I have the following named list output from a analysis. The reproducible code is as follows:
list(structure(c(-213.555409754509, -212.033637890131, -212.029474755074,
-211.320398316741, -211.158815833294, -210.470525157849), .Names = c("wasn",
"chappal", "mummyji", "kmph", "flung", "movie")), structure(c(-220.119433774144,
-219.186901747536, -218.743319709963, -218.088361753899, -217.338920075687,
-217.186050877079), .Names = c("crazy", "wired", "skanndtyagi",
"andr", "unveiled", "contraption")))
I want to convert this to a data frame. I have tried unlist to data frame options using reshape2, dplyr and other solutions given for converting a list to a data frame but without much success. The output that I am looking for is something like this:
Col1 Val1 Col2 Val2
1 wasn -213.55 crazy -220.11
2 chappal -212.03 wired -219.18
3 mummyji -212.02 skanndtyagi -218.74
so on and so forth. The actual out put has multiple columns with paired values and runs into many rows. I have tried the following codes already:
do.call(rbind, lapply(df, data.frame, stringsAsFactors = TRUE))
works partially provides all the character values in a column and numeric values in the second.
data.frame(Reduce(rbind, df))
didn't work - provides the names in the first list and numbers from both the lists as tow different rows
colNames <- unique(unlist(lapply(df, names)))
M <- matrix(0, nrow = length(df), ncol = length(colNames),
dimnames = list(names(df), colNames))
matches <- lapply(df, function(x) match(names(x), colNames))
M[cbind(rep(sequence(nrow(M)), sapply(matches, length)),
unlist(matches))] <- unlist(df)
M
didn't work correctly.
Can someone help?
Since the list elements are all of the same length, you should be able to stack them and then combine them by columns.
Try:
do.call(cbind, lapply(myList, stack))
Here's another way:
as.data.frame( c(col = lapply(x, names), val = lapply(x,unname)) )
How it works. lapply returns a list; two lists combined with c make another list; and a list is easily coerced to a data.frame, since the latter is just a list of vectors having the same length.
Better than coercing to a data.frame is just modifying its class, effectively telling the list "you're a data.frame now":
L = c(col = lapply(x, names), val = lapply(x,unname))
library(data.table)
setDF(L)
The result doesn't need to be assigned anywhere with = or <- because L is modified "in place."
Related
I have a list containing named elements. I am iterating over the list names, performing the computation for each corresponding element, "encapsulating" the results and the name in a vector and finally adding the vector to a table. The row or vector after each iteration contains a mix of characters and numbers.
The first row is getting added but from the second row onwards there is a problem.
In this example, there is supposed to be one column (first) containing alphanumeric names. All rows after the first one contain NAs.
x <- list(a_1=c(1,2,3), b_2=c(3,4,5), c_3=c(5,1,9))
df <- data.frame()
for(name in names(x))
{
tmp <- x[[name]]
m <- mean(tmp)
s <- sum(tmp)
df <- rbind(df, c(name,m,s))
}
df <- as.data.frame(df)
I know there are possibly more efficient ways but for the moment this is more intuitive for me as it is assuring that each computation is associated with a particular name. There can be several columns and rows and the names are extremely helpful to join tables, query, compare etc. They make it easier to trace back results to a particular element in my original list.
Additionally, I would be glad to know other ways in which the element names are always retained while transforming.
Thankyou!
You have to set stringsAsFactors = FALSE in rbind. With stringsAsFactors = TRUE the first iteration in the loop converts the string variables into factors (with the factor levels being the values).
x <- list(a_1=c(1,2,3), b_2=c(3,4,5), c_3=c(5,1,9))
df <- data.frame()
for(name in names(x))
{
tmp <- x[[name]]
m <- mean(tmp)
s <- sum(tmp)
df <- rbind(df, c(name,m,s), stringsAsFactors = FALSE)
}
An easier solution would be to utilize sapply().
x <- list(a_1=c(1,2,3), b_2=c(3,4,5), c_3=c(5,1,9))
df <- data.frame(name = names(x), m = sapply(x, mean), s = sapply(x, sum))
I'm trying to add different suffixes to my data frames so that I can distinguish them after I've merge them. I have my data frames in a list and created a vector for the suffixes but so far I have not been successful.
data2016 is the list containing my 7 data frames
new_names <- c("june2016", "july2016", "aug2016", "sep2016", "oct2016", "nov2016", "dec2016")
data2016v2 <- lapply(data2016, paste(colnames(data2016)), new_names)
Your query is not quite clear. Therefore two solutions.
