I am working in R with a long table stored as a data.table containing values obtained in value changes for variables of numeric and character type. When I want to perform some functions like correlations, regressions, etc. I have to convert the table into wide format and homogenise the timestamp frequency.
I found a way to convert the long table to wide, but I think is not really efficient and I would like to know if there is a better more data.table native approach.
In the reproducible example below, I include the two options I found to perform the wide low transformation and in the comments I indicate what parts I believe are not optimal.
library(zoo)
library(data.table)
dt<-data.table(time=1:6,variable=factor(letters[1:6]),numeric=c(1:3,rep(NA,3)),
character=c(rep(NA,3),letters[1:3]),key="time")
print(dt)
print(dt[,lapply(.SD,typeof)])
#option 1
casted<-dcast(dt,time~variable,value.var=c("numeric","character"))
# types are correct, but I got NA filled columns,
# is there an option like drop
# available for columns instead of rows?
print(casted)
print(casted[,lapply(.SD,typeof)])
# This drop looks ugly but I did not figure out a better way to perform it
casted[,names(casted)[unlist(casted[,lapply(lapply(.SD,is.na),all)])]:=NULL]
# I perform a LOCF, I do not know if I could benefit of
# data.table's roll option somehow and avoid
# the temporal memory copy of my dataset (this would be the second
# and minor issue)
casted<-na.locf(casted)
#option2
# taken from http://stackoverflow.com/questions/19253820/how-to-implement-coalesce-efficiently-in-r
coalesce2 <- function(...) {
Reduce(function(x, y) {
i <- which(is.na(x))
x[i] <- y[i]
x},
list(...))
}
casted2<-dcast(dt[,coalesce2(numeric,character),by=c("time","variable")],
time~variable,value.var="V1")
# There are not NA columns but types are incorrect
# it takes more space in a real table (more observations, less variables)
print(casted2)
print(casted2[,lapply(.SD,typeof)])
# Again, I am pretty sure there is a prettier way to do this
numericvars<-names(casted2)[!unlist(casted2[,lapply(
lapply(lapply(.SD,as.numeric),is.na),all)])]
casted2[,eval(numericvars):=lapply(.SD,as.numeric),.SDcols=numericvars]
# same as option 1, is there a data.table native way to do it?
casted2<-na.locf(casted2)
Any advice/improvement in the process is welcome.
I'd maybe do the char and num tables separately and then rbind:
k = "time"
typecols = c("numeric", "character")
res = rbindlist(fill = TRUE,
lapply(typecols, function(tc){
cols = c(k, tc, "variable")
dt[!is.na(get(tc)), ..cols][, dcast(.SD, ... ~ variable, value.var=tc)]
})
)
setorderv(res, k)
res[, setdiff(names(res), k) := lapply(.SD, zoo::na.locf, na.rm = FALSE), .SDcols=!k]
which gives
time a b c d e f
1: 1 1 NA NA NA NA NA
2: 2 1 2 NA NA NA NA
3: 3 1 2 3 NA NA NA
4: 4 1 2 3 a NA NA
5: 5 1 2 3 a b NA
6: 6 1 2 3 a b c
Note that OP's final result casted2, differs in that it has all cols as char.
Related
I am trying to apply a function to columns of a tibble, or data.frame, depending on the index of columns. It appears to me several time, and I give just one MWE
library(tidyverse)
test <- data.frame(a = c(1,2,3), b = c(7,8,9), c = c(3,5,6))
test <- test %>% as_tibble() %>% mutate_all( ~lead(., 2))
This will lead by 2 every columns (just an example). But what I want is to lead the first column by 1, the second by 2, and so on. Doing something like mutate_all(~lead(., col_number()).
For this little example, I know one way to do it, like:
test <- as.matrix(test)
for (i in 1:ncol(test)){ test[,i] <- lead(test[,i], i) }
There might be other way to do it too, haven't thought about it much (one needs to convert as a matrix first, otherwise it doesn't produce the right result, I don't really know why).
But I'd like to do it with a mutate or apply, being able to get the index of column in general. With a more complex example.
Any idea?
