I have the following data.table (DT):
DT <- data.table(V1 = 1:3, V2 = 4:6, V3 = 7:9)
I would like to select a subset of the variables programmatically (dynamically), by using an object where the relevant variable names are stored. For example, I want to select the two columns "V1" and "V3" stored in a variable "keep"
keep <- c("V1", "V3")
If we were to select the "keep" columns from a data.frame, the following would work:
DT[keep]
Unfortunately, this is not working when this is a data.table. I thought the data.frame and data.table are identical with this kind of behavior, but apperently they aren't. Anybody able to advise on the correct syntax?
This is covered in FAQ 1.1, 1.2 and 2.17.
Some possibilities:
DT[, keep, with = FALSE]
DT[, c('V1', 'V3'), with = FALSE]
DT[, c(1, 3), with = FALSE]
DT[, list(V1, V3)]
The reason DF[c('V1','V3')] works as it does for a data.frame is covered in ?`[.data.frame`
Data frames can be indexed in several modes. When [ and [[ are used
with a single vector index (x[i] or x[[i]]), they index the data frame
as if it were a list. In this usage a drop argument is ignored, with a
warning.
From data.table 1.10.2, you may use the .. prefix when subsetting columns programmatically:
When j is a symbol prefixed with .. it will be looked up in calling scope and its value taken to be column names or numbers [...] It is experimental.
Thus:
DT[ , ..keep]
# V1 V3
# 1: 1 7
# 2: 2 8
# 3: 3 9
Some more possibilities:
DT[, .SD, .SDcols = keep]
DT[, mget(keep)]
Related
I have the following data.table (DT):
DT <- data.table(V1 = 1:3, V2 = 4:6, V3 = 7:9)
I would like to select a subset of the variables programmatically (dynamically), by using an object where the relevant variable names are stored. For example, I want to select the two columns "V1" and "V3" stored in a variable "keep"
keep <- c("V1", "V3")
If we were to select the "keep" columns from a data.frame, the following would work:
DT[keep]
Unfortunately, this is not working when this is a data.table. I thought the data.frame and data.table are identical with this kind of behavior, but apperently they aren't. Anybody able to advise on the correct syntax?
This is covered in FAQ 1.1, 1.2 and 2.17.
Some possibilities:
DT[, keep, with = FALSE]
DT[, c('V1', 'V3'), with = FALSE]
DT[, c(1, 3), with = FALSE]
DT[, list(V1, V3)]
The reason DF[c('V1','V3')] works as it does for a data.frame is covered in ?`[.data.frame`
Data frames can be indexed in several modes. When [ and [[ are used
with a single vector index (x[i] or x[[i]]), they index the data frame
as if it were a list. In this usage a drop argument is ignored, with a
warning.
From data.table 1.10.2, you may use the .. prefix when subsetting columns programmatically:
When j is a symbol prefixed with .. it will be looked up in calling scope and its value taken to be column names or numbers [...] It is experimental.
Thus:
DT[ , ..keep]
# V1 V3
# 1: 1 7
# 2: 2 8
# 3: 3 9
Some more possibilities:
DT[, .SD, .SDcols = keep]
DT[, mget(keep)]
Here I met some problems in R about replacement coding.
Here is the original data.table. There are two datatables:
dt1 <- data.table(V1 = c(1,"A"))
dt2 <- data.table("1" = c(4,5,6), "A" = c("c","d","e"))
Now I want to replace values in dt1 with value in dt2 by matching relationship.
The desired output should be:
dt3 <- data.table(V1 = c("4,5,6", "c,d,e"))
That is, I want to replace values in dt1 with all values in the corresponding column in dt2. And this is a simple example, I want to apply it to the whole data.table in R.
I met so big trouble in dealing with this, so please help me.
Here is a way from your input to desired output.
dt1[, V1 := sapply(dt2, paste, collapse = ',')[V1]]
# Test
all.equal(dt1, dt3)
[1] TRUE
PS. Are you sure that storing the values separated by a comma in a string is the best?
We may do
dt1[, V1 := unlist(lapply(V1, function(x) toString(dt2[[x]])))]
dt1
V1
1: 4, 5, 6
2: c, d, e
I want to convert a subset of data.table cols to a new class. There's a popular question here (Convert column classes in data.table) but the answer creates a new object, rather than operating on the starter object.
Take this example:
dat <- data.frame(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value=rnorm(10))
cols <- c('ID', 'Quarter')
How best to convert to just the cols columns to (e.g.) a factor? In a normal data.frame you could do this:
dat[, cols] <- lapply(dat[, cols], factor)
but that doesn't work for a data.table, and neither does this
dat[, .SD := lapply(.SD, factor), .SDcols = cols]
A comment in the linked question from Matt Dowle (from Dec 2013) suggests the following, which works fine, but seems a bit less elegant.
for (j in cols) set(dat, j = j, value = factor(dat[[j]]))
Is there currently a better data.table answer (i.e. shorter + doesn't generate a counter variable), or should I just use the above + rm(j)?
