I'm trying to use lapply on a list of data frames; but failing at passing the parameters correctly (I think).
List of data frames:
df1 <- data.frame(A = 1:10, B= 11:20)
df2 <- data.frame(A = 21:30, B = 31:40)
listDF <- list(df1, df2,df3) #multiple data frames w. way less columns than the length of vector todos
Vector with columns names:
todos <-c('col1','col2', ......'colN')
I'd like to change the column names using lapply:
lapply (listDF, function(x) { colnames(x)[2:length(x)] <-todos[1:length(x)-1] } )
but this doesn't change the names at all. Am I not passing the data frames themselves, but something else? I just want to change names, not to return the result to a new object.
Thanks in advance, p.
To change multiple column names by name and by index use rename() function of the dplyr package and to rename by just name use setnames() from data. table . From R base functionality, we have colnames() and names() functions that can be used to rename a data frame column by a single index or name.
You can also use a setNames() function within a Map() function to rename each column for your list of dataframes. To keep your code clean as possible, it is best to avoid naming objects with the same names as functions. (Your example code uses a vector called "names" but also uses names() function.)
If, instead of a list, you had a data frame of stock returns, could you still use lapply() ? Yes! Perhaps surprisingly, data frames are actually lists under the hood, and an lapply() call would apply the function to each column of the data frame.
In R Programming Language to apply a function to every integer type value in a data frame, we can use lapply function from dplyr package. And if the datatype of values is string then we can use paste() with lapply.
You can also use setNames
if you want to replace all columns
df1 <- data.frame(A = 1:10, B= 11:20)
df2 <- data.frame(A = 21:30, B = 31:40)
listDF <- list(df1, df2)
new_col_name <- c("C", "D")
lapply(listDF, setNames, nm = new_col_name)
## [[1]]
## C D
## 1 1 11
## 2 2 12
## 3 3 13
## 4 4 14
## 5 5 15
## 6 6 16
## 7 7 17
## 8 8 18
## 9 9 19
## 10 10 20
## [[2]]
## C D
## 1 21 31
## 2 22 32
## 3 23 33
## 4 24 34
## 5 25 35
## 6 26 36
## 7 27 37
## 8 28 38
## 9 29 39
## 10 30 40
If you need to replace only a subset of column names, then you can use the solution of @Jogo
lapply(listDF, function(df) {
names(df)[-1] <- new_col_name[-ncol(df)]
df
})
A last point, in R there is a difference between a:b - 1 and a:(b - 1)
1:10 - 1
## [1] 0 1 2 3 4 5 6 7 8 9
1:(10 - 1)
## [1] 1 2 3 4 5 6 7 8 9
EDIT
If you want to change the column names of the data.frame
in global environment from a list, you can use list2env
but I'm not sure it is the best way to achieve want you want. You also need to modify your list and use named list, the name should be the same as name of the data.frame
you need to replace.
listDF <- list(df1 = df1, df2 = df2)
new_col_name <- c("C", "D")
listDF <- lapply(listDF, function(df) {
names(df)[-1] <- new_col_name[-ncol(df)]
df
})
list2env(listDF, envir = .GlobalEnv)
str(df1)
## 'data.frame': 10 obs. of 2 variables:
## $ A: int 1 2 3 4 5 6 7 8 9 10
## $ C: int 11 12 13 14 15 16 17 18 19 20
try this:
lapply (listDF, function(x) {
names(x)[-1] <- todos[-length(x)]
x
})
you will get a new list with changed dataframes. If you want to manipulate the listDF
directly:
for (i in 1:length(listDF)) names(listDF[[i]])[-1] <- todos[-length(listDF[[i]])]
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