Access data.table columns through vector indexes?












2















i'm stucked with a problem but i can find no satisfying answers on the web. I would like to valorize a data.frame(also a data.table it's good for me) using start:end vectors. An example will clarify what i'm asking.



Suppose i have a data.framelike the following:



df <- data.frame(col_1 = rep(0, 3), col_2 = rep(0, 3), col_3 = rep(0, 3), col_4 = rep(0,3))

df

  col_1 col_2 col_3 col_4

1     0     0     0     0

2     0     0     0     0

3     0     0     0     0


And suppose i have two vectors:



indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)


I would like to valorize to 1 all values in the range indicated by the vectors by row. The output should be the following:



  col_1 col_2 col_3 col_4

1     1     1     0     0

2     0     1     1     1

3     1     1     1     0


I tried something like this:



df[ , indexesStart:indexesEnd] <- 1


But it doesn't work, it just takes indexesStart[1]:indexesEnd[1] and repeat it for all rows.



I must avoid loop cycles because my real data frame has millions rows and it is too slow. Any help is appreciated (a data.table solution would be even better)



Thank you






r dataframe data.table dynamic-columns







share|improve this question

























  • I think that can not be done without a loop (in one or the other form) because for every row you have another set of values to change.

    – jogo
    Nov 21 '18 at 10:57
















2















i'm stucked with a problem but i can find no satisfying answers on the web. I would like to valorize a data.frame(also a data.table it's good for me) using start:end vectors. An example will clarify what i'm asking.



Suppose i have a data.framelike the following:



df <- data.frame(col_1 = rep(0, 3), col_2 = rep(0, 3), col_3 = rep(0, 3), col_4 = rep(0,3))

df

  col_1 col_2 col_3 col_4

1     0     0     0     0

2     0     0     0     0

3     0     0     0     0


And suppose i have two vectors:



indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)


I would like to valorize to 1 all values in the range indicated by the vectors by row. The output should be the following:



  col_1 col_2 col_3 col_4

1     1     1     0     0

2     0     1     1     1

3     1     1     1     0


I tried something like this:



df[ , indexesStart:indexesEnd] <- 1


But it doesn't work, it just takes indexesStart[1]:indexesEnd[1] and repeat it for all rows.



I must avoid loop cycles because my real data frame has millions rows and it is too slow. Any help is appreciated (a data.table solution would be even better)



Thank you






r dataframe data.table dynamic-columns







share|improve this question

























  • I think that can not be done without a loop (in one or the other form) because for every row you have another set of values to change.

    – jogo
    Nov 21 '18 at 10:57














2












2








2








i'm stucked with a problem but i can find no satisfying answers on the web. I would like to valorize a data.frame(also a data.table it's good for me) using start:end vectors. An example will clarify what i'm asking.



Suppose i have a data.framelike the following:



df <- data.frame(col_1 = rep(0, 3), col_2 = rep(0, 3), col_3 = rep(0, 3), col_4 = rep(0,3))

df

  col_1 col_2 col_3 col_4

1     0     0     0     0

2     0     0     0     0

3     0     0     0     0


And suppose i have two vectors:



indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)


I would like to valorize to 1 all values in the range indicated by the vectors by row. The output should be the following:



  col_1 col_2 col_3 col_4

1     1     1     0     0

2     0     1     1     1

3     1     1     1     0


I tried something like this:



df[ , indexesStart:indexesEnd] <- 1


But it doesn't work, it just takes indexesStart[1]:indexesEnd[1] and repeat it for all rows.



I must avoid loop cycles because my real data frame has millions rows and it is too slow. Any help is appreciated (a data.table solution would be even better)



Thank you






r dataframe data.table dynamic-columns







share|improve this question
















i'm stucked with a problem but i can find no satisfying answers on the web. I would like to valorize a data.frame(also a data.table it's good for me) using start:end vectors. An example will clarify what i'm asking.



Suppose i have a data.framelike the following:



df <- data.frame(col_1 = rep(0, 3), col_2 = rep(0, 3), col_3 = rep(0, 3), col_4 = rep(0,3))

df

  col_1 col_2 col_3 col_4

1     0     0     0     0

2     0     0     0     0

3     0     0     0     0


And suppose i have two vectors:



indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)


I would like to valorize to 1 all values in the range indicated by the vectors by row. The output should be the following:



  col_1 col_2 col_3 col_4

1     1     1     0     0

2     0     1     1     1

3     1     1     1     0


I tried something like this:



df[ , indexesStart:indexesEnd] <- 1


But it doesn't work, it just takes indexesStart[1]:indexesEnd[1] and repeat it for all rows.



