Mixing
Replacing an element within a vector in a list
I have a list of vectors that looks like
[[1]][1] 1 1 2
[[2]]
[1] 1 1 2
[[3]]
[1] 2 1 1
[[4]]
[1] 2 2 2
I would like the replace the first component of each of the vectors with a 9. I have tried
out <- append(vecs2T2[[1]], y, after=0)
but this just adds an 9 in at the start and does not replace it (see below).
[1] 9 1 1 2
I would like this entry to read 912.
r
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
asked Nov 21 '18 at 14:23
Sarah Sarah
65
add a comment |
I have a list of vectors that looks like
[[1]][1] 1 1 2
[[2]]
[1] 1 1 2
[[3]]
[1] 2 1 1
[[4]]
[1] 2 2 2
I would like the replace the first component of each of the vectors with a 9. I have tried
out <- append(vecs2T2[[1]], y, after=0)
but this just adds an 9 in at the start and does not replace it (see below).
[1] 9 1 1 2
I would like this entry to read 912.
r
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
asked Nov 21 '18 at 14:23
Sarah Sarah
65
add a comment |
I have a list of vectors that looks like
[[1]][1] 1 1 2
[[2]]
[1] 1 1 2
[[3]]
[1] 2 1 1
[[4]]
[1] 2 2 2
I would like the replace the first component of each of the vectors with a 9. I have tried
out <- append(vecs2T2[[1]], y, after=0)
but this just adds an 9 in at the start and does not replace it (see below).
[1] 9 1 1 2
I would like this entry to read 912.
r
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
asked Nov 21 '18 at 14:23
Sarah Sarah
65
I have a list of vectors that looks like
[[1]][1] 1 1 2
[[2]]
[1] 1 1 2
[[3]]
[1] 2 1 1
[[4]]
[1] 2 2 2
I would like the replace the first component of each of the vectors with a 9. I have tried
out <- append(vecs2T2[[1]], y, after=0)
but this just adds an 9 in at the start and does not replace it (see below).
[1] 9 1 1 2
I would like this entry to read 912.
r
r
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
asked Nov 21 '18 at 14:23
Sarah Sarah
65
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
asked Nov 21 '18 at 14:23
Sarah Sarah
65
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
edited Nov 21 '18 at 14:35
Sotos
29.8k51640
29.8k51640
asked Nov 21 '18 at 14:23
Sarah Sarah
65
asked Nov 21 '18 at 14:23
Sarah Sarah
65
asked Nov 21 '18 at 14:23
Sarah Sarah
65
65
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
lapply(ll, replace, 1, 9)
This goes vector by vector, and replaces the 1st item with 9. (Replace's arguments are: (data, list-of-indexes, list-of-values), with the list of values recycled to be as long as the list of indexes.)
replace() is just defined as:
replace <- function (x, list, values) {
x[list] <- values
x
}
so you can also use that method.
lapply(ll, function(x) { x[1] <- 9 ; x })
You can use either with purrr::map(), too:
purrr::map(ll, ~{ .x[1] <- 9 ; .x })
purrr::map(ll, replace, 1, 9)
Head-to-head (not the best microbenchmark setup in the world tho):
microbenchmark::microbenchmark(
purr_repl = purrr::map(ll, replace, 1, 9),
purr_op = purrr::map(ll, ~{ .x[1] <- 9 ; .x }),
lapp_repl = lapply(ll, replace, 1, 9),
lapp_op = lapply(ll, function(x) { x[1] <- 9 ; x }),
Map = Map(function(x, y)c(x, y[-1]), 9, ll)
)
## Unit: microseconds
## expr min lq mean median uq max neval
## purr_repl 27.510 29.7555 49.98242 31.4735 33.4805 1506.400 100
## purr_op 84.415 86.9550 125.07364 90.0665 98.9465 2423.406 100
## lapp_repl 4.422 4.8350 5.94472 5.1965 5.5930 34.947 100
## lapp_op 4.672 5.4250 19.14590 5.9045 6.5015 1215.477 100
## Map 10.670 12.2490 28.94712 13.5935 14.7170 1238.311 100
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
answered Nov 21 '18 at 14:29
iodiod
3,8432722
That worked perfectly. Thank you
– Sarah
Nov 21 '18 at 14:31
Happy to hear. Don't forget to accept! :)
– iod
Nov 21 '18 at 14:33
@hrbrmstr - thanks for all the edits, but the bottom line from your comparison at the bottom is that my original solution is the fastest one, by a lot, isn't it?