One option is using purrr::map2_df to sequentially lead every column based on column number.
purrr::map2_df(test, seq_along(test), dplyr::lead)
# A tibble: 3 x 3
# a b c
# <dbl> <dbl> <dbl>
#1 2 9 NA
#2 3 NA NA
#3 NA NA NA
We can also use base R Map
test[] <- Map(function(x, y) c(tail(x, -y), rep(NA, y)), test, seq_along(test))
We can use data.table shift
library(data.table)
setDT(test)[, Map(shift, .SD, n = 1:3, type = 'lead')]
# a b c
#1: 2 9 NA
#2: 3 NA NA
#3: NA NA NA
Or using purrr
library(purrr)
map2_dfr(test, 1:3, ~shift(.x, type = 'lead'))
I have a pairwise table of values, and I'm trying to find the fastest way to apply some function to various subsets of this table. I'm experimenting with data.table to see if it will suit my needs.
For example, I start with this vector of data points, which I convert to a pairwise distance matrix.
dat <- c(spA = 4, spB = 10, spC = 8, spD = 1, spE = 5, spF = 9)
pdist <- as.matrix(dist(dat))
pdist[upper.tri(pdist, diag = TRUE)] <- NA
It looks like this:
> pdist
spA spB spC spD spE spF
spA NA NA NA NA NA NA
spB 6 NA NA NA NA NA
spC 4 2 NA NA NA NA
spD 3 9 7 NA NA NA
spE 1 5 3 4 NA NA
spF 5 1 1 8 4 NA
Converting this table to a data.table
library(data.table)
pdist <- as.data.table(pdist, keep.rownames=TRUE)
setkey(pdist, rn)
> pdist
rn spA spB spC spD spE spF
1: spA NA NA NA NA NA NA
2: spB 6 NA NA NA NA NA
3: spC 4 2 NA NA NA NA
4: spD 3 9 7 NA NA NA
5: spE 1 5 3 4 NA NA
6: spF 5 1 1 8 4 NA
If I have some subset that I want to extract the values for,
sub <- c('spB', 'spF', 'spD')
I can do the following, which yields the submatrix that I am interested in:
> pdist[.(sub), sub, with=FALSE]
spB spF spD
1: NA NA NA
2: 1 NA 8
3: 9 NA NA
Now, how can I apply a function, for example taking the mean (but potentially a custom function), of all values in this subset? I can do it this way, but I wonder if there are better ways in line with data.table manipulation.
> mean(unlist(pdist[.(sub), sub, with=FALSE]), na.rm=TRUE)
[1] 6
UPDATE
Following up on this, I decided to see how different in performance a matrix vs a data.table approach would be:
dat <- runif(1000)
names(dat) <- paste0('sp', 1:1000)
spSub <- replicate(10000, sample(names(dat), 100), simplify=TRUE)
# calculate pairwise distance matrix
pdist <- as.matrix(dist(dat))
pdist[upper.tri(pdist, diag = TRUE)] <- NA
# convert to data.table
pdistDT <- as.data.table(pdist, keep.rownames='sp')
setkey(pdistDT, sp)
matMethod <- function(pdist, sub) {
return(mean(pdist[sub, sub], na.rm=TRUE))
}
dtMethod <- function(pdistDT, sub) {
return(mean(unlist(pdistDT[.(sub), sub, with=FALSE]), na.rm=TRUE))
}
> system.time(q1 <- lapply(spSub, function(x) matMethod(pdist, x)))
user system elapsed
18.116 0.154 18.317
> system.time(q2 <- lapply(spSub, function(x) dtMethod(pdistDT, x)))
user system elapsed
795.456 13.357 806.820
It appears that going through the data.table step here is leading to a big performance cost.