I must avoid loop cycles because my real data frame has millions rows and it is too slow. Any help is appreciated (a data.table solution would be even better)



Thank you






r dataframe data.table dynamic-columns




r dataframe data.table dynamic-columns






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 '18 at 13:38









Henrik

41.5k994109




41.5k994109










asked Nov 21 '18 at 10:39









Andrea RivoltaAndrea Rivolta

133




133













  • I think that can not be done without a loop (in one or the other form) because for every row you have another set of values to change.

    – jogo
    Nov 21 '18 at 10:57



















  • I think that can not be done without a loop (in one or the other form) because for every row you have another set of values to change.

    – jogo
    Nov 21 '18 at 10:57

















I think that can not be done without a loop (in one or the other form) because for every row you have another set of values to change.

– jogo
Nov 21 '18 at 10:57





I think that can not be done without a loop (in one or the other form) because for every row you have another set of values to change.

– jogo
Nov 21 '18 at 10:57












2 Answers
2






active

oldest

votes


















2














This will do it:



df <- data.frame(col_1=rep(0,3),col_2=rep(0,3),col_3=rep(0,3),col_4=rep(0,3))

indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)



for (i in 1:nrow(df)) df[i, indexesStart[i]:indexesEnd[i]] <- 1



df


Here is another technique using a twocolumn matrix as index:



I <- do.call(rbind, lapply(1:length(indexesStart), function(i) cbind(i, indexesStart[i]:indexesEnd[i])))

df[I] <- 1


In the second variant I hided the loop (and the hidden loop is in another place).






share|improve this answer


























  • Thanks, but i need to avoid for loops

    – Andrea Rivolta
    Nov 21 '18 at 10:51











  • In my second variant I hided the loop (and the hidden loop is in another place).

    – jogo
    Nov 21 '18 at 11:51






  • 1





    The second variant works very fast (less than a second). I had to make some adjustment because (i did not specify this aspect in the post to have it easier) because for each row i have multiple range of columns (length(indexes) > dim(df)[1]), but i have been able to manage this through indexes mapping. Now, if i View (or print) df[I] everything is ok, but when i do the assignment it says: Error in [<-.data.frame(*tmp*`, I, value = 1) : 'value' is the wrong length

    – Andrea Rivolta
    Nov 21 '18 at 13:17








  • 1





    Ok, i solved it: the mapping introduced coordinates repetitions. Using unique(I) everything works fine. From 1600 to 2 seconds. Thank you very much

    – Andrea Rivolta
    Nov 21 '18 at 13:30



















0














Try this, it avoids any looping or lapply and is vectorized. This takes advantage of the fact that a data.frame is really a list. 



impute <- function(lst, start, end){ lst[start:end] <- 1; lst }



fill <- function(df, start, end){

  cols <- names(df)

  lst <- as.list(as.data.frame(t(df)))

  res <- as.data.frame(t(Vectorize(impute)(lst, start, end)))

  names(res) <- names(df)

  row.names(res) <- row.names(df)

  res

}



res <- fill(df, indexesStart, indexesEnd)


Takes around 5 seconds to do 1 million rows on my MacBook Pro.






share|improve this answer


























  • This solution works as well, but jogo's way is faster! Thank you anyway :)

    – Andrea Rivolta
    Nov 21 '18 at 13:32











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2 Answers
2






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oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














This will do it:



df <- data.frame(col_1=rep(0,3),col_2=rep(0,3),col_3=rep(0,3),col_4=rep(0,3))

indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)



for (i in 1:nrow(df)) df[i, indexesStart[i]:indexesEnd[i]] <- 1



df


Here is another technique using a twocolumn matrix as index:



I <- do.call(rbind, lapply(1:length(indexesStart), function(i) cbind(i, indexesStart[i]:indexesEnd[i])))

df[I] <- 1


In the second variant I hided the loop (and the hidden loop is in another place).






share|improve this answer


























  • Thanks, but i need to avoid for loops

    – Andrea Rivolta
    Nov 21 '18 at 10:51











  • In my second variant I hided the loop (and the hidden loop is in another place).