– iod
Nov 21 '18 at 14:42
Yep, hence why I added it. Your solution was 👍🏼 (some folks have their heads stuck in the tidyverse so they'll appreciate the purrr additions despite the serious performance hit). Adding the alternates outside your answer seemed wrong since it's all the same thing (just less performant)
– hrbrmstr
Nov 21 '18 at 14:45
add a comment |
Another idea is to use Map and concatenate 9 with the each vector minus its first element
Map(function(x, y)c(x, y[-1]), 9, l1)
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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oldest
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oldest
votes
lapply(ll, replace, 1, 9)
This goes vector by vector, and replaces the 1st item with 9. (Replace's arguments are: (data, list-of-indexes, list-of-values), with the list of values recycled to be as long as the list of indexes.)
replace() is just defined as:
replace <- function (x, list, values) {
x[list] <- values
x
}
so you can also use that method.
lapply(ll, function(x) { x[1] <- 9 ; x })
You can use either with purrr::map(), too:
purrr::map(ll, ~{ .x[1] <- 9 ; .x })
purrr::map(ll, replace, 1, 9)
Head-to-head (not the best microbenchmark setup in the world tho):
microbenchmark::microbenchmark(
purr_repl = purrr::map(ll, replace, 1, 9),
purr_op = purrr::map(ll, ~{ .x[1] <- 9 ; .x }),
lapp_repl = lapply(ll, replace, 1, 9),
lapp_op = lapply(ll, function(x) { x[1] <- 9 ; x }),
Map = Map(function(x, y)c(x, y[-1]), 9, ll)
)
## Unit: microseconds
## expr min lq mean median uq max neval
## purr_repl 27.510 29.7555 49.98242 31.4735 33.4805 1506.400 100
## purr_op 84.415 86.9550 125.07364 90.0665 98.9465 2423.406 100
## lapp_repl 4.422 4.8350 5.94472 5.1965 5.5930 34.947 100
## lapp_op 4.672 5.4250 19.14590 5.9045 6.5015 1215.477 100
## Map 10.670 12.2490 28.94712 13.5935 14.7170 1238.311 100
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
answered Nov 21 '18 at 14:29
iodiod
3,8432722
That worked perfectly. Thank you
– Sarah
Nov 21 '18 at 14:31
Happy to hear. Don't forget to accept! :)
– iod
Nov 21 '18 at 14:33
@hrbrmstr - thanks for all the edits, but the bottom line from your comparison at the bottom is that my original solution is the fastest one, by a lot, isn't it?
– iod
Nov 21 '18 at 14:42
Yep, hence why I added it. Your solution was 👍🏼 (some folks have their heads stuck in the tidyverse so they'll appreciate the purrr additions despite the serious performance hit). Adding the alternates outside your answer seemed wrong since it's all the same thing (just less performant)
– hrbrmstr
Nov 21 '18 at 14:45
add a comment |
lapply(ll, replace, 1, 9)
This goes vector by vector, and replaces the 1st item with 9. (Replace's arguments are: (data, list-of-indexes, list-of-values), with the list of values recycled to be as long as the list of indexes.)
replace() is just defined as:
replace <- function (x, list, values) {
x[list] <- values
x
}
so you can also use that method.
