I have a data set in Excel with a lot of vlookup formulas that I am trying to transpose in R using the data.table package.
In my example below I am saying, for each row, find the value in column y within column x and return the value in column z.
The first row results in na because the value 6 doesn't exist in column x.
On the second row the value 5 appears twice in column x but returning the first match is fine, which is e in this case
I've added in the result column which is the expected outcome.
library(data.table)
dt <- data.table(x = c(1,2,3,4,5,5),
y = c(6,5,4,3,2,1),
z = c("a", "b", "c", "d", "e", "f"),
Result = c("na", "e", "d", "c", "b", "a"))
Many thanks
You can do this with a join, but need to change the order first:
setorder(dt, y)
dt[.(x = x, z = z), result1 := i.z, on = .("y" = x)]
setorder(dt, x)
# x y z Result result1
#1: 1 6 a na NA
#2: 2 5 b e e
#3: 3 4 c d d
#4: 4 3 d c c
#5: 5 1 f a a
#6: 5 2 e b b
I haven't tested if this is faster than match for a big data.table, but it might be.
We can just use match to find the index of those matching elements of 'y' with that of 'x' and use that to index to get the corresponding 'z'
dt[, Result1 := z[match(y,x)]]
dt
# x y z Result Result1
#1: 1 6 a na NA
#2: 2 5 b e e
#3: 3 4 c d d
#4: 4 3 d c c
#5: 5 2 e b b
#6: 5 1 f a a
Related
I was trying to create a function where I aggregate data table values on the basis of one column but I can't pass the argument in toString() for column names. The following example can show it better:
t1 <- data.table(P = c("a", "b", "c", "d", "a", "b"), Q =
c("1","2","3","4","5","6"))
t1[ ,toString(Q), by = P] # this works
t1[ ,toString(colnames(t1)[2]), by = P] # this does not give me the desired result
I get the following result with the above:
P V1
1: a Q
2: b Q
3: c Q
4: d Q
As compared to the expected:
P V1
1: a 1, 5
2: b 2, 6
3: c 3
4: d 4
I have tried removing the quotes using noquotes() but nothing works out for me.
Can anyone point out where I might be making a mistake?
We can use get to return the value
t1[ , toString(get(colnames(t1)[2])), by = P]
# P V1
#1: a 1, 5
#2: b 2, 6
#3: c 3
#4: d 4
Or with eval/as.symbol
t1[, toString(eval(as.symbol(names(t1)[2]))), by = P]
Or the standard way would be to specify in .SDcols
t1[, toString(.SD[[1]]), by = P, .SDcols = names(t1)[2]]
I have been searching for a solution for subsetting a data table using matching values for certain columns in another data table.
Here is in example:
set.seed(2)
dt <-
data.table(a = 1:10,
b = rnorm(10),
c = runif(10),
d = letters[1:10])
dt2 <-
data.table(a = 5:20,
b = rnorm(16),
c = runif(16),
d = letters[5:20])
This is the result I need:
> dt2
1: 5 -2.311069085 0.62512173 e
2: 6 0.878604581 0.26030004 f
3: 7 0.035806718 0.85907312 g
4: 8 1.012828692 0.43748800 h
5: 9 0.432265155 0.38814476 i
6: 10 2.090819205 0.46150111 j
where I have the rows returned from the second data table where a and d match even though b and c may not. The real data are mutually exclusive, and I need to match on three columns.
We can use %in% to match the columns and subset accordingly.
dt2[a %in% dt$a & d %in% dt$d]
# a b c d
#1: 5 -2.31106908 0.6251217 e
#2: 6 0.87860458 0.2603000 f
#3: 7 0.03580672 0.8590731 g
#4: 8 1.01282869 0.4374880 h
#5: 9 0.43226515 0.3881448 i
#6: 10 2.09081921 0.4615011 j
Here is an option using join and specifying the on
na.omit(dt2[dt[, c("a", "d"), with = FALSE], on = c("a", "d")])
# a b c d
#1: 5 -2.31106908 0.6251217 e
#2: 6 0.87860458 0.2603000 f
#3: 7 0.03580672 0.8590731 g
#4: 8 1.01282869 0.4374880 h
#5: 9 0.43226515 0.3881448 i
#6: 10 2.09081921 0.4615011 j
I Would like to extract the next 'n' rows after I find a string in R.
For example, let's say I have the following data frame:
df<-as.data.frame(rep(c("a","b","c","d","e","f"),10))
I would like to extract every row that includes "b", as well as the next two rows (in this example, I would like to extract rows with "b", or "c", or "d")
BUT, please, I don't want to specify "c" and "d", I just want the next two rows after "b" as well (in my real data the next two rows are not consistent).