Please see the solution posted here for an every more general solution. It may also help:
data.table: transforming subset of columns with a function, row by row
To apply the function, you can do the following:
Part 1. A Step-by-Step Solution
(1.a) Get the data into Data.Table format:
library(data.table)
library(magrittr) #for access to pipe operator
pdist <- as.data.table(pdist, keep.rownames=TRUE)
setkey(pdist, rn)
(1.b) Then, Get the list of Column Names:
# Get the list of names
sub <- c('spB', 'spF', 'spD')
(1.c) Define the function you want to apply
#Define the function you wish to apply
# Where, normalize is just a function as defined in the question:
normalize <- function(X, X.mean = mean(X, na.rm=T), X.sd = sd(X, na.rm=T)){
X <- (X - X.mean) / X.sd
return(X)}
(1.d) Apply the function:
# Voila:
pdist[, unlist(.SD, use.names = FALSE), .SDcols = sub] %>% normalize()
#Or, you can apply the function inside the [], as below:
pdist[, unlist(.SD, use.names = FALSE) %>% normalize(), .SDcols = sub]
# Or, if you prefer to do it without the pipe operator:
pdist[, normalize(unlist(.SD, use.names = FALSE)), .SDcols = sub]
Part 2. Some Advantages for Data.Table approach
Since you seem familiar with matrix approach, I just wanted to point out some advantages of keeping the data.table approach
(2.a) Apply functions within group by using the "by ="
One advantage over matrix is that you can still apply functions within group by using the "by =" argument.
In the example here, I assume you have a variable called "Grp."
With the by=Grp line, the normalization is within group now.
pdist[, unlist(.SD) %>% normalize(), .SDcols = sub, by=Grp]
(2.b) Another advantage is that you can keep other identifying information, for example, if each row has a "participant identifier" P.Id that you wish to keep and repeat:
pdist[, .(Combined.Data = unlist(.SD)), .SDcols = sub, by=P.Id][order(P.Id),.(P.Id, Transformed = normalize(Combined.Data), Combined.Data)]
In the first step, done in this portion of the code: pdist[, .(Combined.Data = unlist(.SD)), .SDcols = sub, by=P.Id]
First, we create a new column called Combined.Data for data in all three columns identified in "sub"
Next to each row of the combined data, the appropriate Participant Id will repeat in column P.Id
In the second step, done in this portion of the code:
[,.(P.Id, Normalized = normalize(Combined.Data), Combined.Data)]
We can create a new column called Normalized to store the normalized values that result from applying the function normalize()
In addition, we can also include the Combined.Data column as well
So, with this single line:
pdist[, .(Combined.Data = unlist(.SD)), .SDcols = sub, by=P.Id][order(P.Id),.(P.Id, Transformed = normalize(Combined.Data), Combined.Data)]
we subset columns,
collapse data across the subset,
keep track of the identifier for each datum (P.Id) even when collapsed,
apply a transformation on the entire collapsed data, and
end-up with a neat output in the form of a data table with 3 columns: (1) P.Id, (2) Transformed, & (3) Combined.Data (original values).
and, the order(P.Id) allows the output to appear meaningfully ordered.
The same would be possible with matrix approach, but would be much more cumbersome and take more lines of code.
Data table allows for powerful manipulation and management of data, especially when you start chaining operations together.
(2.c) Finally, if you just wish to keep row information as simple row.numbers, you can use the .I feature of the data.table package:
pdist[, .(.I, normalize(unlist(.SD)), .SDcols = sub]
This feature can be quite helpful, especially if you dont have a participant or row identifier that is inherently meaningful.
Part 3. Disadvantage: Time Cost
I recreated the corrected time cost shown above and the solution for Data Table does take significantly longer
dat <- runif(1000)
names(dat) <- paste0('sp', 1:1000)
spSub <- replicate(10000, sample(names(dat), 100), simplify=TRUE)
# calculate pairwise distance matrix
pdist <- as.matrix(dist(dat))
pdist[upper.tri(pdist, diag = TRUE)] <- NA
# convert to data.table
pdistDT <- as.data.table(pdist, keep.rownames='sp')
# pdistDT$sp %<>% as.factor()
setkey(pdistDT, sp)
matMethod <- function(pdist, sub) {
return(mean(pdist[sub, sub], na.rm=TRUE))
}
dtMethod <- function(pdistDT, sub) {
return(pdistDT[sub, sub, with = FALSE] %>%
unlist(., recursive = FALSE, use.names = FALSE) %>%
mean(., na.rm = TRUE))
}
dtMethod1 <- function(pdistDT, sub) {
return(pdistDT[sub, sub, with = FALSE] %>%
melt.data.table(., measure.vars = sub, na.rm=TRUE) %$%
mean(value))
}
system.time(q1 <- apply(spSub, MARGIN = 2, function(x) matMethod(pdist, x)))
# user system elapsed
# 2.86 0.00 3.27
system.time(q2 <- apply(spSub, MARGIN = 2, function(x) dtMethod(pdistDT, x)))
# user system elapsed
# 57.20 0.02 57.23
system.time(q3 <- apply(spSub, MARGIN = 2, function(x) dtMethod1(pdistDT, x)))
# user system elapsed
# 62.78 0.06 62.91
Here is what my data look like.