lapply(ll, function(x) { x[1] <- 9 ; x })
You can use either with purrr::map(), too:
purrr::map(ll, ~{ .x[1] <- 9 ; .x })
purrr::map(ll, replace, 1, 9)
Head-to-head (not the best microbenchmark setup in the world tho):
microbenchmark::microbenchmark(
purr_repl = purrr::map(ll, replace, 1, 9),
purr_op = purrr::map(ll, ~{ .x[1] <- 9 ; .x }),
lapp_repl = lapply(ll, replace, 1, 9),
lapp_op = lapply(ll, function(x) { x[1] <- 9 ; x }),
Map = Map(function(x, y)c(x, y[-1]), 9, ll)
)
## Unit: microseconds
## expr min lq mean median uq max neval
## purr_repl 27.510 29.7555 49.98242 31.4735 33.4805 1506.400 100
## purr_op 84.415 86.9550 125.07364 90.0665 98.9465 2423.406 100
## lapp_repl 4.422 4.8350 5.94472 5.1965 5.5930 34.947 100
## lapp_op 4.672 5.4250 19.14590 5.9045 6.5015 1215.477 100
## Map 10.670 12.2490 28.94712 13.5935 14.7170 1238.311 100
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
answered Nov 21 '18 at 14:29
iodiod
3,8432722
That worked perfectly. Thank you
– Sarah
Nov 21 '18 at 14:31
Happy to hear. Don't forget to accept! :)
– iod
Nov 21 '18 at 14:33
@hrbrmstr - thanks for all the edits, but the bottom line from your comparison at the bottom is that my original solution is the fastest one, by a lot, isn't it?
– iod
Nov 21 '18 at 14:42
Yep, hence why I added it. Your solution was 👍🏼 (some folks have their heads stuck in the tidyverse so they'll appreciate the purrr additions despite the serious performance hit). Adding the alternates outside your answer seemed wrong since it's all the same thing (just less performant)
– hrbrmstr
Nov 21 '18 at 14:45
add a comment |
lapply(ll, replace, 1, 9)
This goes vector by vector, and replaces the 1st item with 9. (Replace's arguments are: (data, list-of-indexes, list-of-values), with the list of values recycled to be as long as the list of indexes.)
replace() is just defined as:
replace <- function (x, list, values) {
x[list] <- values
x
}
so you can also use that method.
lapply(ll, function(x) { x[1] <- 9 ; x })
You can use either with purrr::map(), too:
purrr::map(ll, ~{ .x[1] <- 9 ; .x })
purrr::map(ll, replace, 1, 9)
Head-to-head (not the best microbenchmark setup in the world tho):
microbenchmark::microbenchmark(
purr_repl = purrr::map(ll, replace, 1, 9),
purr_op = purrr::map(ll, ~{ .x[1] <- 9 ; .x }),
lapp_repl = lapply(ll, replace, 1, 9),
lapp_op = lapply(ll, function(x) { x[1] <- 9 ; x }),
Map = Map(function(x, y)c(x, y[-1]), 9, ll)
)
## Unit: microseconds
## expr min lq mean median uq max neval
## purr_repl 27.510 29.7555 49.98242 31.4735 33.4805 1506.400 100
## purr_op 84.415 86.9550 125.07364 90.0665 98.9465 2423.406 100
## lapp_repl 4.422 4.8350 5.94472 5.1965 5.5930 34.947 100
## lapp_op 4.672 5.4250 19.14590 5.9045 6.5015 1215.477 100
## Map 10.670 12.2490 28.94712 13.5935 14.7170 1238.311 100
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
answered Nov 21 '18 at 14:29
iodiod
3,8432722
lapply(ll, replace, 1, 9)
This goes vector by vector, and replaces the 1st item with 9. (Replace's arguments are: (data, list-of-indexes, list-of-values), with the list of values recycled to be as long as the list of indexes.)
replace() is just defined as:
replace <- function (x, list, values) {
x[list] <- values
x
}
so you can also use that method.
