I currently have the below list and am trying to develop a data frame from the results. Essentially, I would like to take the first list and add it to a new data frame, creating columns with variable names, and storing the results in the first row. Then move to the next list and create a new y column so that any variables with y results would be added to that list.
list
[[1]]
x xy
1.000365 1.000365
[[2]]
x y
1.007184 1.007184
[[3]]
x y
1.020803 1.020803
[[4]]
NA
Is this possible to do? I've been trying to figure out how a for loop or lapply might work in this scenario but am unsure.
Thanks.
You can use [ in lappy on unique names of the vectors:
i <- unique(unlist(lapply(x, names)))
setNames(as.data.frame(do.call(rbind, lapply(x, `[`, i))), i)
# x xy y
#1 1.000365 1.000365 NA
#2 1.007184 NA 1.007184
#3 1.020803 NA 1.020803
#4 NA NA NA
Data:
x <- list(c(x=1.000365, xy=1.000365), c(x=1.007184, y=1.007184),
c(x=1.020803, y=1.020803), NA)
lapply, or even better bind_rows would be good for this:
library(dplyr)
d <- bind_rows( your.list )
Note, I assume the xy name of the 2nd element of the first list entry is a typo?
Related
HEADLINE: Is there a way to get R to recognize data.frame column names contained within lists in the same way that it can recognize free-floating vectors?
SETUP: Say I have a vector named varA:
(varA <- 1:6)
# [1] 1 2 3 4 5 6
To get the length of varA, I could do:
length(varA)
#[1] 6
and if the variable was contained within a larger list, the variable and its length could still be found by doing:
list <- list(vars = "varA")
length(get(list$vars[1]))
#[1] 6
PROBLEM:
This is not the case when I substitute the vector for a dataframe column and I don't know how to work around this:
rows <- 1:6
cols <- c("colA")
(df <- data.frame(matrix(NA,
nrow = length(rows),
ncol = length(cols),
dimnames = list(rows, cols))))
# colA
# 1 NA
# 2 NA
# 3 NA
# 4 NA
# 5 NA
# 6 NA
list <- list(vars = "varA",
cols = "df$colA")
length(get(list$vars[1]))
#[1] 6
length(get(list$cols[1]))
#Error in get(list$cols[1]) : object 'df$colA' not found
Though this contrived example seems inane, because I could always use the simple length(variable) approach, I'm actually interested in writing data from hundreds of variables varying in lengths onto respective dataframe columns, and so keeping them in a list that I could iterate through would be very helpful. I've tried everything I could think of, but it may be the case that it's just not possible in R, especially given that I cannot find any posts with solutions to the issue.
You could try:
> length(eval(parse(text = list$cols[1])))
[1] 6
Or:
list <- list(vars = "varA",
cols = "colA")
length(df[, list$cols[1]])
[1] 6
Or with regex:
list <- list(vars = "varA",
cols = "df$colA")
length(df[, sub(".*\\$", "", list$cols[1])])
[1] 6
If you are truly working with a data frame d, then nrow(d) is the length of all of the variables in d. There should be no reason to use length in this case.
If you are actually working with a list x containing variables of potentially different lengths, then you should use the [[ operator to extract those variables by name (see ?Extract):
x <- list(a = 1:10, b = rnorm(20L))
l <- list(vars = "a")
length(d[[l$vars[1L]]]) # 10
If you insist on using get (you shouldn't), then you need to supply a second argument telling it where to look for the variable (see ?get):
length(get(l$vars[1L], x)) # 10
After searching for some time, I cannot find a smooth R-esque solution.
I have a list of vectors that I want to convert to dataframes and add a column with the names of the vectors. I cant do this with cbind() and melt() to a single dataframe b/c there are vectors with different number of rows.
Basic example would be:
list<-list(a=c(1,2,3),b=c(4,5,6,7))
var<-"group"
What I have come up with and works is:
list<-lapply(list, function(x) data.frame(num=x,grp=""))
for (j in 1:length(list)){
list[[j]][,2]<-names(list[j])
names(list[[j]])[2]<-var
}
But I am trying to better use lapply() and have cleaner coding practices. Right now I rely so heavily on for and if statements, which a lot of the base functions do already and much more efficiently than I can code at this point.
The psuedo code I would like is something like:
list<-lapply(list, function(x) data.frame(num=x,get(var)=names(x))
Is there a clean way to get this done?
