using data.table to replace multiple columns on single condition
I want to change the default value (which is 255) to NA.
dt <- data.table(x = c(1,5,255,0,NA), y = c(1,7,255,0,0), z = c(4,2,7,8,255))
coords <- c('x', 'y')
Which gives the following code:
x y z
1: 1 1 4
2: 5 7 2
3: 255 255 7
4: 0 0 8
5: NA 0 255
I the furthest I came up with is this:
dt[.SD == 255, (.SD) := NA, .SDcols = coords]
Please note that column z stays the same. So just the columns which are specified and not all columns.
But that doesn't help me to get the sollution:
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
I am looking for a sustainable solution because the original dataset is a couple of million rows.
EDIT:
I have found a solution but it is quite ugly and is definately too slow as it takes almost 10 seconds to get through a dataframe of 22009 x 86. Does anyone have a better solution?
The code:
dt[, replace(.SD, .SD == 255, NA), .SDcols = coords, by = c(colnames(dt)[!colnames(dt) %in% coords])]
r data.table
asked Nov 13 at 13:44
Tunder250
1618
|
show 6 more comments
I want to change the default value (which is 255) to NA.
dt <- data.table(x = c(1,5,255,0,NA), y = c(1,7,255,0,0), z = c(4,2,7,8,255))
coords <- c('x', 'y')
Which gives the following code:
x y z
1: 1 1 4
2: 5 7 2
3: 255 255 7
4: 0 0 8
5: NA 0 255
I the furthest I came up with is this:
dt[.SD == 255, (.SD) := NA, .SDcols = coords]
Please note that column z stays the same. So just the columns which are specified and not all columns.
But that doesn't help me to get the sollution:
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
I am looking for a sustainable solution because the original dataset is a couple of million rows.
EDIT:
I have found a solution but it is quite ugly and is definately too slow as it takes almost 10 seconds to get through a dataframe of 22009 x 86. Does anyone have a better solution?
The code:
dt[, replace(.SD, .SD == 255, NA), .SDcols = coords, by = c(colnames(dt)[!colnames(dt) %in% coords])]
r data.table
asked Nov 13 at 13:44
Tunder250
1618
You can trydt[, replace(.SD, .SD == 255, NA)]
– Sotos
Nov 13 at 13:46
Thank you for your reply, Sotos. I edited my post. I am looking for a solution that is easily upscalable when the amount of rows heavily increase. I am not sure if the function replace is that friendly.
– Tunder250
Nov 13 at 13:51
couple of million rows is not very big.replacewill do just fine
– Sotos
Nov 13 at 13:55
okay, thank you. But it doesn't include the other columns.
– Tunder250
Nov 13 at 13:59
2
You can do this when you read in the table:fread("path/to/file", na.strings=c("NA", "255"))
– Scott Ritchie
Nov 13 at 14:02
|
show 6 more comments
I want to change the default value (which is 255) to NA.
dt <- data.table(x = c(1,5,255,0,NA), y = c(1,7,255,0,0), z = c(4,2,7,8,255))
coords <- c('x', 'y')
Which gives the following code:
x y z
1: 1 1 4
2: 5 7 2
3: 255 255 7
4: 0 0 8
5: NA 0 255
I the furthest I came up with is this:
dt[.SD == 255, (.SD) := NA, .SDcols = coords]
Please note that column z stays the same. So just the columns which are specified and not all columns.
But that doesn't help me to get the sollution:
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
I am looking for a sustainable solution because the original dataset is a couple of million rows.
EDIT:
I have found a solution but it is quite ugly and is definately too slow as it takes almost 10 seconds to get through a dataframe of 22009 x 86. Does anyone have a better solution?
The code:
dt[, replace(.SD, .SD == 255, NA), .SDcols = coords, by = c(colnames(dt)[!colnames(dt) %in% coords])]
r data.table
asked Nov 13 at 13:44
Tunder250
1618
I want to change the default value (which is 255) to NA.
dt <- data.table(x = c(1,5,255,0,NA), y = c(1,7,255,0,0), z = c(4,2,7,8,255))
coords <- c('x', 'y')
Which gives the following code:
x y z
1: 1 1 4
2: 5 7 2
3: 255 255 7
4: 0 0 8
5: NA 0 255
I the furthest I came up with is this:
dt[.SD == 255, (.SD) := NA, .SDcols = coords]
Please note that column z stays the same. So just the columns which are specified and not all columns.
But that doesn't help me to get the sollution:
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
I am looking for a sustainable solution because the original dataset is a couple of million rows.
EDIT:
I have found a solution but it is quite ugly and is definately too slow as it takes almost 10 seconds to get through a dataframe of 22009 x 86. Does anyone have a better solution?
The code:
dt[, replace(.SD, .SD == 255, NA), .SDcols = coords, by = c(colnames(dt)[!colnames(dt) %in% coords])]
r data.table
r data.table
asked Nov 13 at 13:44
Tunder250
1618
asked Nov 13 at 13:44
Tunder250
1618
edited Nov 13 at 14:29
asked Nov 13 at 13:44
Tunder250
1618
asked Nov 13 at 13:44
Tunder250
1618
asked Nov 13 at 13:44
Tunder250
1618
1618
You can trydt[, replace(.SD, .SD == 255, NA)]
– Sotos
Nov 13 at 13:46
Thank you for your reply, Sotos. I edited my post. I am looking for a solution that is easily upscalable when the amount of rows heavily increase. I am not sure if the function replace is that friendly.