The beginning is the same for either solution. Suppose you have these four dataframes:
df1x <- data.frame(v1 = rnorm(50),
v2 = runif(50))
df2x <- data.frame(v3 = rnorm(60),
v4 = runif(60))
df3x <- data.frame(v1 = rnorm(50),
v2 = runif(50))
df4x <- data.frame(v3 = rnorm(60),
v4 = runif(60))
Suppose further you assemble them in a list, something akin to your data2016using mgetand ls and describing a pattern to match them:
my_list <- mget(ls(pattern = "^df\\d+x$"))
The names of the dataframes in this list are the following:
names(my_list)
[1] "df1x" "df2x" "df3x" "df4x"
Solution 1:
Suppose you want to change the names of the dataframes thus:
new_names <- c("june2016", "july2016","aug2016", "sep2016")
Then you can simply assign new_namesto names(my_list):
names(my_list) <- new_names
And the result is:
names(my_list)
[1] "june2016" "july2016" "aug2016" "sep2016"
Solution 2:
You want to add the new_names literally as suffixes to the 'old' names, in which case you would use pasteor paste0 thus:
names(my_list) <- paste0(names(my_list), "_", new_names)
And the result is:
names(my_list)
[1] "df1x_june2016" "df2x_july2016" "df3x_aug2016" "df4x_sep2016"
You could use an index number within lapply to reference both the list and your vector of suffixes. Because there are a couple steps, I'll wrap the process in a function(). (Called an anonymous function because we aren't assigning a name to it.)
data2016v2 <- lapply(1:7, function(i) {
this_data <- data2016[[i]] # Double brackets for a list
names(this_data) <- paste0(names(this_data), new_names[i]) # Single bracket for vector
this_data # The renamed data frame to be placed into data2016v2
})
Notice in the paste0() line we are recycling the term in new_names[i], so for example if new_names[i] is "june2016" and your first data.frame has columns "A", "B", and "C" then it would give you this:
> paste0(c("A", "B", "C"), "june2016")
[1] "Ajune2016" "Bjune2016" "Cjune2016"
(You may want to add an underscore in there?)
As an aside, it sounds like you might be better served by adding the "june2016" as a column in your data (like say a variable named month with "june2016" as the value in each row) and combining your data using something like bind_rows() from the dplyr package, running it "long" instead of "wide".
I have a list similar to this one:
set.seed(1602)
l <- list(data.frame(subst_name = sample(LETTERS[1:10]), perc = runif(10), crop = rep("type1", 10)),
data.frame(subst_name = sample(LETTERS[1:7]), perc = runif(7), crop = rep("type2", 7)),
data.frame(subst_name = sample(LETTERS[1:4]), perc = runif(4), crop = rep("type3", 4)),
NULL,
data.frame(subst_name = sample(LETTERS[1:9]), perc = runif(9), crop = rep("type5", 9)))
Question: How can I extract the subst_name-column of each data.frame and combine them with cbind() (or similar functions) to a new data.frame without messing up the order of each column? Additionally the columns should be named after the corresponding crop type (this is possible 'cause the crop types are unique for each data.frame)
EDIT: The output should look as follows:
Having read the comments I'm aware that within R it doesn't make much sense but for the sake of having alook at the output the data.frame's View option is quite handy.
