I'm a total noob at R and I've tried (and retried) to search for an answer to the following problem, but I've not been able to get any of the proposed solutions to do what I'm interested in.
I have two lists of named elements, with each element pointing to data frames with identical layouts:
(EDIT)
df1 <- data.frame(A=c(1,2,3),B=c("A","B","C"))
df2 <- data.frame(A=c(98,99),B=c("Y","Z"))
lst1 <- c(X=df1,Y=df2)
df3 <- data.frame(A=c(4,5),B=c("D","E"))
lst2 <- c(X=df3)
(EDIT 2)
So it seems like storing multiple data frames in a list is a bad idea, as it will convert the data frames to lists. So I'll go out looking for an alternative way to store a set of named data frames.
In general the names of the elements in the two elements might overlap partially, completely, or not at all.
I'm looking for a way to merge the two lists into a single list:
<some-function-sequence>(lst1, lst2)
->
c(X=rbind(df1,df3),Y=df2)
-resulting in something like this:
[EDIT: Syntax changed to correctly reflect desired result (list-of-data frames)]
$X
A B
1 1 A
2 2 B
3 3 C
4 4 D
5 5 E
$X.B
A B
1 98 Y
2 99 Z
I.e:
IF the lists contain identical element names, each pointing to a data frame, THEN I want to 'rbind' the rows from these two data frames and assign the resulting data frame to the same element name in the resulting list.
Otherwise the element names and data frames from both lists should just be copied into the resulting list.
I've tried the solutions from a number of discussions such as:
Can I combine a list of similar dataframes into a single dataframe?
Combine/merge lists by elements names
Simultaneously merge multiple data.frames in a list
Combine/merge lists by elements names (list in list)
Convert a list of data frames into one data frame
-but I've not been able to find the right solution. A general problem seems to be that the data frame ends up being converted into a list by the application of 'mapply/sapply/merge/...' - and usually also sliced and/or merged in ways which I am not interested in. :)
Any help with this will be much appreciated!
[SOLUTION]
The solution seems to be to change the use of c(...) when collecting data frames to list(...) after which the solution proposed by Pierre seems to give the desired result.
Here is a proposed solution using split and c to combine like terms. Please read the caveat at the bottom:
s <- split(c(lst1, lst2), names(c(lst1,lst2)))
lapply(s, function(lst) do.call(function(...) unname(c(...)), lst))
# $X.A
# [1] 1 2 3 4 5
#
# $X.B
# [1] "A" "B" "C" "D" "E"
#
# $Y.A
# [1] 98 99
#
# $Y.B
# [1] "Y" "Z"
This solution is based on NOT having factors as strings. It will not throw an error but the factors will be converted to numbers. Below I show how I transformed the data to remove factors. Let me know if you require factors:
df1 <- data.frame(A=c(1,2,3),B=c("A","B","C"), stringsAsFactors=FALSE)
df2 <- data.frame(A=c(98,99),B=c("Y","Z"), stringsAsFactors=FALSE)
lst1 <- c(X=df1,Y=df2)
df3 <- data.frame(A=c(4,5),B=c("D","E"), stringsAsFactors=FALSE)
lst2 <- c(X=df3)
If the data is stored in lists we can use:
lapply(split(c(lst1, lst2), names(c(lst1,lst2))), function(lst) do.call(rbind, lst))
The following solution is probably not the most efficient way. However, if I got your problem right this should work ;)
# Example data
# Some vectors
a <- 1:5
b <- 3:7
c <- rep(5, 5)
d <- 5:1
# Some dataframes, data1 and data3 have identical column names
data1 <- data.frame(a, b)
data2 <- data.frame(c, b)
data3 <- data.frame(a, b)
data4 <- data.frame(c, d)
# 2 lists
list1 <- list(data1, data2)
list2 <- list(data3, data4)
# Loop, wich checks for the dataframe names and rbinds dataframes with the same column names
final_list <- list1
used_lists <- numeric()
for(i in 1:length(list1)) {
for(j in 1:length(list2)) {
if(sum(colnames(list1[[i]]) == colnames(list2[[j]])) == ncol(list1[[i]])) {
final_list[[i]] <- rbind(list1[[i]], list2[[j]])
used_lists <- c(used_lists, j)
}
}
}
# Adding the other dataframes, which did not have the same column names
for(i in 1:length(list2)) {
if((i %in% used_lists) == FALSE) {
final_list[[length(final_list) + 1]] <- list2[[i]]
}
}
# Final list, which includes all other lists
final_list
Related
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?