– Tunder250
Nov 13 at 13:51
couple of million rows is not very big.replacewill do just fine
– Sotos
Nov 13 at 13:55
okay, thank you. But it doesn't include the other columns.
– Tunder250
Nov 13 at 13:59
2
You can do this when you read in the table:fread("path/to/file", na.strings=c("NA", "255"))
– Scott Ritchie
Nov 13 at 14:02
|
show 6 more comments
You can trydt[, replace(.SD, .SD == 255, NA)]
– Sotos
Nov 13 at 13:46
Thank you for your reply, Sotos. I edited my post. I am looking for a solution that is easily upscalable when the amount of rows heavily increase. I am not sure if the function replace is that friendly.
– Tunder250
Nov 13 at 13:51
couple of million rows is not very big.replacewill do just fine
– Sotos
Nov 13 at 13:55
okay, thank you. But it doesn't include the other columns.
– Tunder250
Nov 13 at 13:59
2
You can do this when you read in the table:fread("path/to/file", na.strings=c("NA", "255"))
– Scott Ritchie
Nov 13 at 14:02
You can try
dt[, replace(.SD, .SD == 255, NA)]– Sotos
Nov 13 at 13:46
You can try
dt[, replace(.SD, .SD == 255, NA)]– Sotos
Nov 13 at 13:46
Thank you for your reply, Sotos. I edited my post. I am looking for a solution that is easily upscalable when the amount of rows heavily increase. I am not sure if the function replace is that friendly.
– Tunder250
Nov 13 at 13:51
Thank you for your reply, Sotos. I edited my post. I am looking for a solution that is easily upscalable when the amount of rows heavily increase. I am not sure if the function replace is that friendly.
– Tunder250
Nov 13 at 13:51
couple of million rows is not very big.
replace will do just fine– Sotos
Nov 13 at 13:55
couple of million rows is not very big.
replace will do just fine– Sotos
Nov 13 at 13:55
okay, thank you. But it doesn't include the other columns.
– Tunder250
Nov 13 at 13:59
okay, thank you. But it doesn't include the other columns.
– Tunder250
Nov 13 at 13:59
2
2
You can do this when you read in the table:
fread("path/to/file", na.strings=c("NA", "255"))– Scott Ritchie
Nov 13 at 14:02
You can do this when you read in the table:
fread("path/to/file", na.strings=c("NA", "255"))– Scott Ritchie
Nov 13 at 14:02
|
show 6 more comments
1 Answer
1
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oldest
votes
Here is how you can keep the columns outside .SDcols,
library(data.table)
dt[, (coords) := replace(.SD, .SD == 255, NA), .SDcols = coords]
which gives,
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
answered Nov 13 at 14:49
Sotos
28k51640
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
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active
oldest
votes
Here is how you can keep the columns outside .SDcols,
library(data.table)
dt[, (coords) := replace(.SD, .SD == 255, NA), .SDcols = coords]
which gives,
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
answered Nov 13 at 14:49
Sotos
28k51640
add a comment |
Here is how you can keep the columns outside .SDcols,
library(data.table)
dt[, (coords) := replace(.SD, .SD == 255, NA), .SDcols = coords]
which gives,
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
answered Nov 13 at 14:49
Sotos
28k51640
add a comment |
Here is how you can keep the columns outside .SDcols,
library(data.table)
dt[, (coords) := replace(.SD, .SD == 255, NA), .SDcols = coords]
which gives,
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
answered Nov 13 at 14:49
Sotos
28k51640
Here is how you can keep the columns outside .SDcols,
library(data.table)
dt[, (coords) := replace(.SD, .SD == 255, NA), .SDcols = coords]
which gives,
x y z
1: 1 1 4
2: 5 7 2
3: NA NA 7
4: 0 0 8
5: NA 0 255
answered Nov 13 at 14:49
Sotos
28k51640
answered Nov 13 at 14:49
Sotos
28k51640
answered Nov 13 at 14:49
Sotos
28k51640
answered Nov 13 at 14:49
Sotos
28k51640
28k51640
add a comment |
add a comment |
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You can try
dt[, replace(.SD, .SD == 255, NA)]– Sotos
Nov 13 at 13:46
Thank you for your reply, Sotos. I edited my post. I am looking for a solution that is easily upscalable when the amount of rows heavily increase. I am not sure if the function replace is that friendly.
– Tunder250
Nov 13 at 13:51
couple of million rows is not very big.
replacewill do just fine– Sotos
Nov 13 at 13:55
okay, thank you. But it doesn't include the other columns.
– Tunder250
Nov 13 at 13:59
2
You can do this when you read in the table:
fread("path/to/file", na.strings=c("NA", "255"))– Scott Ritchie
Nov 13 at 14:02