id interest_string
1 YI{Z0{ZI{
2 ZO{
3 <NA>
4 ZT{
As you can see, can be multiple codes concatenated into a single column, seperated by {. It is also possible for a row to have no interest_string values at all.
How can I manipulate this data frame to extract the values into a format like this:
id interest
1 YI
1 Z0
1 ZI
2 Z0
3 <NA>
4 ZT
I need to complete this task with R.
Thanks in advance.
This is one solution
out <- with(dat, strsplit(as.character(interest_string), "\\{"))
## or
# out <- with(dat, strsplit(as.character(interest_string), "{", fixed = TRUE))
out <- cbind.data.frame(id = rep(dat$id, times = sapply(out, length)),
interest = unlist(out, use.names = FALSE))
Giving:
R> out
id interest
1 1 YI
2 1 Z0
3 1 ZI
4 2 ZO
5 3 <NA>
6 4 ZT
Explanation
The first line of solution simply splits each element of the interest_string factor in data object dat, using \\{ as the split indicator. This indicator has to be escaped and in R that requires two \. (Actually it doesn't if you use fixed = TRUE in the call to strsplit.) The resulting object is a list, which looks like this for the example data
R> out
[[1]]
[1] "YI" "Z0" "ZI"
[[2]]
[1] "ZO"
[[3]]
[1] "<NA>"
[[4]]
[1] "ZT"
We have almost everything we need in this list to form the output you require. The only thing we need external to this list is the id values that refer to each element of out, which we grab from the original data.
Hence, in the second line, we bind, column-wise (specifying the data frame method so we get a data frame returned) the original id values, each one repeated the required number of times, to the strsplit list (out). By unlisting this list, we unwrap it to a vector which is of the required length as given by your expected output. We get the number of times we need to replicate each id value from the lengths of the components of the list returned by strsplit.
A nice and tidy data.table solution:
library(data.table)
DT <- data.table( read.table( textConnection("id interest_string
1 YI{Z0{ZI{
2 ZO{
3 <NA>
4 ZT{"), header=TRUE))
DT$interest_string <- as.character(DT$interest_string)
DT[, {
list(interest=unlist(strsplit( interest_string, "{", fixed=TRUE )))
}, by=id]
gives me
id interest
1: 1 YI
2: 1 Z0
3: 1 ZI
4: 2 ZO
5: 3 <NA>
6: 4 ZT
I have read in a large data file into R using the following command
data <- as.data.set(spss.system.file(paste(path, file, sep = '/')))
The data set contains columns which should not belong, and contain only blanks. This issue has to do with R creating new variables based on the variable labels attached to the SPSS file (Source).
Unfortunately, I have not been able to determine the options necessary to resolve the problem. I have tried all of: foreign::read.spss, memisc:spss.system.file, and Hemisc::spss.get, with no luck.
Instead, I would like to read in the entire data set (with ghost columns) and remove unnecessary variables manually. Since the ghost columns contain only blank spaces, I would like to remove any variables from my data.table where the number of unique observations is equal to one.
My data are large, so they are stored in data.table format. I would like to determine an easy way to check the number of unique observations in each column, and drop columns which contain only one unique observation.
require(data.table)
### Create a data.table
dt <- data.table(a = 1:10,
b = letters[1:10],
c = rep(1, times = 10))
### Create a comparable data.frame
df <- data.frame(dt)
### Expected result
unique(dt$a)
### Expected result
length(unique(dt$a))
However, I wish to calculate the number of obs for a large data file, so referencing each column by name is not desired. I am not a fan of eval(parse()).
