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
Related
I am just trying to fill gaps but in a loop. It is a monthly data, and fill_gaps produces NAs for every day. I am not sure why.
for (x in 2:length(differencing)){
for(micky in 1:length(differencing$`d_ BA`)){
if(is.na(differencing[micky,x])== T){
differencing[micky,x] = differencing[micky-1,x]
}
}
}
here is the error that I am getting:
Error: Assigned data `differencing[(micky - 1), x]` must be compatible with row subscript `micky`.
x 1 row must be assigned.
x Assigned data has 0 rows.
i Row updates require a list value. Do you need `list()` or `as.list()`?
Run `rlang::last_error()` to see where the error occurred.
This can be easily done using fill
library(tidyr)
library(dplyr)
differencing %>%
fill(everything())
Or we can use na.locf from zoo
library(zoo)
na.locf(differencing)
In the OP's loop, in the first line, it would be
for (x in 2:length(differencing$`d_ BA`)
...
as length of a data.frame will be the number of columns (as mentioned in the comments) and is different from length of a column i.e. vector
As the OP mentioned none of them works (OP didn't provide any example), using a small reproducible example ('tmp')
tmp %>%
fill(everything())
# 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 6 258 110 3.15 3.440 17.02 0 0 3 2
#Valiant 18.1 6 258 110 2.76 3.460 20.22 1 0 3 1
or using na.locf
na.locf(tmp)
# 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 6 258 110 3.15 3.440 17.02 0 0 3 2
#Valiant 18.1 6 258 110 2.76 3.460 20.22 1 0 3 1
data
tmp <- head(mtcars)
tmp[c(2, 5, 6), c(3, 4, 2)] <- NA
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
I'm new to writing functions in R, but want to write a function to add 1% of the median of a variable to itself, using dplyr, and replace the variable with this transformation.
x is a numeric variable.
add_median <- function(df, x) {
x <- enquo(x)
x <- quo_name(x)
mutate(x=x+.01*median(x, na.rm=T))
}
When I run newDF <- DF %>% add_median(variable_of_interest), I get the following error:
Error in 0.01 * median(x, na.rm = T) : non-numeric argument to binary operator
What am I doing wrong here?
We could change the function to evaluate with {{}} and then use assign (:=) instead of = in mutate
library(dplyr)
add_median <- function(df, x) {
df %>%
mutate({{x}} := {{x}} + .01 * median({{x}}, na.rm = TRUE))
}
If we need to change multiple columns, use mutate_at
add_median_multiple <- function(df, vec){
df %>%
mutate_at(vars(vec), ~ . + .01 * median(., na.rm = TRUE))
}
-testing
data(mtcars)
head(mtcars) %>%
add_median(mpg)
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.21 6 160 110 3.90 2.620 16.46 0 1 4 4
#Mazda RX4 Wag 21.21 6 160 110 3.90 2.875 17.02 0 1 4 4
#Datsun 710 23.01 4 108 93 3.85 2.320 18.61 1 1 4 1
#Hornet 4 Drive 21.61 6 258 110 3.08 3.215 19.44 1 0 3 1
#Hornet Sportabout 18.91 8 360 175 3.15 3.440 17.02 0 0 3 2
#Valiant 18.31 6 225 105 2.76 3.460 20.22 1 0 3 1
comparison with original 'mpg' column
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
add_median_multiple(head(mtcars), c('mpg', 'wt'))
# mpg cyl disp hp drat wt qsec vs am gear carb
#Mazda RX4 21.21 6 160 110 3.90 2.65045 16.46 0 1 4 4
#Mazda RX4 Wag 21.21 6 160 110 3.90 2.90545 17.02 0 1 4 4
#Datsun 710 23.01 4 108 93 3.85 2.35045 18.61 1 1 4 1
#Hornet 4 Drive 21.61 6 258 110 3.08 3.24545 19.44 1 0 3 1
#Hornet Sportabout 18.91 8 360 175 3.15 3.47045 17.02 0 0 3 2
#Valiant 18.31 6 225 105 2.76 3.49045 20.22 1 0 3 1
I'm trying to calculate a new column with a user defined function that needs data from same row and a fixed value valid for all rows:
myfunc <- function(ds,colname,val1,col1,col2){
# content of new column <colname> should be computed from:
ds[colname] = val1 + ds[col1] * ds[col2] # for each row of ds
return(ds)
}
v1 = 2
data(mtcars)
mt = head(mtcars)
mt
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
apply(mt,'newcol',v1,mt$wt,mt$qsec)
mt
What I would like to see in mt$newcol in first row is: 2 + 2.620 * 16.46 (-> 45.12) and all other rows similiar.
So, how can I send a fixed value (v1) and two values from each row to my function and store returned value in this row in a new column?
Thanks
dplyr approach:
library(dplyr)
data(mtcars)
myfunc <- function(ds, new_column, val1, col1, col2){
name <- rownames(ds)
ds <- ds %>%
mutate(!!as.name(new_column) := val1 + !!as.name(col1) + !!as.name(col2),
car_name = name) %>%
select(car_name, mpg:!!as.name(new_column))
return(ds)
}
head(
myfunc(ds = mtcars,
new_column = "new_column",
val1 = 2,
col1 = "hp",
col2 = "vs")
)
output
car_name mpg cyl disp hp drat wt qsec vs am gear carb new_column
1 Mazda RX4 21.0 6 160 110 3.90 2.620 16.46 0 1 4 4 112
2 Mazda RX4 Wag 21.0 6 160 110 3.90 2.875 17.02 0 1 4 4 112
3 Datsun 710 22.8 4 108 93 3.85 2.320 18.61 1 1 4 1 96
4 Hornet 4 Drive 21.4 6 258 110 3.08 3.215 19.44 1 0 3 1 113
5 Hornet Sportabout 18.7 8 360 175 3.15 3.440 17.02 0 0 3 2 177
6 Valiant 18.1 6 225 105 2.76 3.460 20.22 1 0 3 1 108
Can you sort a df based on object class? Say
data("mtcars")
mtcars$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.