Get data frame from character variable - r

I pass data frame name as string into a function. How do I get content of referenced data frame from the string? Suppose I have string 'mtcars' and I want to print data frame mtcars:
printdf <- function(dataframe) {
print(dataframe)
}
printdf('mtcars');

I think you'll need a get in there if the input is a string. Also, depending on your usage of the function, the explicit print might not be necessary:
printdf <- function(dataframe) {
get(dataframe)
# print(get(dataframe))
}
head(printdf("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

Related

RStudio: colnames() function not showing name of very first column

When I run colnames(), it never shows the name of this first column.
For example, after wasting a lot of time researching online, I discovered the name of the first column in mtcars is das_Auto.
Why doesn't this name show when I run this code?
[colnames(mtcars)][1]
What's the easiest way to determine the name of the first column in a data set?
This is because the first 'column' of mtcars is not actually a column but an index. If you want to convert it to a column you can run the below:
df <- cbind(das_Auto = rownames(mtcars), mtcars)
rownames(df) <- 1:nrow(mtcars)
head(df)
das_Auto mpg cyl disp hp drat wt qsec vs am gear carb
1 Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4
2 Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4
3 Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1
4 Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1
5 Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2
6 Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1

How to create function to use regular expressions to replace column names in a data frame?

I am feeling lost with how to create a helper function in R that takes the following 3 arguments:
a data frame,
a string pattern, and
a string "replacement pattern".
The function is supposed to replace occurrences of the string pattern in the names of the variables in the data frame with the replacement pattern.
Any guidance, tips or help would be greatly appreciated.
func <- function(x, nm1, nm2, ...) {
names(x) <- gsub(nm1, nm2, names(x), ...)
x
}
head(func(mtcars, "c", "C"))
# 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

Convert LIst To Dataframe Using For Loop And Saving Under Different Names In R

I am trying to convert my list consisting of 52 components to a dataframe for each of the components.
Without using the for loop will look something like this which is tedious:
df1 = as.data.frame(list[1])
df2 = as.data.frame(list[2])
df3 = as.data.frame(list[3])
.
.
.
df50 = as.data.frame(list[50])
How do I achieve this using the for loop? My attempt:
for (i in seq_along(list)) {
noquote(paste0("df", i)) = as.data.frame(list[i])
}
Error: target of assignment expands to non-language objec
I think I'll have to invovle assign.
If you have list of dataframes in list, you can name them and then use list2env to have them as separate dataframes in the environment.
names(list) <- paste0('df', seq_along(list))
list2env(list, .GlobalEnv)
Using a reproducible exmaple,
temp <- list(mtcars, mtcars)
names(temp) <- paste0('df', seq_along(temp))
list2env(temp, .GlobalEnv)
head(df1)
# 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
head(df2)
# 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
However, note that
list is an internal function in R, so it is better to name your variables something else.
As #MrFlick suggested try to keep your data in a list as lists are easier to manage rather than creating numerous objects in your global environment.
We can use assign instead of noquote from the OP's function
for (i in seq_along(list)) {
assign(paste0("df", i), value = list[[i]])
}

R: Sort columns by object class

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.

how to define the name of a new object with a string?

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

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