### I want to determine the number of unique obs in
# each variable, for a large list of vars
lapply(names(df), function(x) {
length(unique(df[, x]))
})
### Unexpected result
length(unique(dt[, 'a', with = F])) # Returns 1
It seems to me the problem is that
dt[, 'a', with = F]
returns an object of class "data.table". It makes sense that the length of this object is 1, since it is a data.table containing 1 variable. We know that data.frames are really just lists of variables, and so in this case the length of the list is just 1.
Here's pseudo code for how I would remedy the solution, using the data.frame way:
for (x in names(data)) {
unique.obs <- length(unique(data[, x]))
if (unique.obs == 1) {
data[, x] <- NULL
}
}
Any insight as to how I may more efficiently ask for the number of unique observations by column in a data.table would be much appreciated. Alternatively, if you can recommend how to drop observations if there is only one unique observation within a data.table would be even better.
Update: uniqueN
As of version 1.9.6, there is a built in (optimized) version of this solution, the uniqueN function. Now this is as simple as:
dt[ , lapply(.SD, uniqueN)]
If you want to find the number of unique values in each column, something like
dt[, lapply(.SD, function(x) length(unique(x)))]
## a b c
## 1: 10 10 1
To get your function to work you need to use with=FALSE within [.data.table, or simply use [[ instead (read fortune(312) as well...)
lapply(names(df) function(x) length(unique(dt[, x, with = FALSE])))
or
lapply(names(df) function(x) length(unique(dt[[x]])))
will work
In one step
dt[,names(dt) := lapply(.SD, function(x) if(length(unique(x)) ==1) {return(NULL)} else{return(x)})]
# or to avoid calling `.SD`
dt[, Filter(names(dt), f = function(x) length(unique(dt[[x]]))==1) := NULL]
The approaches in the other answers are good. Another way to add to the mix, just for fun :
for (i in names(DT)) if (length(unique(DT[[i]]))==1) DT[,(i):=NULL]
or if there may be duplicate column names :
for (i in ncol(DT):1) if (length(unique(DT[[i]]))==1) DT[,(i):=NULL]
NB: (i) on the LHS of := is a trick to use the value of i rather than a column named "i".
Here is a solution to your core problem (I hope I got it right).
require(data.table)
### Create a data.table
dt <- data.table(a = 1:10,
b = letters[1:10],
d1 = "",
c = rep(1, times = 10),
d2 = "")
dt
a b d1 c d2
1: 1 a 1
2: 2 b 1
3: 3 c 1
4: 4 d 1
5: 5 e 1
6: 6 f 1
7: 7 g 1
8: 8 h 1
9: 9 i 1
10: 10 j 1
First, I introduce two columns d1 and d2 that have no values whatsoever. Those you want to delete, right? If so, I just identify those columns and select all other columns in the dt.
only_space <- function(x) {
length(unique(x))==1 && x[1]==""
}
bolCols <- apply(dt, 2, only_space)
dt[, (1:ncol(dt))[!bolCols], with=FALSE]
Somehow, I have the feeling that you could further simplify it...
Output:
a b c
1: 1 a 1
2: 2 b 1
3: 3 c 1
4: 4 d 1
5: 5 e 1
6: 6 f 1
7: 7 g 1
8: 8 h 1
9: 9 i 1
10: 10 j 1
There is an easy way to do that using "dplyr" library, and then use select function as follow:
library(dplyr)
newdata <- select(old_data, first variable,second variable)
Note that, you can choose as many variables as you like.
Then you will get the type of data that you want.
Many thanks,
Fadhah
I am working with a large (millions of rows) data table with a list column containing deeply nested lists, which do not have uniform structure, size or order of elements (list(x=1,y=2) and list(y=2,x=1) may both be present and should be treated as identical). I need to repeatedly perform arbitrary groupings that include some columns from the data table as well as a subset of the data in the list column. Not all rows have values that will match the subset.
The approach I've come up with feels overly complicated. Here are the key points:
Identifying values in a nested list structure. My approach is to use ul <- unlist(list_col), which "flattens" nested data structures and builds hierarchical names for direct access to each element, e.g., address.country.code.
Ensuring that permutations of the same unlisted data are considered equal from a grouping standpoint. My approach is to order the unlisted vectors by the names of their values via ul[order(names(ul))] and assign the result as a new character vector column by reference.
