Is it possible to set all column names to upper or lower within a dplyr or magrittr chain?
In the example below I load the data and then, using a magrittr pipe, chain it through to my dplyr mutations. In the 4th line I use the tolower function , but this is for a different purpose: to create a new variable with lowercase observations.
mydata <- read.csv('myfile.csv') %>%
mutate(Year = mdy_hms(DATE),
Reference = (REFNUM),
Event = tolower(EVENT)
I'm obviously looking for something like colnames = tolower but know this doesn't work/exist.
I note the dplyr rename function but this isn't really helpful.
In magrittr the colname options are:
set_colnames instead of base R's colnames<-
set_names instead of base R's names<-
I've tried numerous permutations with these but no dice.
Obviously this is very simple in base r.
names(mydata) <- tolower(names(mydata))
However it seems incongruous with the dplyr/magrittr philosophies that you'd have to do that as a clunky one liner, before moving on to an elegant chain of dplyr/magrittr code.
with {dplyr} we can do :
mydata %>% rename_all(tolower)
or
mydata %>% rename(across(everything(), tolower))
iris %>% setNames(tolower(names(.))) %>% head
Or equivalently use replacement function in non-replacement form:
iris %>% `names<-`(tolower(names(.))) %>% head
iris %>% `colnames<-`(tolower(names(.))) %>% head # if you really want to use `colnames<-`
Using magrittr's "compound assignment pipe-operator" %<>% might be, if I understand your question correctly, an even more succinct option.
library("magrittr")
names(iris) %<>% tolower
?`%<>%` # for more
mtcars %>%
set_colnames(value = casefold(colnames(.), upper = FALSE)) %>%
head
casefold is available in base R and can convert in both direction, i.e. can convert to either all upper case or all lower case by using the flag upper, as need might be.
Also colnames() will use only column headers for case conversion.
You could also define a function:
upcase <- function(df) {
names(df) <- toupper(names(df))
df
}
library(dplyr)
mtcars %>% upcase %>% select(MPG)
Related
I'm trying as per
dplyr mutate using variable columns
&
dplyr - mutate: use dynamic variable names
to use dynamic names in mutate. What I am trying to do is to normalize column data by groups subject to a minimum standard deviation. Each column has a different minimum standard deviation
e.g. (I omitted loops & map statements for convenience)
require(dplyr)
require(magrittr)
data(iris)
iris <- tbl_df(iris)
minsd <- c('Sepal.Length' = 0.8)
varname <- 'Sepal.Length'
iris %>% group_by(Species) %>% mutate(!!varname := mean(pluck(iris,varname),na.rm=T)/max(sd(pluck(iris,varname)),minsd[varname]))
I got the dynamic assignment & variable selection to work as suggested by the reference answers. But group_by() is not respected which, for me at least, is the main benefit of using dplyr here
desired answer is given by
iris %>% group_by(Species) %>% mutate(!!varname := mean(Sepal.Length,na.rm=T)/max(sd(Sepal.Length),minsd[varname]))
Is there a way around this?
I actually did not know much about pluck, so I don't know what went wrong, but I would go for this and this works:
iris %>%
group_by(Species) %>%
mutate(
!! varname :=
mean(!!as.name(varname), na.rm = T) /
max(sd(!!as.name(varname)),
minsd[varname])
)
Let me know if this isn't what you were looking for.
The other answer is obviously the best and it also solved a similar problem that I have encountered. For example, with !!as.name(), there is no need to use group_by_() (or group_by_at or arrange_() (or arrange_at()).
However, another way is to replace pluck(iris,varname) in your code with .data[[varname]]. The reason why pluck(iris,varname) does not work is that, I suppose, iris in pluck(iris,varname) is not grouped. However, .data refer to the tibble that executes mutate(), and so is grouped.
An alternative to as.name() is rlang::sym() from the rlang package.
Consider a data.frame with a mix of data types.
For a weird purpose, a user needs to convert all columns to characters.
How is it best done? A tidyverse attempt at solution is this:
map(mtcars,as.character) %>% map_df(as.list) %>% View()
c2<-map(mtcars,as.character) %>% map_df(as.list)
when I call str(c2) it should say a tibble or data.frame with all characters.
The other option would be some parameter settings for write.csv() or in write_csv() to achieve the same thing in the resulting file output.
EDIT: 2021-03-01
Beginning with dplyr 1.0.0, the _all() function variants are superceded. The new way to accomplish this is using the new across() function.
library(dplyr)
mtcars %>%
mutate(across(everything(), as.character))
With across(), we choose the set of columns we want to modify using tidyselect helpers (here we use everything() to choose all columns), and then specify the function we want to apply to each of the selected columns. In this case, that is as.character().
Original answer:
You can also use dplyr::mutate_all.
library(dplyr)
mtcars %>%
mutate_all(as.character)
In base R:
x[] <- lapply(x, as.character)
This converts the columns to character class in place, retaining the data.frame's attributes. A call to data.frame() would cause them to be lost.
Attribute preservation using dplyr: Attributes seem to be preserved during dplyr::mutate(across(everything(), as.character)). Previously they were destroyed by dplyr::mutate_all.