With the help of this SO-Question I came up with the following sollution. (There's probably room for improvement)
a <- lapply(l, '[[', 1) # extract the first element of the dfs in the list
a <- Filter(function(x) !is.null(unlist(x)), a) # remove NULLs
a <- lapply(a, as.character)
max.length <- max(sapply(a, length))
## Add NA values to list elements
b <- lapply(a, function(v) { c(v, rep(NA, max.length-length(v)))})
e <- as.data.frame(do.call(cbind, d))
names(e) <- unlist(lapply(lapply(lapply(l, '[[', "crop"), '[[', 2), as.character))
It is not really correct to do this with the given example because the number of rows is not the same in each one of the list's data frames . But if you don't care you can do:
nullElements = unlist(sapply(l,is.null))
l = l[!nullElements] #delete useless null elements in list
columns=lapply(l,function(x) return(as.character(x$subst_name)))
newDf = as.data.frame(Reduce(cbind,columns))
If you don't want recycled elements in the columns you can do
for(i in 1:ncol(newDf)){
colLength = nrow(l[[i]])
newDf[(colLength+1):nrow(newDf),i] = NA
}
newDf = newDf[1:max(unlist(sapply(l,nrow))),] #remove possible extra NA rows
Note that I edited my previous code to remove NULL entries from l to simplify things
I have a list resembling the one below:
# Initial object
vec <- c("levelA-1", "levelA-2", "levelA-3",
"levelB-1", "levelB-2", "levelB-3")
lstVec <- strsplit(x = vec, split = "-")
I would like to arrive at a list of the following structure:
lstRes <- list(levelA = list(1:3),
lvelB = list(1:3))
Notes
The list has the following characteristics:
First level elements are transformed into distinct lists
Second level elements created via strsplit are elements of those lists
this suffices:
mat <- do.call(rbind, lstVec)
result <- split(mat[,2], mat[,1])
the do.call and rbind stack the result of lstVec by row into a matrix (thanks to G. Grothendieck for pointing out this is not a data frame), then the split split mat[,2] by mat[,1].
as Aaron says, ti is a little odd that you want nested list. but you can get it
lapply(result, as.list)
i am not sure how good rbind is. but another way to obtain mat is
mat <- matrix(unlist(lstVec), ncol = 2, byrow = TRUE)
merger <- cbind(as.character(Date),weather1$High,weather1$Low,weather1$Avg..High,weather1$Avg.Low,sale$Scanned.Movement[a])
After cbind the data, the new DF has column names automatically V1, V2......
I want rename the column by
colnames(merger)[,1] <- "Date"
but failed. And when I use merger$V1 ,
Error in merger$V1 : $ operator is invalid for atomic vectors
You can also name columns directly in the cbind call, e.g.
cbind(date=c(0,1), high=c(2,3))
Output:
date high
[1,] 0 2
[2,] 1 3
Try:
colnames(merger)[1] <- "Date"
Example
Here is a simple example:
a <- 1:10
b <- cbind(a, a, a)
colnames(b)
# change the first one
colnames(b)[1] <- "abc"
# change all colnames
colnames(b) <- c("aa", "bb", "cc")
you gave the following example in your question:
colnames(merger)[,1]<-"Date"
the problem is the comma: colnames() returns a vector, not a matrix, so the solution is:
colnames(merger)[1]<-"Date"
If you pass only vectors to cbind() it creates a matrix, not a dataframe. Read ?data.frame.
A way of producing a data.frame and being able to do this in one line is to coerce all matrices/data frames passed to cbind into a data.frame while setting the column names attribute using setNames:
a = matrix(rnorm(10), ncol = 2)
b = matrix(runif(10), ncol = 2)
cbind(setNames(data.frame(a), c('n1', 'n2')),
setNames(data.frame(b), c('u1', 'u2')))
which produces:
n1 n2 u1 u2
1 -0.2731750 0.5030773 0.01538194 0.3775269
2 0.5177542 0.6550924 0.04871646 0.4683186
3 -1.1419802 1.0896945 0.57212043 0.9317578
4 0.6965895 1.6973815 0.36124709 0.2882133
5 0.9062591 1.0625280 0.28034347 0.7517128
Unfortunately, there is no setColNames function analogous to setNames for data frames that returns the matrix after the column names, however, there is nothing to stop you from adapting the code of setNames to produce one:
setColNames <- function (object = nm, nm) {
colnames(object) <- nm
object
}
See this answer, the magrittr package contains functions for this.
If you offer cbind a set of arguments all of whom are vectors, you will get not a dataframe, but rather a matrix, in this case an all character matrix. They have different features. You can get a dataframe if some of your arguments remain dataframes, Try:
merger <- cbind(Date =as.character(Date),
weather1[ , c("High", "Low", "Avg..High", "Avg.Low")] ,
ScnMov =sale$Scanned.Movement[a] )
It's easy just add the name which you want to use in quotes before adding
vector
a_matrix <- cbind(b_matrix,'Name-Change'= c_vector)