Related
Suppose I have two datasets that I want to left-join,
i <- data.table(id=1:3, k=7:9, l=7:9, m=7:9, n=7:9)
and
x <- data.table(id=c(1,2), x=c(10,20))
To left-join, keeping all lines in i, I execute
x[i, .(id=i.id, k=i.k, l=i.l, m=i.m, n=i.n, x=x.x), on=.(id=id)]
but I wonder whether there is an easier and more efficient way that makes it unnecessary to spell out all the columns from i.
For example, in the reverse case (that is, when I want to keep all columns from i), I could use the := operator, as in x[i, k:=i.k, on=.(id=id)]. My understanding is that this makes things also more efficient because columns do not need to be copied. Is there something comparable for this case?
you can use data-tables setcolorder() after the join..
setcolorder( x[i, on = "id"], c( names(i), "x" ) )
# id k l m n x
# 1: 1 7 7 7 7 10
# 2: 2 8 8 8 8 20
# 3: 3 9 9 9 9 NA
What's wrong with merge?
y <- merge(i, x, all.x = TRUE, by = "id")
I would like to join the two data frames :
a <- data.frame(x=c(1,3,5))
b <- data.frame(start=c(0,4),end=c(2,6),y=c("a","b"))
with a condition like (x>start)&(x<end) in order to get such a result:
# x y
#1 1 a
#2 2 <NA>
#3 3 b
I don't want to make a potentially large cartesian product and then select only the few rows matching the condition and I'd like a solution using the tidyverse (I am not interested in a solution using SQL which would be a confession of failure). I thought of the 'fuzzyjoin' package but I cannot find examples fitting my need : the function to apply for the condition has only two arguments. I also tried to put 'start' and 'end' into a single argument with data.frame(z=I(purrr::map2(b$start,b$end,list)),y=b$y)
# z y
#1 0, 2 a
#2 4, 6 b
but although the data looks fine fuzzy_left_join doesn't accept it.
I search for solutions working in more general cases (n variables on the LHS, m on the RHS, not necessarily numeric with arbitrary conditions).
UPDATE
I also want to be able to express conditions like (x=start+1)|(x=end+1) giving here:
# x y
#1 1 a
#2 3 a
#3 5 b
For this case you don't need multi_by or multy_match_fun, this works :
library(fuzzyjoin)
fuzzy_left_join(a, b, by = c(x = "start", x = "end"), match_fun = list(`>`, `<`))
# x start end y
# 1 1 0 2 a
# 2 3 NA NA <NA>
# 3 5 4 6 b
I eventually went to the code of fuzzy_join and found a way to make what I want even without proper documentation. fuzzy_let_join doesn't work but there is the following way (not really pretty and it actually does a cartesian product):
g <- function(x,y) (x>y[,"start"])&(x<y[,"end"])
fuzzy_join(a,b, multi_by = list(x="x",y=c("start","end"))
, multi_match_fun = g, mode = "left") %>% select(x,y)
data.table approach could be
library(data.table)
name1 <- setdiff(names(setDT(b)), names(setDT(a)))
#perform left outer join and then select required columns
a[b, (name1) := mget(name1), on = .(x > start, x < end)][, .(x, y)]
which gives
x y
1: 1 a
2: 3 <NA>
3: 5 b
Sample data:
a <- data.frame(x = c(1, 3, 5))
b <- data.frame(start = c(0, 4), end = c(2, 6), y = c("a", "b"))
Update: In case you want to join both dataframes on (x=start+1)|(x=end+1) condition then you can try
library(data.table)
DT1 <- as.data.table(a)
DT2 <- as.data.table(b)
#Perform 1st join on "x = start+1" and then another on "x = end+1". Finally row-bind both results.