Example
x <- mtcars
attr(x, "example") <- "1"
In the second case below, the example attribute is retained:
# Destroys attributes
data.frame(lapply(x, as.character)) %>%
attributes()
# Preserves attributes
x[] <- lapply(x, as.character)
attributes(x)
This might work, but not sure if it's the best.
df = data.frame(lapply(mtcars, as.character))
str(df)
Most efficient way using data.table-
data.table::setDT(mtcars)
mtcars[, (colnames(mtcars)) := lapply(.SD, as.character), .SDcols = colnames(mtcars)]
Note: You can use this to convert few columns of a data table to your desired column type.
If we want to convert all columns to character then we can also do something like this-
to_col_type <- function(col_names,type){
get(paste0("as.", type))(dt[[col_names]])
}
mtcars<- rbindlist(list(Map(to_col_type ,colnames(mtcars),"character")))
mutate_all in the accepted answer is superseded.
You can use mutate() function with across():
library(dplyr)
mtcars %>%
mutate(across(everything(), as.character))
I recently discovered the pipe operator %>%, which can make code more readable. Here is my MWE.
library(dplyr) # for the pipe operator
library(lsr) # for the cohensD function
set.seed(4) # make it reproducible
dat <- data.frame( # create data frame
subj = c(1:6),
pre = sample(1:6, replace = TRUE),
post = sample(1:6, replace = TRUE)
)
dat %>% select(pre, post) %>% sapply(., mean) # works as expected
However, I struggle using the pipe operator in this particular case
dat %>% select(pre, post) %>% cohensD(.$pre, .$post) # piping returns an error
cohensD(dat$pre, dat$post) # classical way works fine
Why is it not possible to subset columns using the placeholder .in combination with $? Is it worthwhile to write this line using a pipe operator %>%, or does it complicate syntax? The classical way of writing this seems more concise.
This would work:
dat %>% select(pre, post) %>% {cohensD(.$pre, .$post)}
Wrapping the last call into curly braces makes it be treated like an expression and not a function call. When you pipe something into an expression, the . gets replaced as expected. I often use this trick to call a function which does not interface well with piping.
What is inside the braces happens to be a function call but could really be any expression of . .
Since you're going from a bunch of data into one (row of) value(s), you're summarizing. in a dplyr pipeline you can then use the summarize function, within the summarize function you don't need to subset and can just call pre and post
Like so:
dat %>% select(pre, post) %>% summarize(CD = cohensD(pre, post))
(The select statement isn't actually necessary in this case, but I left it in to show how this works in a pipeline)
It doesn't work because the . operator has to be used directly as an argument, and not inside a nested function (like $...) in your call.
If you really want to use piping, you can do it with the formula interface, but with a little reshaping before (melt is from reshape2 package):
dat %>% select(pre, post) %>% melt %>% cohensD(value~variable, .)
#### [1] 0.8115027
In dplyr, you can imply the data frame and pass it down to subsequent functions like so:
df <- df %>% select(one_of(c("Species", "Genus"))) %>%
mutate(newcol = sum(length+width)
Is there a way in the syntax to make it so that I don't have to repeat the df on the right side of the arrow?
It is not available in dplyr but is in magrittr library.
library(magrittr)
library(dplyr)
df %<>% ...
The %<>% operator says "pass to the functions on the right and than save to the object with the same name". For example:
mtcars %<>% mutate(cyl_plus_gear = cyl + gear) %>% arrange(cyl_plus_gear)
Thanks; I also found that this can be done by simply passing in the data frame as the first argument to the first function call in deployer.
Ex:
df <- select(df,1:10)%>%filter(value > 10)
I have a dplyr question: How do I use transmute over each column without writing each column out by hand? I.e. is there something like transmute_each()?
I want to do the following: Using dplyr I want to get the z-score of each column for a MWE below:
tickers <- c(rep(1,10),rep(2,10))
df <- data.frame(cbind(tickers,rep(1:20),rep(2:21),rep(2:21),rep(4:23),rep(3:22)))
colnames(df) <- c("tickers","col1","col2","col3","col4","col5")
df %>% group_by(tickers)
Is there a simple way to then use transmute to achieve the following:
for(i in 2:ncol(df)){
df[,i] <- df[,i] - mean(df[,i])/sd(df[,i])
}
Many thanks
Now that there is a transmute_at() function (as of dplyr 0.7), you can do the following:
df %>%
group_by(tickers) %>%
transmute_at(.vars = vars(starts_with("col")),
.funs = funs(scale(.))) %>%
ungroup
Note that this uses the scale() function from base R, which by default converts a numeric vector into a z-score.
Also, the use of vars() in the .vars argument allows you to use all the helper functions that are available for dplyr's select(), such as one_of(), ends_with(), etc.
Finally, instead of writing funs(scale(.)) here, since you're using a simple function in the .funs argument, you can just write .funs = scale.
I solved this using the following:
df %>%
group_by(tickers) %>%
mutate_at(.funs = funs((. - mean(.))/sd(.)),
.cols = vars(matches("col")))