Reputation: 408
I have (what I think) is a really simple question, but I can't figure out how to do it. I'm fairly new to lists, loops, etc.
I have a small dataset:
df <- c("one","two","three","four")
df <- as.data.frame(df)
df
I need to loop through this dataset and create a list of datasets, such that this is the outcome:
[[1]]
one
[[2]]
one
two
[[3]]
one
two
three
This is more or less as far as I've gotten:
blah <- list()
for(i in 1:3){
blah[[i]]<- i
}
The length will be variable when I use this in the future, so I need to automate it in a loop. Otherwise, I would just do
one <- df[1,]
two <- df[2,]
list(one, rbind(one, two))
Any ideas?
Upvotes: 2
Views: 673
Reputation: 389355
You can try using lapply
:
result <- lapply(seq(nrow(df)), function(x) df[seq_len(x), , drop = FALSE])
result
#[[1]]
# df
#1 one
# [[2]]
# df
#1 one
#2 two
#[[3]]
# df
#1 one
#2 two
#3 three
#[[4]]
# df
#1 one
#2 two
#3 three
#4 four
seq(nrow(df))
creates a sequence from 1 to number of rows in your data (which is 4 in this case). function(x)
part is called as anonymous function where each value from 1 to 4 is passed to one by one. seq_len(x)
creates a sequence from 1 to x
i.e 1 to 1 in first iteration, 1 to 2 in second and so on. We use this sequence to subset the rows from dataframe (df[seq_len(x), ]
). Since the dataframe has only 1 column when we subset it , it changes it to a vector. To avoid that we add drop = FALSE
.
Upvotes: 2
Reputation: 5798
Base R solution:
# Coerce df vector of data.frame to character, store as new data.frame: str_df => data.frame
str_df <- transform(df, df = as.character(df))
# Allocate some memory in order to split data into a list: df_list => empty list
df_list <- vector("list", nrow(str_df))
# Split the string version of the data.frame into a list as required:
# df_list => list of character vectors
df_list <- lapply(seq_len(nrow(str_df)), function(i){
str_df[if(i == 1){1}else{1:i}, grep("df", names(str_df))]
}
)
Data:
df <- c("one","two","three","four")
df <- as.data.frame(df)
df
Upvotes: 1