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.
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
I have N lists that have some identical column names. Here is a MWE with two list:
ls <- list()
ls[[1]] <- list("a"=1:2,
"b"=20,
"c"=numeric(0))
names(ls[[1]]$a) <- c("a1", "a2")
ls[[2]] <- list("a"=3:4,
"b"=30,
"c"=1:4,
"d"="f")
names(ls[[2]]$a) <- c("a1", "a2")
Is it possible merge these into a resulting list lsRes, where lsRes has the following properties:
lsRes$a contains two elements, where the first is the named vector
c(1,2) (with names c(a1, a2)) and the second a named vector c(3,4)
(with names (c(a1.a2)))
lsRes$b contains two elements, where the first is 20 and the second is 30
lsRes$c contains two elements, where the first is numeric(0) and the second is 1:4
lsRes$d contains
two elements, where the first is NA and the second is "f"
I looked at this and this, but they describe different cases
Assuming that we need to have the output also as a list, we create the common names and then assign those doesn't any of the common names to NA
nm1 <- unique(unlist(sapply(ls, names)))
lsRes <- lapply(ls, function(x) {x[setdiff(nm1, names(x))] <- NA; x})
lengths(lsRes)
#[1] 4 4
If we need to have a list of 4 elements, then use transpose
library(purrr)
lsRes %>%
transpose
I used the following for loop to read 7 csv files and add them to a list.
list <- list()
l <- 1
for(i in 1:7){
data <- read.csv(paste("file",i,".csv",sep=""),header=FALSE)
list[[l]] <- data
l <- l + 1
}
So now I have a list named "list" containing 7 dataframes, right?
Each of the 8 dataframes contain the same three columns (NAME, SURNAME, AGE).
I now want to add:
df <- dataframe(NAME,SURNAME,AGE) ## to each dataframe in the list.
Did that help at all? My question is, how can I achieve that for all 7 objects in the list automatically!
If the 'lst' has seven data.frames and want to 'rbind' the 8th dataset to each of the datasets in the list, we can use Map
Map(rbind, lst, list(d1))
Or using lapply
lapply(lst, rbind, d1)
Update
If the 'lst' is of length 8, and wants to rbind the first 7 elements with the dataset in the 8th element, then you can just do
Map(rbind, lst[-8], lst[8])
data
set.seed(24)
lst <- lapply(1:7, function(i) as.data.frame(matrix(sample(0:10, 3*10,
replace=TRUE), ncol=3)))
set.seed(49)
d1 <- as.data.frame(matrix(sample(1:20, 3*10, replace=TRUE), ncol=3))
Or, if the ultimate goal is to just ensure all 8 CSV files make it into one data.frame:
# generate some sample files
files <- sprintf("iris%d.csv", i)
for (i in 1:8) { write.csv(iris, files, row.names=FALSE) }
# make one happy data frame
do.call(rbind, lapply(files, read.csv))
I have a for loop that generates a dataframe every time it loops through. I am trying to create a list of data frames but I cannot seem to figure out a good way to do this.
For example, with vectors I usually do something like this:
my_numbers <- c()
for (i in 1:4){
my_numbers <- c(my_numbers,i)
}
This will result in a vector c(1,2,3,4). I want to do something similar with dataframes, but accessing the list of data frames is quite difficult when i use:
my_dataframes <- list(my_dataframes,DATAFRAME).