Besides using the option as suggested by Matt Dowle, another way of changing the column classes is as follows:
dat[, (cols) := lapply(.SD, factor), .SDcols = cols]
By using the := operator you update the datatable by reference. A check whether this worked:
> sapply(dat,class)
ID Quarter value
"factor" "factor" "numeric"
As suggeted by #MattDowle in the comments, you can also use a combination of for(...) set(...) as follows:
for (col in cols) set(dat, j = col, value = factor(dat[[col]]))
which will give the same result. A third alternative is:
for (col in cols) dat[, (col) := factor(dat[[col]])]
On a smaller datasets, the for(...) set(...) option is about three times faster than the lapply option (but that doesn't really matter, because it is a small dataset). On larger datasets (e.g. 2 million rows), each of these approaches takes about the same amount of time. For testing on a larger dataset, I used:
dat <- data.table(ID=c(rep("A", 1e6), rep("B",1e6)),
Quarter=c(1:1e6, 1:1e6),
value=rnorm(10))
Sometimes, you will have to do it a bit differently (for example when numeric values are stored as a factor). Then you have to use something like this:
dat[, (cols) := lapply(.SD, function(x) as.integer(as.character(x))), .SDcols = cols]
WARNING: The following explanation is not the data.table-way of doing things. The datatable is not updated by reference because a copy is made and stored in memory (as pointed out by #Frank), which increases memory usage. It is more an addition in order to explain the working of with = FALSE.
When you want to change the column classes the same way as you would do with a dataframe, you have to add with = FALSE as follows:
dat[, cols] <- lapply(dat[, cols, with = FALSE], factor)
A check whether this worked:
> sapply(dat,class)
ID Quarter value
"factor" "factor" "numeric"
If you don't add with = FALSE, datatable will evaluate dat[, cols] as a vector. Check the difference in output between dat[, cols] and dat[, cols, with = FALSE]:
> dat[, cols]
[1] "ID" "Quarter"
> dat[, cols, with = FALSE]
ID Quarter
1: A 1
2: A 2
3: A 3
4: A 4
5: A 5
6: B 1
7: B 2
8: B 3
9: B 4
10: B 5
You can use .SDcols:
dat[, cols] <- dat[, lapply(.SD, factor), .SDcols=cols]
I have a data frame with two columns. I want to add an additional two columns to the data set with counts based on aggregates.
df <- structure(list(ID = c(1045937900, 1045937900),
SMS.Type = c("DF1", "WCB14"),
SMS.Date = c("12/02/2015 19:51", "13/02/2015 08:38"),
Reply.Date = c("", "13/02/2015 09:52")
), row.names = 4286:4287, class = "data.frame")
I want to simply count the number of Instances of SMS.Type and Reply.Date where there is no null. So in the toy example below, i will generate the 2 for SMS.Type and 1 for Reply.Date
I then want to add this to the data frame as total counts (Im aware they will duplicate out for the number of rows in the original dataset but thats ok)
I have been playing around with aggregate and count function but to no avail
mytempdf <-aggregate(cbind(testtrain$SMS.Type,testtrain$Response.option)~testtrain$ID,
train,
function(x) length(unique(which(!is.na(x)))))
mytempdf <- aggregate(testtrain$Reply.Date~testtrain$ID,
testtrain,
function(x) length(which(!is.na(x))))
Can anyone help?
Thank you for your time
Using data.table you could do (I've added a real NA to your original data).
I'm also not sure if you really looking for length(unique()) or just length?
library(data.table)
cols <- c("SMS.Type", "Reply.Date")
setDT(df)[, paste0(cols, ".count") :=
lapply(.SD, function(x) length(unique(na.omit(x)))),
.SDcols = cols,
by = ID]
# ID SMS.Type SMS.Date Reply.Date SMS.Type.count Reply.Date.count
# 1: 1045937900 DF1 12/02/2015 19:51 NA 2 1
# 2: 1045937900 WCB14 13/02/2015 08:38 13/02/2015 09:52 2 1
In the devel version (v >= 1.9.5) you also could use uniqueN function
Explanation
This is a general solution which will work on any number of desired columns. All you need to do is to put the columns names into cols.
lapply(.SD, is calling a certain function over the columns specified in .SDcols = cols
paste0(cols, ".count") creates new column names while adding count to the column names specified in cols
:= performs assignment by reference, meaning, updates the newly created columns with the output of lapply(.SD, in place
by argument is specifying the aggregator columns
After converting your empty strings to NAs:
library(dplyr)
mutate(df, SMS.Type.count = sum(!is.na(SMS.Type)),
Reply.Date.count = sum(!is.na(Reply.Date)))
While reading a data set using fread, I've noticed that sometimes I'm getting duplicated column names, for example (fread doesn't have check.names argument)
> data.table( x = 1, x = 2)
x x
1: 1 2
The question is: is there any way to remove 1 of 2 columns if they have the same name?
How about
dt[, .SD, .SDcols = unique(names(dt))]
This selects the first occurrence of each name (I'm not sure how you want to handle this).
As #DavidArenburg suggests in comments above, you could use check.names=TRUE in data.table() or fread()
.SDcols approaches would return a copy of the columns you're selecting. Instead just remove those duplicated columns using :=, by reference.
dt[, which(duplicated(names(dt))) := NULL]
# x
# 1: 1
Different approaches:
Indexing
my.data.table <- my.data.table[ ,-2]
Subsetting
my.data.table <- subset(my.data.table, select = -2)
Making unique names if 1. and 2. are not ideal (when having hundreds of columns, for instance)
setnames(my.data.table, make.names(names = names(my.data.table), unique=TRUE))
Optionnaly systematize deletion of variables which names meet some criterion (here, we'll get rid of all variables having a name ending with ".X" (X being a number, starting at 2 when using make.names)
my.data.table <- subset(my.data.table,
select = !grepl(pattern = "\\.\\d$", x = names(my.data.table)))