DT <- rbindlist(list(DT1[DT2[, start_temp := start+1], on = c(x = "start_temp"), .(x, y), nomatch = 0],
DT1[DT2[, end_temp := end+1], on = c(x = "end_temp"), .(x, y), nomatch = 0]))
DT
# x y
#1: 1 a
#2: 5 b
#3: 3 a
A possible answer to explain what I am trying to do : extending dplyr in some way. And I will be happy to know if there are ways to improve this solution or some problems I didn't see.
The solution avoids the cartesian product, but duplicates into lists of data frames both one of the input data frame and the result. I didn't include the final column selection of x and y that is easy to code.
my_left_join <- function(.DATA1,.DATA2,.WHERE)
{
call = as.list(match.call())
df1 <- .DATA1
df1$._row_ <- 1:nrow(df1)
dfl1 <- replyr::replyr_split(df1,"._row_")
eval(substitute(
dfl2 <- mapply(function(.x)
{filter(.DATA2,with(.x,WHERE)) %>%
mutate(._row_=.x$._row_)}
, dfl1, SIMPLIFY=FALSE)
,list(WHERE=call$.WHERE)))
df2 <- replyr::replyr_bind_rows(dfl2)
left_join(df1,df2,by="._row_") %>% select(-._row_)
}
my_left_join(a,b,(x>start)&(x<end))
# x start end y
#1 1 0 2 a
#2 3 NA NA <NA>
#3 5 4 6 b
my_left_join(a,b,(x==(start+1))|(x==(end+1)))
# x start end y
#1 1 0 2 a
#2 3 0 2 a
#3 5 4 6 b
You can try a GenomicRanges solution
library(GenomicRanges)
# setup GRanges objects
a_gr <- GRanges(1, IRanges(a$x,a$x))
b_gr <- GRanges(1, IRanges(b$start, b$end))
# find overlaps between the two data sets
res <- as.data.frame(findOverlaps(a_gr,b_gr))
# create the expected output
a$y <- NA
a$y[res$queryHits] <- as.character(b$y)[res$subjectHits]
a
x y
1 1 a
2 3 <NA>
3 5 b
I have a huge data frame. I am stuck with if function. Let me first present the simple example and then I lay down my problem:
z <- c(0,1,2,3,4,5)
y <- c(2,2,2,3,3,3)
a <- c(1,1,1,2,2,2)
x <- data.frame(z,y,a)
Problem: I want to run if function which sums column z values based for row which has same y and a only if the second row of each group has corresponding z equals 1
I am sorry but I am quite new in R so not able to present any reasonable codes which I have done by my own.
Any help would be highly appreciated.
As mentioned, your problem isn't clearly stated.
Perhaps you are looking to do something like this:
x$new <- with(x, ave(z, y, a, FUN = function(k)
ifelse(k[2] == 1, sum(k), NA)))
x
# z y a new
# 1 0 2 1 3
# 2 1 2 1 3
# 3 2 2 1 3
# 4 3 3 2 NA
# 5 4 3 2 NA
# 6 5 3 2 NA
Here, I've created a new column "new" which sums the values of "z" grouped by "y" and "a", but only if the second value in the group is equal to 1.
Since you say your data frame is quite large, you might want to convert your data frame to a data.table object using the data.table package. You will likely find that the required operations are much faster if you have a great many rows. However, the construction of the code for your case is not straight forward with data.table.
If I understnad what you want to do (which is not entirely clear to me) you could try the following:
library(data.table)
z <- c(0,1,2,3,4,5)
y <- c(2,2,2,3,3,3)
a <- c(1,1,1,2,2,2)
x <- data.frame(z,y,a)
xx <- as.data.table(x) # Make a data.table object
setkey(xx, z) # Make the z column a key
xx[1, sum(a)] # Sum all values in column a where the key z = 1
[1] 1
# Now try the other sum you mention
xx[, sum(z), by = list(z = y)] # A column sum over groups defined by z = y
z V1
1: 2 2
2: 3 3
sum(xx[, sum(z), by = list(z = y)][, V1]) # Summing over the sums for each group should do it
[1] 5
To create the sum over the column a where z = 1, I made the z column a key. The syntax xx[1, sum(a)] sums a where the key value (z value) is 1.
