce1a <- function(...) {
ans <- ..1
for (elt in list(...)[-1]) {
i <- which(is.na(ans))
ans[i] <- elt[i]
}
ans
}
setDT(d1)
setDT(d2)
#melt into long formats and full outer join the 2
mdt <- merge(melt(d1, id.vars="id"), melt(d2, id.vars="id"), by=c("id","variable"), all=TRUE)
#perform a coalesce on vectors
mdt[, value := do.call(coalesce1a, .SD), .SDcols=grep("value", names(mdt), value=TRUE)]
#pivot into original format and subset to those in d1
dcast.data.table(mdt, id ~ variable, value.var="value")[
d1, .SD, on=.(id)]
Here is a possibility using dplyr::left_join:
left_join(d1, d2, by = "id") %>%
mutate(
x = ifelse(!is.na(x.x), x.x, x.y),
y = ifelse(!is.na(y.x), y.x, y.y)) %>%
select(id, x, y, z)
# id x y z
#1 1 1 97 d
#2 2 14 3 c
#3 3 4 2 b
We can use data.table with coalesce from dplyr. Create a vector of column names that are common ('nm1') and difference ('nm2') in both datasets. Convert the first dataset to 'data.table' (setDT(d1)), join on the 'id' column, assign (:=) the coalesced columns of the first and second (with prefix i. - if there are common columns) to update the values in the first dataset
library(data.table)
nm1 <- setdiff(intersect(names(d1), names(d2)), 'id')
nm2 <- setdiff(names(d2), names(d1))
setDT(d1)[d2, c(nm1, nm2) := c(Map(dplyr::coalesce, mget(nm1),
mget(paste0("i.", nm1))), mget(nm2)), on = .(id)]
d1
# id x y z
#1: 1 1 97 d
#2: 2 14 3 c
#3: 3 4 2 b
I want to create a for loop by variable names.
Each time, I calculte the max between each two variables, and define a new one in data df. New variables look like this:var1_1, var1_2... Here is my code:
df=data.frame(matrix(c(1:6), nrow = 2))
colnames(df) = c("x", "y", "z")
for(i in length(names(df))-1){
df = df %>% mutate(paste0("var", i, "_", i+1) = max(names(df)[i], names(df)[i+1]))
}
But there gives error.
Expected output:
>df
x y z var1_2 var1_3 var2_3
1 3 5 3 5 5
2 4 6 4 6 6
One way via base R,
m1 <- sapply(combn(names(df),2, simplify = FALSE), function(i) do.call(pmax, df[i]))
nms <- combn(ncol(m1), 2, function(i) paste0('Var', i[1], '_', i[2]))
cbind(df, setNames(data.frame(m1), nms))
# x y z Var1_2 Var1_3 Var2_3
#1 1 3 5 3 5 5
#2 2 4 6 4 6 6
If you really want to use a Loop you can try:
ind<-combn(3,2)
for(i in 1:dim(df)[2]){
i <- ind[,i]
name <- paste0("var", i[1], "_", i[2])
val <- names(df)[i[ifelse(sum(df[,i[1]]) > sum(df[,i[2]]),1,2)]]
df <- mutate_(df, .dots= setNames(list(val),name))
}
I have a big data set (roughly 10 000 rows), and want to create a function that counts the number of complete cases (not NAs) per group. I tried various functions (aggregate, table, sum(complete.cases), group_by, etc), but somehow I miss one - probably little - trick. Thanks for any help!
A little sample data set to explain, the result I need.
x <- data.frame(group = c(1:4),
age = c(4:1, c(11, NA,13, NA)),
speed = c(12, NA,15,NA))
print(x)
# group age speed
#1 1 4 12
#2 2 3 NA
#3 3 2 15
#4 4 1 NA
#5 1 11 12
#6 2 NA NA
#7 3 13 15
#8 4 NA NA
One function I wrote reads as follows:
CountPerGroup <- function(group) {
data.set <- subset(x,group %in% group)
vect <- vector()
for (i in 1:length(group)) {
vect[i] <- sum(complete.cases(data.set))
}
output <- data.frame(cbind(group,count=vect))
return(output)
}
The result of
CountPerGroup(2:1)
is
group count
1 2 4
2 1 4
Unfortunately, this is wrong. Instead the outcome should look like
group count
1 2 1
2 1 4
What am I missing? How can I tell R to count of complete.cases per Group?
Thank you very much for any help on this!
Something like should do the trick if you wish to maintain your functionality:
x <- data.frame(group = c(1:4),
age = c(4:1, c(11, NA,13, NA)),
speed = c(12, NA,15,NA))
CountPerGroup <- function(x, groups) {
data.set <- subset(x, group %in% groups)
ans <- sapply(split(data.set, data.set$group),
function(y) sum(complete.cases(y)))
return(data.frame(group = names(ans), count = unname(ans)))
}
CountPerGroup(x, 1:2)
# group count
#1 1 2
#2 2 0
Which is correct from what I can count. But it does not agree with your suggested outcome.
EDIT
It seems that you want the number of non-NA instead and correctly sorted. Use this function instead:
CountPerGroup2 <- function(x, groups) {
data.set <- subset(x, group %in% groups)
ans <- sapply(split(data.set, data.set$group),
function(y) sum(!is.na(y[, !grepl("group", names(y))])))[groups]
return(data.frame(group = names(ans), count = unname(ans)))
}
CountPerGroup2(x, 2:1)
# group count
#1 2 1
#2 1 4
If you are just looking for a way to get the full count of non-NA values per group, you could use something like:
library(plyr)
x <- data.frame(group = c(1:4),
age = c(4:1, c(11, NA,13, NA)),
speed = c(12, NA,15,NA))
counts <- ddply(x, "group", summarize, count=sum(!is.na(c(age, speed))))
## group count
## 1 1 4
## 2 2 1
## 3 3 4
## 4 4 1
You do miss out on having a function that lets you query a subset of the groups, but you get a one-line way to calculate the full solution.
Here is a way with data.table
library(data.table)
library(functional)
countPerGroup = function(x, vec)
{
dt = data.table(x)
d1 = setkey(dt, group)[group %in% vec]
d2 = d1[,lapply(.SD, Compose(Negate(is.na), sum)),by=group]
transform(d2, count=age+speed, speed=NULL, age=NULL)
}
countPerGroup(x, 1:2)
# group count
#1: 1 4
#2: 2 1
countPerGroup(x, c(1,2))
# group count
#1: 1 4
#2: 2 1
If you have a high number of lines in your data.table, it is particularly efficient!
I just had the same problem and found an easier solution
library(data.table)
x <- data.table(group = c(1:4),
age = c(4:1, c(11, NA,13, NA)),
speed = c(12, NA,15,NA))
x[,sum(complete.cases(.SD)), by=group]