Second closely related question, if I already have a list of dataframes, why is it so hard to reassign column values and names using lapply()?
So using something like:
list<-list(a=data.frame(num=c(1,2,3),grp=""),b=data.frame(num=c(4,5,6,7),grp=""))
var<-"group"
#pseudo code
list<-lapply(list, function(x) x[,2]<-names(x)) #populate second col with name of df[x]
list<-lapply(list, function(x) names[[x]][2]<-var) #set 2nd col name to 'var'
The first line of pseudo code throws an error about matching row lengths. Why does lapply() not just loop over and repeat names(x) like the same function on a single dataframe does in a for loop?
For the second line, as I understand it I can use setNames() to reassign all the column names, but how do I make this work for just one of the col names?
Many thanks for any ideas or pointing to other threads that cover this and helping me understand the behavior of lapply() in this context.
A full R base approach without using loops
> l<-list(a=c(1,2,3),b=c(4,5,6,7))
> data.frame(grp=rep(names(l), lengths(l)), num=unlist(l), row.names = NULL)
grp num
1 a 1
2 a 2
3 a 3
4 b 4
5 b 5
6 b 6
Related to your first/main question you can use the function enframe from package tibble for this purpose
library(tibble)
library(tidyr)
library(dplyr)
l<-list(a=c(1,2,3),b=c(4,5,6,7))
l %>%
enframe(name = "group", value="value") %>%
unnest(value) %>%
group_split(group)
Try this:
library(dplyr)
mylist <- list(a = c(1,2,3), b = c(4,5,6,7))
bind_rows(lapply(names(mylist), function(x) tibble(grp = x, num = mylist[[x]])))
# A tibble: 7 x 2
grp num
<chr> <dbl>
1 a 1
2 a 2
3 a 3
4 b 4
5 b 5
6 b 6
7 b 7
This is essentially a lapply-based solution where you iterate over the names of your list, and not the individual list elements themselves. If you prefer to do everything in base R, note that the above is equivalent to
do.call(rbind, lapply(names(mylist), function(x) data.frame(grp = x, num = mylist[[x]], stringsAsFactors = F)))
Having said that, tibbles as modern implementation of data.frames are preferred, as is bind_rows over the do.call(rbind... construct.
As to the second question, note the following:
lapply(mylist, function(x) str(x))
num [1:3] 1 2 3
num [1:4] 4 5 6 7
....
lapply(mylist, function(x) names(x))
$a
NULL
$b
NULL
What you see here is that the function inside of lapply gets the elements of mylist. In this case, it get's to work with the numeric vector. This does not have any name as far as the function that is called inside lapply is concerned. To highlight this, consider the following:
names(c(1,2,3))
NULL
Which is the same: the vector c(1,2,3) does not have a name attribute.
I have a list of data.frames and I want to change the class of one column of the data.frame (from factor to date). I was trying to use lapply but then, the original list only contains that column (not the whole data.frame). I don't understand this behaviour..to solve this I use a common loop, but I was wondering if anyone could have any suggestion.
Let's say, as an example, I have this simple data:
m1 <- data.frame("date"=c("2010-02-03","2010-01-05"),"value"=c(5,3))
m2 <- data.frame("date"=c("2010-02-03","2010-01-05"),"value"=c(1,3))
mylist <- list(m1,m2)
#change date
newlist <- lapply(mylist, function(x) as.Date(x$date))
newlist will have only the date..
Is there any way to use lapply for that..I am working with large dataset, and usually lapply works fine, but in this case, I don't know what I'm doing wrong.
Many thanks.
We can use transform
lapply(mylist, transform, date = as.Date(date))
In your function you have to first change the date column and then return x
newlist <- lapply(mylist, function(x) {
x$date <- as.Date(x$date)
x
})
newlist
# [[1]]
# date value
# 1 2010-02-03 5
# 2 2010-01-05 3
#
# [[2]]
# date value
# 1 2010-02-03 1
# 2 2010-01-05 3
I am trying to train a data that's converted from a document term matrix to a dataframe. There are separate fields for the positive and negative comments, so I wanted to add a string to the column names to serve as a "tag", to differentiate the same word coming from the different fields - for example, the word hello can appear both in the positive and negative comment fields (and thus, represented as a column in my dataframe), so in my model, I want to differentiate these by making the column names positive_hello and negative_hello.