I've tried many things, but no success.. Thanks in advance! Nick
You can find the indices of rows with b and then use those and the next two of each, something like this:
df <- data.frame(col1=rep(c("a","b","c","d","e","f"),3), col2=letters[1:18], stringsAsFactors = FALSE)
df
col1 col2
1 a a
2 b b
3 c c
4 d d
5 e e
6 f f
7 a g
8 b h
9 c i
10 d j
11 e k
12 f l
13 a m
14 b n
15 c o
16 d p
17 e q
18 f r
bs <- which(df$col1=="b")
df[sort(bs+rep(0:2, each=length(bs)),] #2 is the number of rows you want after your desired match (b).
col1 col2
2 b b
3 c c
4 d d
8 b h
9 c i
10 d j
14 b n
15 c o
16 d p
I added a second column to illustrate the dataframe better, otherwise a vector would be returned.
My "SOfun" package has a function called getMyRows which does what you ask for, with the exception of returning a list instead of a data.frame.
I had left the result as a list to make it easier to handle some edge cases, like where the requests for rows would overlap. For example, in the following sample data, there are two consecutive "b" values. There's also a "b" value in the final row.
df <- data.frame(col1 = c("a", "b", "b",
rep(c("a", "b", "c", "d", "e", "f"), 3), "b"),
col2 = letters[1:22])
library(SOfun)
getMyRows(df, which(df$col1 == "b"), 0:2, TRUE)
# [[1]]
# col1 col2
# 2 b b
# 3 b c
# 4 a d
#
# [[2]]
# col1 col2
# 3 b c
# 4 a d
# 5 b e
#
# [[3]]
# col1 col2
# 5 b e
# 6 c f
# 7 d g
#
# [[4]]
# col1 col2
# 11 b k
# 12 c l
# 13 d m
#
# [[5]]
# col1 col2
# 17 b q
# 18 c r
# 19 d s
#
# [[6]]
# col1 col2
# 22 b v
The usage is essentially:
Specify the data.frame.
Specify the index positions to use as the base. Here, we want all rows where "col1" equals "b" to be our base index position.
Specify the range of rows interested in. -1:3, for example, would give you one row before to three rows after the base.
TRUE means that you are specifying the starting points by their numeric indices.
I have two named vectors
v1 <- 1:4
v2 <- 3:5
names(v1) <- c("a", "b", "c", "d")
names(v2) <- c("c", "e", "d")
I want to add them up by the names, i.e. the expected result is
> v3
a b c d e
1 2 6 9 4
Is there a way to programmatically do this in R? Note the names may not necessarily be in a sorted order, like in v2 above.
Just combine the vectors (using c, for example) and use tapply:
v3 <- c(v1, v2)
tapply(v3, names(v3), sum)
# a b c d e
# 1 2 6 9 4
Or, for fun (since you're just doing sum), continuing with "v3":
xtabs(v3 ~ names(v3))
# names(v3)
# a b c d e
# 1 2 6 9 4
I suppose with "data.table" you could also do something like:
library(data.table)
as.data.table(Reduce(c, mget(ls(pattern = "v\\d"))),
keep.rownames = TRUE)[, list(V2 = sum(V2)), by = V1]
# V1 V2
# 1: a 1
# 2: b 2
# 3: c 6
# 4: d 9
# 5: e 4
(I shared the latter not so much for "data.table" but to show an automated way of capturing the vectors of interest.)
I am having a brain cramp. Below is a toy dataset:
df <- data.frame(
id = 1:6,
v1 = c("a", "a", "c", NA, "g", "h"),
v2 = c("z", "y", "a", NA, "a", "g"),
stringsAsFactors=F)
I have a specific value that I want to find across a set of defined columns and I want to identify the position it is located in. The fields I am searching are characters and the trick is that the value I am looking for might not exist. In addition, null strings are also present in the dataset.
Assuming I knew how to do this, the variable position indicates the values I would like returned.
> df
id v1 v2 position
1 1 a z 1
2 2 a y 1
3 3 c a 2
4 4 <NA> <NA> 99
5 5 g a 2
6 6 h g 99
The general rule is that I want to find the position of value "a", and if it is not located or if v1 is missing, then I want 99 returned.
In this instance, I am searching across v1 and v2, but in reality, I have 10 different variables. It is also worth noting that the value I am searching for can only exist once across the 10 variables.
What is the best way to generate this recode?
Many thanks in advance.
Use match:
> df$position <- apply(df,1,function(x) match('a',x[-1], nomatch=99 ))
> df
id v1 v2 position
1 1 a z 1
2 2 a y 1
3 3 c a 2
4 4 <NA> <NA> 99
5 5 g a 2
6 6 h g 99
Firstly, drop the first column:
df <- df[, -1]
Then, do something like this (disclaimer: I'm feeling terribly sleepy*):
( df$result <- unlist(lapply(apply(df, 1, grep, pattern = "a"), function(x) ifelse(length(x) == 0, 99, x))) )
v1 v2 result
1 a z 1
2 a y 1
3 c a 2
4 <NA> <NA> 99
5 g a 2
6 h g 99
* sleepy = code is not vectorised
EDIT (slightly different solution, I still feel sleepy):
df$result <- rapply(apply(df, 1, grep, pattern = "a"), function(x) ifelse(length(x) == 0, 99, x))