I can create groups with the data.table object with by, which is analogous to a SQL WHERE clause if you are familiar with SQL. However, the result is the sum of the column z for each of groups created. This may be inefficient if you have a great many possible matching values where z = y. The outer sum adds the values for each group in the sub-selected V1 column of the inner result.
If you are going to use data.table in a serious way study the informative vignettes available for that package.
M Dowle, T Short, S Lianoglou, A Srinivasan with contributions from R Saporta and E Antonyan (2014). data.table: Extensions of data.frame. R package version 1.9.2. http://CRAN.R-project.org/package=data.table
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 looking for patterns for manipulating data.table objects whose structure resembles that of dataframes created with melt from the reshape2 package. I am dealing with data tables with millions of rows. Performance is critical.
The generalized form of the question is whether there is a way to perform grouping based on a subset of values in a column and have the result of the grouping operation create one or more new columns.
A specific form of the question could be how to use data.table to accomplish the equivalent of what dcast does in the following:
input <- data.table(
id=c(1, 1, 1, 2, 2, 2, 3, 3, 3, 3),
variable=c('x', 'y', 'y', 'x', 'y', 'y', 'x', 'x', 'y', 'other'),
value=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))
dcast(input,
id ~ variable, sum,
subset=.(variable %in% c('x', 'y')))
the output of which is
id x y
1 1 1 5
2 2 4 11
3 3 15 9
Quick untested answer: seems like you're looking for by-without-by, a.k.a. grouping-by-i :
setkey(input,variable)
input[c("x","y"),sum(value)]
This is like a fast HAVING in SQL. j gets evaluated for each row of i. In other words, the above is the same result but much faster than :
input[,sum(value),keyby=variable][c("x","y")]
The latter subsets and evals for all the groups (wastefully) before selecting only the groups of interest. The former (by-without-by) goes straight to the subset of groups only.
The group results will be returned in long format, as always. But reshaping to wide afterwards on the (relatively small) aggregated data should be relatively instant. That's the thinking anyway.
The first setkey(input,variable) might bite if input has a lot of columns not of interest. If so, it might be worth subsetting the columns needed :
DT = setkey(input[ , c("variable","value")], variable)
DT[c("x","y"),sum(value)]
In future when secondary keys are implemented that would be easier :
set2key(input,variable) # add a secondary key
input[c("x","y"),sum(value),key=2] # syntax speculative
To group by id as well :
setkey(input,variable)
input[c("x","y"),sum(value),by='variable,id']
and including id in the key might be worth setkey's cost depending on your data :
setkey(input,variable,id)
input[c("x","y"),sum(value),by='variable,id']
If you combine a by-without-by with by, as above, then the by-without-by then operates just like a subset; i.e., j is only run for each row of i when by is missing (hence the name by-without-by). So you need to include variable, again, in the by as shown above.
Alternatively, the following should group by id over the union of "x" and "y" instead (but the above is what you asked for in the question, iiuc) :
input[c("x","y"),sum(value),by=id]
> setkey(input, "id")
> input[ , list(sum(value)), by=id]
id V1
1: 1 6
2: 2 15
3: 3 34
> input[ variable %in% c("x", "y"), list(sum(value)), by=id]
id V1
1: 1 6
2: 2 15
3: 3 24
The last one:
> input[ variable %in% c("x", "y"), list(sum(value)), by=list(id, variable)]
id variable V1
1: 1 x 1
2: 1 y 5
3: 2 x 4
4: 2 y 11
5: 3 x 15
6: 3 y 9
I'm not sure if this is the best way, but you can try:
input[, list(x = sum(value[variable == "x"]),
y = sum(value[variable == "y"])), by = "id"]
# id x y
# 1: 1 1 5
# 2: 2 4 11
# 3: 3 15 9