Performing grouping on subsets of the flattened values. I was not able to get by= to work in any way with a column whose values are lists or vectors. Therefore, I had to find a way to map unique character vectors to simple values. I did this with digest.
Here are the two workhorse functions:
# Flatten list column in a data.table
flatten_list_col <- function(dt, col_name, flattened_col_name='props') {
flatten_props <- function(d) {
if (length(d) > 0) {
ul <- unlist(d)
names <- names(ul)
if (length(names) > 0) {
ul[order(names)]
} else {
NA
}
} else {
NA
}
}
flattened <- lapply(dt[[col_name]], flatten_props)
dt[, as.character(flattened_col_name) := list(flattened), with=F]
}
# Group by properties in a flattened list column
group_props <- function(prop_group, prop_col_name='props') {
substitute({
l <- lapply(eval(as.name(prop_col_name)), function(x) x[names(x) %in% prop_group])
as.character(lapply(l, digest))
}, list(prop_group=prop_group, prop_col_name=prop_col_name))
}
Here is a reproducible example:
library(data.table)
dt <- data.table(
id=c(1,1,1,2,2,2),
count=c(1,1,2,2,3,3),
d=list(
list(x=1, y=2),
list(y=2, x=1),
list(x=1, y=2, z=3),
list(y=5, abc=list(a=1, b=2, c=3)),
NA,
NULL
)
)
flatten_list_col(dt, 'd')
dt[, list(total=sum(count)), by=list(id, eval(group_props(c('x', 'y'))))]
The output is:
> flatten_list_col(dt, 'd')
id count d props
1: 1 1 <list> 1,2
2: 1 1 <list> 1,2
3: 1 2 <list> 1,2,3
4: 2 2 <list> 1,2,3,5
5: 2 3 NA NA
6: 2 3 NA
> dt[, list(total=sum(count)), by=list(id, eval(group_props(c('x', 'y'))))]
id group_props total
1: 1 325c6bbb2c33456d0301cf3909dd1572 4
2: 2 7aa1e567cd0d6920848d331d3e49fb7e 2
3: 2 ee7aa3b9ffe6bffdee83b6ecda90faac 6
This approach works but is pretty inefficient because of the need to flatten & order the lists and because of the need to calculate digests. I'm wondering about the following:
Can this be done without having to create a flattened column by instead retrieving values directly from the list column? This will probably require specifying selected properties as expressions as opposed to simple names.
Is there a way to get around the need for digest?
There are a number of issues here. The most important (and one you haven't come to yet due to others) is that that you are assigning by reference but trying to replace with more values than you have space to do so by reference.
Take this very simple example
DT <- data.table(x=1, y = list(1:5))
DT[,new := unlist(y)]
Warning message:
In `[.data.table`(DT, , `:=`(new, unlist(y))) :
Supplied 5 items to be assigned to 1 items of column 'new' (4 unused)
You will lose all but the firstnrow(DT) items in the newly created list. They wont correspond to the rows of the data.table
Therefore you will have to create a new data.table that will be large enough for you to explode these list variables. This won't be possible by reference.
newby <- dt[,list(x, props = as.character(unlist(data))), by = list(newby = seq_len(nrow(dt)))][,newby:=NULL]
newby
x props
1: 1 1
2: 1 2
3: 1 2
4: 1 1
5: 1 10
6: 2 1
7: 2 2
8: 2 3
9: 2 5
10: 2 1
11: 2 2
12: 2 3
13: 3 NA
14: 3 NA
Note that as.character is required to ensure that all values are the same type, and a type that won't lose data in the conversion. At the momemnt you have a logical NA value amongst lists of numeric / integer data.
Another edit to force all components to be character (even the NA). props is now a list with 1 character vector for each row.
flatten_props <- function(data) {
if (is.list(data)){
ul <- unlist(data)
if (length(ul) > 1) {
ul <- ul[order(names(ul))]
}
as.character(ul) } else {
as.character(unlist(data))}}
dt[, props := lapply(data, flatten_props)]
dt
x data props
1: 1 <list> 1,2
2: 1 <list> 10,1,2
3: 2 <list> 1,2,3
4: 2 <list> 1,2,3,5
5: 3 NA NA
6: 3
dt[,lapply(props,class)]
V1 V2 V3 V4 V5 V6
1: character character character character character character