I am looking for a way to rename columns in such a way that a specific string will be appended to all columns in the dataframe. Say, for mtcars, I want to rename all of the columns to have "_sample" at the end, so that the column names would become mpg_sample, cyl_sample, disp_sample and so on, which were originally mpg, cyl, and disp.
I'm considering using sapplyor lapply, but I haven't had any progress on it. Any help would be greatly appreciated.
Use colnames and paste0 functions:
df = data.frame(x = 1:2, y = 2:1)
colnames(df)
[1] "x" "y"
colnames(df) <- paste0('tag_', colnames(df))
colnames(df)
[1] "tag_x" "tag_y"
If you want to prefix each item in a column with a string, you can use paste():
# Generate sample data
df <- data.frame(good=letters, bad=LETTERS)
# Use the paste() function to append the same word to each item in a column
df$good2 <- paste('positive', df$good, sep='_')
df$bad2 <- paste('negative', df$bad, sep='_')
# Look at the results
head(df)
good bad good2 bad2
1 a A positive_a negative_A
2 b B positive_b negative_B
3 c C positive_c negative_C
4 d D positive_d negative_D
5 e E positive_e negative_E
6 f F positive_f negative_F
Edit:
Looks like I misunderstood the question. But you can rename columns in a similar way:
colnames(df) <- paste(colnames(df), 'sample', sep='_')
colnames(df)
[1] "good_sample" "bad_sample" "good2_sample" "bad2_sample"
Or to rename one specific column (column one, in this case):
colnames(df)[1] <- paste('prefix', colnames(df)[1], sep='_')
colnames(df)
[1] "prefix_good_sample" "bad_sample" "good2_sample" "bad2_sample"
You can use setnames from the data.table package, it doesn't create any copy of your data.
library(data.table)
df <- data.frame(a=c(1,2),b=c(3,4))
# a b
# 1 1 3
# 2 2 4
setnames(df,paste0(names(df),"_tag"))
print(df)
# a_tag b_tag
# 1 1 3
# 2 2 4
I have two dataframes which look like follows:
df1 <- data.frame(V1 = 1:4, V2 = rep(2, 4), V3 = 7:4)
df2 <- data.frame(V2 = rep(NA, 4), V1 = rep(NA, 4), V3 = rep(NA, 4))
I need to write a function which assigns the values of df1 to df2, if the columnnames of both dataframes are the same. The structure of the function should look like this:
fun <- function(x){
if(# If the name of x is the same like the name of a column in df1)
out <- df1$? # Here I need to assign df1$"x" somehow
out
}
fun(df2$V1)
The output should look like this:
[1] 1 2 3 4
Unfortunately I couldnt find a solution by myself. Is there a way how I could do this? Thank you very much in advance!
I need to write a function which assigns the values of df1 to df2, if
the columnnames of both dataframes are the same.
Are you sure you need a function?
names_in_common <- intersect(names(df1),names(df2))
df2[,names_in_common] <- df1[,names_in_common]
Using Joachim Schork's code:
names_in_common <- intersect(names(df1),names(df2))
df2[,names_in_common] <- df1[,names_in_common]
and if you want to change a single column of df2:
names_in_common <- intersect(names(df1), names(df2[, "V1", drop=FALSE]))
df2[,names_in_common] <- df1[,names_in_common]
This is impossible, because when you access a column of a data.frame using the dollar syntax you lose the column name. There's no way for fun() to determine the column name of the vector that was passed in as an argument.
Instead, you can simply call fun() using the column name itself as the argument, rather than the vector of NAs, which are not useful and not used at all inside the function. In other words, the call becomes
fun('V1');
Then you can write the function as follows:
fun <- function(name) df1[[name]];
Demo:
fun('V1');
## [1] 1 2 3 4
Although now that I think about it, you might as well just index df1 directly, since that's all the function does now:
df1$V1;
## [1] 1 2 3 4
Rereading your question, you said you want to assign the column from df1 to df2, although your example code doesn't do that. Assuming you did want to carry out this assignment inside the function, you could do this:
fun <- function(name) df2[[name]] <<- df1[[name]];
Demo:
fun('V1');
df2;
## V2 V1 V3
## 1 NA 1 NA
## 2 NA 2 NA
## 3 NA 3 NA
## 4 NA 4 NA
This makes use of the superassignment operator <<-.