I'm trying to assign something to the output of Sys.Date(). For example, say I want to name the object with today's date:
format(Sys.Date(), "%b%d") <- mtcars
I get the error:
invalid (NULL) left side of assignment.
I tried:
eval(parse(text = format(Sys.Date(), "%b%d"))) <- mtcars
Gets the same error msg. What am I missing / is there a solution?
We can use assign
assign(format(Sys.Date(), "%b%d"), mtcars)
head(Sep11)
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
Related
I want to rename() some variables in my data programmatically, so I can to it via map at some point.
I'm looking for the equivalent of,
library(tidyverse)
mtcars %>% rename(
"MPG" = "mpg"
)
but using environment variables instead. I tried !!sym() by doing the following,
library(tidyverse)
new_name <- "MPG"
old_name <- "mpg"
mtcars %>% rename(
!!sym(new_name) = !!sym(old_name)
)
However, I get the error Error: unexpected ')' in ")". I am not sure what I am missing here!
We could use setNames and evaluate (!!!)
head(mtcars %>%
rename(!!! setNames(old_name, new_name)))
-output
MPG cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 6 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
You can use {{}} -
library(dplyr)
new_name <- "MPG"
old_name <- "mpg"
mtcars %>% rename({{new_name}} := {{old_name}}) %>% head
# MPG cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
I have a set a variables say Var1, Var2 to Varn. They all take three possible values 0, 1, and 2. I want to replace all 2 as 1
like so
df$Var1[df$Var1 >= 1] <- 1
This does the job. But when I try to write a function to do this
MakeBinary <- function(varName dfName){dfName$varName[dfName$varNAme > = 1] <- 1}
and use this function like:
MakeBinary(Var2, df)
I got an error message: Error in $<-.data.frame(*tmp*, "varName", value = numeric(0)) :
replacement has 0 rows, data has 512.
I just want to know why I got this message. Thanks. My sample size is 512.
If we are passing column name as string, then use [[ instead of $ and return the dataset
MakeBinary <- function(varName, dfName){
dfName[[varName]][dfName[[varName]] >= 1] <- 1
dfName
}
MakeBinary("Var2", df)
example with mtcars
MakeBinary("carb", head(mtcars))
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 1
#Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 1
#Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 1
#Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
Unquoted arguments for variable names can be passed as well, but it needs to be converted to string
MakeBinary <- function(varName, dfName){
varName <- deparse(substitute(varName))
dfName[[varName]][dfName[[varName]] >= 1] <- 1
dfName
}
MakeBinary(Var2, df)
Using a reproducible example with mtcars
MakeBinary(carb, head(mtcars))
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 1
#Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 1
#Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 1
#Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
i want to swap a specific column with the last column, and then delete the last column after swapping. After delete ncol(testFrame) will decrease by 1
Usually a reproducible example is expected but your description is clear enough to understand what you want to do.
Using mtcars as sample data
df <- mtcars
head(df)
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
#Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
#Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
#Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
#Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
#Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
swap_column <- 3
cols <- seq_len(ncol(df))
df1 <- df[replace(cols, cols == swap_column, ncol(df))][-ncol(df)]
head(df1)
# mpg cyl carb hp drat wt qsec vs am gear
#Mazda RX4 21.0 6 4 110 3.90 2.620 16.46 0 1 4
#Mazda RX4 Wag 21.0 6 4 110 3.90 2.875 17.02 0 1 4
#Datsun 710 22.8 4 1 93 3.85 2.320 18.61 1 1 4
#Hornet 4 Drive 21.4 6 1 110 3.08 3.215 19.44 1 0 3
#Hornet Sportabout 18.7 8 2 175 3.15 3.440 17.02 0 0 3
#Valiant 18.1 6 1 105 2.76 3.460 20.22 1 0 3
We replace the column number swap_column with last column number (ncol(df)) and then remove the last column (-ncol(df)).
We can do this conveniently with add_column from tibble. The .after and .before parameters can take either column index or column name. Suppose, we need to shift last column to third position
library(tibble)
data(mtcars)
df1 <- add_column(mtcars[-ncol(mtcars)], mtcars[ncol(mtcars)], .after = 2)
head(df1)
# mpg cyl carb disp hp drat wt qsec vs am gear
#Mazda RX4 21.0 6 4 160 110 3.90 2.620 16.46 0 1 4
#Mazda RX4 Wag 21.0 6 4 160 110 3.90 2.875 17.02 0 1 4
#Datsun 710 22.8 4 1 108 93 3.85 2.320 18.61 1 1 4
#Hornet 4 Drive 21.4 6 1 258 110 3.08 3.215 19.44 1 0 3
#Hornet Sportabout 18.7 8 2 360 175 3.15 3.440 17.02 0 0 3
#Valiant 18.1 6 1 225 105 2.76 3.460 20.22 1 0 3
Can you sort a df based on object class? Say
data("mtcars")
mtcars$cyl <- as.factor(mtcars$cyl)
mtcars$vs <- as.factor(mtcars$vs)
mtcars$am <- as.factor(mtcars$am)
sapply(mtcars,class)
and I want all numeric variables first and then all factors at the end? I want to be able to do this on a much larger dataset so I prefer solutions that do not rely on subsetting by column number. Cheers.
Maybe this one?
head(mtcars)
# mpg cyl disp hp drat wt qsec vs am gear carb
# Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
# Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
# Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
# Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
# Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
# Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
x <- mtcars[,names(sort(unlist(lapply(mtcars, class)), decreasing = T))]
head(x)
# mpg disp hp drat wt qsec gear carb cyl vs am
# Mazda RX4 21.0 160 110 3.90 2.620 16.46 4 4 6 0 1
# Mazda RX4 Wag 21.0 160 110 3.90 2.875 17.02 4 4 6 0 1
# Datsun 710 22.8 108 93 3.85 2.320 18.61 4 1 4 1 1
# Hornet 4 Drive 21.4 258 110 3.08 3.215 19.44 3 1 6 1 0
# Hornet Sportabout 18.7 360 175 3.15 3.440 17.02 3 2 8 0 0
# Valiant 18.1 225 105 2.76 3.460 20.22 3 1 6 1 0
In x, as you see, the columns cyl, vs and am that are of class factor are place at the end and those of class numeric first.
I would like to define a string
string<- "modelName"
That could be used to name an object later. Something like
paste0(string) <- mtcars
cat(string) <- mtcars
print(string) <- mtcars
get(string) <- mtcars
The needed result is the dataset called "modelName". None of the examples above work, obviously.
Question:
How can create one create an object which name is defined by the sourced string?
As #Spacedman notes this is not generally the way things are done but you can use assign
string<- "modelName"
assign(string, mtcars)
> head(modelName)
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1
In general it may be perferable to use sometthing like a list:
x <- list()
x[[string]] <- mtcars
> head(x$modelName)
mpg cyl disp hp drat wt qsec vs am gear carb
Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1