Help please. The main goal is just to create a list of dataframes that I can later on access dataframe by dataframe. Thank you.
I'm sure you've noticed that list does not do what you want it to do, nor should it. c also doesn't work in this case because it flattens data frames, even when recursive=FALSE.
You can use append. As in,
data_frame_list = list()
for( i in 1:5 ){
d = create_data_frame(i)
data_frame_list = append(data_frame_list,)
}
Better still, you can assign directly to indexed elements, even if those elements don't exist yet:
data_frame_list = list()
for( i in 1:5 ){
data_frame_list[[i]] = create_data_frame(i)
}
This applies to vectors, too. But if you want to create a vector c(1,2,3,4) just use 1:4, or its underlying function seq.
Of course, lapply or the *lply functions from plyr are often better than looping depending on your application.
Continuing with your for loop method, here's a little example of creating and accessing.
> my_numbers <- vector('list', 4)
> for (i in 1:4) my_numbers[[i]] <- data.frame(x = seq(i))
And we can access the first column of each data frame with,
> sapply(my_numbers, "[", 1)
# $x
# [1] 1
#
# $x
# [1] 1 2
#
# $x
# [1] 1 2 3
#
# $x
# [1] 1 2 3 4
Other ways of accessing the data is my_numbers[[1]] for the first data set,
lapply(my_numbers, "[", 1,) to access the first row of each data frame, etc.
You can use operator [[ ]] for this purpose.
l <- list()
df1 <- data.frame(name = 'df1', a = 1:5 , b = letters[1:5])
df2 <- data.frame(name = 'df2', a = 6:10 , b = letters[6:10])
df3 <- data.frame(name = 'df3', a = 11:20 , b = letters[11:20])
df <- rbind(df1,df2,df3)
for(df_name in unique(df$name)){
l[[df_name]] <- df[df$name == df_name,]
}
In this example, there are three separate data frames and in order to store them
in a list using a for loop, we place them in one. Using the operator [[ we can even name the data frame in the list as we want and store it in the list normally.
I have a dataframe, df and a function process that returns a list of two dataframes, a and b. I use dlply to split up the df on an id column, and then return a list of lists of dataframes. Here's sample data/code that approximates the actual data and methods:
df <- data.frame(id1=rep(c(1,2,3,4), each=2))
process <- function(df) {
a <- data.frame(d1=rnorm(1), d2=rnorm(1))
b <- data.frame(id1=df$id1, a=rnorm(nrow(df)), b=runif(nrow(df)))
list(a=a, b=b)
}
require(plyr)
output <- dlply(df, .(id1), process)
output is a list of lists of dataframes, the nested list will always have two dataframes, named a and b. In this case the outer list has a length 4.
What I am looking to generate is a dataframe with all the a dataframes, along with an id column indicating their respective value (I believe this is left in the list as the split_labels attribute, see str(output)). Then similarly for the b dataframes.
So far I have in part used this question to come up with this code:
list <- unlist(output, recursive = FALSE)
list.a <- lapply(1:4, function(x) {
list[[(2*x)-1]]
})
all.a <- rbind.fill(list.a)
Which gives me the final a dataframe (and likewise for b with a different subscript into list), however it doesn't have the id column I need and I'm pretty sure there's got to be a more straightforward or elegant solution. Ideally something clean using plyr.
Not very clean but you can try something like this (assuming the same data generation process).
list.aID <- lapply(1:4, function(x) {
cbind(list[[(2*x) - 1]], list[[2*x]][1, 1, drop = FALSE])
})
all.aID <- rbind.fill(list.aID)
all.aID
all.aID
d1 d2 id1
1 0.68103 -0.74023 1
2 -0.50684 1.23713 2
3 0.33795 -0.37277 3
4 0.37827 0.56892 4