data.table does not modify a column as expected
I have a function
delete.all.after.pattern <- function (x,pattern) strsplit(x,pattern)[[1]][1]
and a data.table
a <- c(1:3)
b <- c("a","bn undesired text","c")
dt <- data.table(a=a, b=b)
Thus I expect that dt [, b:=delete.all.after.pattern(b,"\n")] would result in
a b
1: 1 a
2: 2 b
3: 3 c
instead of :
a b
1: 1 a
2: 2 a
3: 3 a
What am I missing?
r string data.table
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
add a comment |
I have a function
delete.all.after.pattern <- function (x,pattern) strsplit(x,pattern)[[1]][1]
and a data.table
a <- c(1:3)
b <- c("a","bn undesired text","c")
dt <- data.table(a=a, b=b)
Thus I expect that dt [, b:=delete.all.after.pattern(b,"\n")] would result in
a b
1: 1 a
2: 2 b
3: 3 c
instead of :
a b
1: 1 a
2: 2 a
3: 3 a
What am I missing?
r string data.table
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
add a comment |
I have a function
delete.all.after.pattern <- function (x,pattern) strsplit(x,pattern)[[1]][1]
and a data.table
a <- c(1:3)
b <- c("a","bn undesired text","c")
dt <- data.table(a=a, b=b)
Thus I expect that dt [, b:=delete.all.after.pattern(b,"\n")] would result in
a b
1: 1 a
2: 2 b
3: 3 c
instead of :
a b
1: 1 a
2: 2 a
3: 3 a
What am I missing?
r string data.table
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
I have a function
delete.all.after.pattern <- function (x,pattern) strsplit(x,pattern)[[1]][1]
and a data.table
a <- c(1:3)
b <- c("a","bn undesired text","c")
dt <- data.table(a=a, b=b)
Thus I expect that dt [, b:=delete.all.after.pattern(b,"\n")] would result in
a b
1: 1 a
2: 2 b
3: 3 c
instead of :
a b
1: 1 a
2: 2 a
3: 3 a
What am I missing?
r string data.table
r string data.table
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
edited Nov 22 '18 at 2:08
Ronak Shah
35.8k103856
35.8k103856
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
asked Nov 22 '18 at 1:44
Fabio CorreaFabio Correa
408
408
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I think you are looking for:
dt [, b := sapply(b, delete.all.after.pattern, pattern="\n")]
Your function is not vectorized and hence it only returns the first element, which is repeated for all rows.
Or you can call strsplit directly in j:
dt [, b := lapply(strsplit(b, "n"), `[[`, 1L)]
You can also put the code into a function and call it as well
fun <- function(x, p) lapply(strsplit(x, p), `[[`, 1L)
dt [, b := fun(b, "n")]
Another way is to use data.table::tstrsplit as follows:
dt[, b := tstrsplit(b, "\n", keep=1L)]
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
your full answer clarified me about other R strucure issues, thanks a lot.
– Fabio Correa
Nov 22 '18 at 10:27
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
I think you are looking for:
dt [, b := sapply(b, delete.all.after.pattern, pattern="\n")]
Your function is not vectorized and hence it only returns the first element, which is repeated for all rows.
Or you can call strsplit directly in j:
dt [, b := lapply(strsplit(b, "n"), `[[`, 1L)]
You can also put the code into a function and call it as well
fun <- function(x, p) lapply(strsplit(x, p), `[[`, 1L)
dt [, b := fun(b, "n")]
Another way is to use data.table::tstrsplit as follows:
dt[, b := tstrsplit(b, "\n", keep=1L)]
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
your full answer clarified me about other R strucure issues, thanks a lot.
– Fabio Correa
Nov 22 '18 at 10:27
add a comment |
I think you are looking for:
dt [, b := sapply(b, delete.all.after.pattern, pattern="\n")]
Your function is not vectorized and hence it only returns the first element, which is repeated for all rows.
Or you can call strsplit directly in j:
dt [, b := lapply(strsplit(b, "n"), `[[`, 1L)]
You can also put the code into a function and call it as well
fun <- function(x, p) lapply(strsplit(x, p), `[[`, 1L)
dt [, b := fun(b, "n")]
Another way is to use data.table::tstrsplit as follows:
dt[, b := tstrsplit(b, "\n", keep=1L)]
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
your full answer clarified me about other R strucure issues, thanks a lot.
– Fabio Correa
Nov 22 '18 at 10:27
add a comment |
I think you are looking for:
dt [, b := sapply(b, delete.all.after.pattern, pattern="\n")]
Your function is not vectorized and hence it only returns the first element, which is repeated for all rows.
Or you can call strsplit directly in j:
dt [, b := lapply(strsplit(b, "n"), `[[`, 1L)]
You can also put the code into a function and call it as well
fun <- function(x, p) lapply(strsplit(x, p), `[[`, 1L)
dt [, b := fun(b, "n")]
Another way is to use data.table::tstrsplit as follows:
dt[, b := tstrsplit(b, "\n", keep=1L)]
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
I think you are looking for:
dt [, b := sapply(b, delete.all.after.pattern, pattern="\n")]
Your function is not vectorized and hence it only returns the first element, which is repeated for all rows.
Or you can call strsplit directly in j:
dt [, b := lapply(strsplit(b, "n"), `[[`, 1L)]
You can also put the code into a function and call it as well
fun <- function(x, p) lapply(strsplit(x, p), `[[`, 1L)
dt [, b := fun(b, "n")]
Another way is to use data.table::tstrsplit as follows:
dt[, b := tstrsplit(b, "\n", keep=1L)]
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
edited Nov 22 '18 at 1:58
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
answered Nov 22 '18 at 1:53
chinsoon12chinsoon12
8,66111219
8,66111219
your full answer clarified me about other R strucure issues, thanks a lot.
– Fabio Correa
Nov 22 '18 at 10:27
add a comment |
your full answer clarified me about other R strucure issues, thanks a lot.
– Fabio Correa
Nov 22 '18 at 10:27
your full answer clarified me about other R strucure issues, thanks a lot.
– Fabio Correa
Nov 22 '18 at 10:27
your full answer clarified me about other R strucure issues, thanks a lot.
– Fabio Correa
Nov 22 '18 at 10:27
add a comment |
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