lapply(ll, function(x) { x[1] <- 9 ; x })
You can use either with purrr::map(), too:
purrr::map(ll, ~{ .x[1] <- 9 ; .x })
purrr::map(ll, replace, 1, 9)
Head-to-head (not the best microbenchmark setup in the world tho):
microbenchmark::microbenchmark(
purr_repl = purrr::map(ll, replace, 1, 9),
purr_op = purrr::map(ll, ~{ .x[1] <- 9 ; .x }),
lapp_repl = lapply(ll, replace, 1, 9),
lapp_op = lapply(ll, function(x) { x[1] <- 9 ; x }),
Map = Map(function(x, y)c(x, y[-1]), 9, ll)
)
## Unit: microseconds
## expr min lq mean median uq max neval
## purr_repl 27.510 29.7555 49.98242 31.4735 33.4805 1506.400 100
## purr_op 84.415 86.9550 125.07364 90.0665 98.9465 2423.406 100
## lapp_repl 4.422 4.8350 5.94472 5.1965 5.5930 34.947 100
## lapp_op 4.672 5.4250 19.14590 5.9045 6.5015 1215.477 100
## Map 10.670 12.2490 28.94712 13.5935 14.7170 1238.311 100
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
answered Nov 21 '18 at 14:29
iodiod
3,8432722
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
edited Nov 21 '18 at 14:35
hrbrmstr
61k688150
61k688150
answered Nov 21 '18 at 14:29
iodiod
3,8432722
answered Nov 21 '18 at 14:29
iodiod
3,8432722
answered Nov 21 '18 at 14:29
iodiod
3,8432722
3,8432722
That worked perfectly. Thank you
– Sarah
Nov 21 '18 at 14:31
Happy to hear. Don't forget to accept! :)
– iod
Nov 21 '18 at 14:33
@hrbrmstr - thanks for all the edits, but the bottom line from your comparison at the bottom is that my original solution is the fastest one, by a lot, isn't it?
– iod
Nov 21 '18 at 14:42
Yep, hence why I added it. Your solution was 👍🏼 (some folks have their heads stuck in the tidyverse so they'll appreciate the purrr additions despite the serious performance hit). Adding the alternates outside your answer seemed wrong since it's all the same thing (just less performant)
– hrbrmstr
Nov 21 '18 at 14:45
add a comment |
That worked perfectly. Thank you
– Sarah
Nov 21 '18 at 14:31
Happy to hear. Don't forget to accept! :)
– iod
Nov 21 '18 at 14:33
@hrbrmstr - thanks for all the edits, but the bottom line from your comparison at the bottom is that my original solution is the fastest one, by a lot, isn't it?
– iod
Nov 21 '18 at 14:42
Yep, hence why I added it. Your solution was 👍🏼 (some folks have their heads stuck in the tidyverse so they'll appreciate the purrr additions despite the serious performance hit). Adding the alternates outside your answer seemed wrong since it's all the same thing (just less performant)
– hrbrmstr
Nov 21 '18 at 14:45
That worked perfectly. Thank you
– Sarah
Nov 21 '18 at 14:31
That worked perfectly. Thank you
– Sarah
Nov 21 '18 at 14:31
Happy to hear. Don't forget to accept! :)
– iod
Nov 21 '18 at 14:33
Happy to hear. Don't forget to accept! :)
– iod
Nov 21 '18 at 14:33
@hrbrmstr - thanks for all the edits, but the bottom line from your comparison at the bottom is that my original solution is the fastest one, by a lot, isn't it?
– iod
Nov 21 '18 at 14:42
@hrbrmstr - thanks for all the edits, but the bottom line from your comparison at the bottom is that my original solution is the fastest one, by a lot, isn't it?
– iod
Nov 21 '18 at 14:42
Yep, hence why I added it. Your solution was 👍🏼 (some folks have their heads stuck in the tidyverse so they'll appreciate the purrr additions despite the serious performance hit). Adding the alternates outside your answer seemed wrong since it's all the same thing (just less performant)
– hrbrmstr
Nov 21 '18 at 14:45
Yep, hence why I added it. Your solution was 👍🏼 (some folks have their heads stuck in the tidyverse so they'll appreciate the purrr additions despite the serious performance hit). Adding the alternates outside your answer seemed wrong since it's all the same thing (just less performant)
– hrbrmstr
Nov 21 '18 at 14:45
add a comment |
Another idea is to use Map and concatenate 9 with the each vector minus its first element
Map(function(x, y)c(x, y[-1]), 9, l1)
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
add a comment |
Another idea is to use Map and concatenate 9 with the each vector minus its first element
Map(function(x, y)c(x, y[-1]), 9, l1)
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
add a comment |
Another idea is to use Map and concatenate 9 with the each vector minus its first element
Map(function(x, y)c(x, y[-1]), 9, l1)
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
Another idea is to use Map and concatenate 9 with the each vector minus its first element
Map(function(x, y)c(x, y[-1]), 9, l1)
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
answered Nov 21 '18 at 14:35
SotosSotos
29.8k51640
29.8k51640
add a comment |
add a comment |
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