    – jogo
    Nov 21 '18 at 11:51






  • 1





    The second variant works very fast (less than a second). I had to make some adjustment because (i did not specify this aspect in the post to have it easier) because for each row i have multiple range of columns (length(indexes) > dim(df)[1]), but i have been able to manage this through indexes mapping. Now, if i View (or print) df[I] everything is ok, but when i do the assignment it says: Error in [<-.data.frame(*tmp*`, I, value = 1) : 'value' is the wrong length

    – Andrea Rivolta
    Nov 21 '18 at 13:17








  • 1





    Ok, i solved it: the mapping introduced coordinates repetitions. Using unique(I) everything works fine. From 1600 to 2 seconds. Thank you very much

    – Andrea Rivolta
    Nov 21 '18 at 13:30
















2














This will do it:



df <- data.frame(col_1=rep(0,3),col_2=rep(0,3),col_3=rep(0,3),col_4=rep(0,3))

indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)



for (i in 1:nrow(df)) df[i, indexesStart[i]:indexesEnd[i]] <- 1



df


Here is another technique using a twocolumn matrix as index:



I <- do.call(rbind, lapply(1:length(indexesStart), function(i) cbind(i, indexesStart[i]:indexesEnd[i])))

df[I] <- 1


In the second variant I hided the loop (and the hidden loop is in another place).






share|improve this answer


























  • Thanks, but i need to avoid for loops

    – Andrea Rivolta
    Nov 21 '18 at 10:51











  • In my second variant I hided the loop (and the hidden loop is in another place).

    – jogo
    Nov 21 '18 at 11:51






  • 1





    The second variant works very fast (less than a second). I had to make some adjustment because (i did not specify this aspect in the post to have it easier) because for each row i have multiple range of columns (length(indexes) > dim(df)[1]), but i have been able to manage this through indexes mapping. Now, if i View (or print) df[I] everything is ok, but when i do the assignment it says: Error in [<-.data.frame(*tmp*`, I, value = 1) : 'value' is the wrong length

    – Andrea Rivolta
    Nov 21 '18 at 13:17








  • 1





    Ok, i solved it: the mapping introduced coordinates repetitions. Using unique(I) everything works fine. From 1600 to 2 seconds. Thank you very much

    – Andrea Rivolta
    Nov 21 '18 at 13:30














2












2








2







This will do it:



df <- data.frame(col_1=rep(0,3),col_2=rep(0,3),col_3=rep(0,3),col_4=rep(0,3))

indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)



for (i in 1:nrow(df)) df[i, indexesStart[i]:indexesEnd[i]] <- 1



df


Here is another technique using a twocolumn matrix as index:



I <- do.call(rbind, lapply(1:length(indexesStart), function(i) cbind(i, indexesStart[i]:indexesEnd[i])))

df[I] <- 1


In the second variant I hided the loop (and the hidden loop is in another place).






share|improve this answer















This will do it:



df <- data.frame(col_1=rep(0,3),col_2=rep(0,3),col_3=rep(0,3),col_4=rep(0,3))

indexesStart <- c(1, 2, 1)

indexesEnd   <- c(2, 4, 3)



for (i in 1:nrow(df)) df[i, indexesStart[i]:indexesEnd[i]] <- 1



df


Here is another technique using a twocolumn matrix as index:



I <- do.call(rbind, lapply(1:length(indexesStart), function(i) cbind(i, indexesStart[i]:indexesEnd[i])))

df[I] <- 1


In the second variant I hided the loop (and the hidden loop is in another place).







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 21 '18 at 12:37

























answered Nov 21 '18 at 10:47









jogojogo

10k92135




10k92135













  • Thanks, but i need to avoid for loops

    – Andrea Rivolta
    Nov 21 '18 at 10:51











  • In my second variant I hided the loop (and the hidden loop is in another place).

    – jogo
    Nov 21 '18 at 11:51






  • 1





    The second variant works very fast (less than a second). I had to make some adjustment because (i did not specify this aspect in the post to have it easier) because for each row i have multiple range of columns (length(indexes) > dim(df)[1]), but i have been able to manage this through indexes mapping. Now, if i View (or print) df[I] everything is ok, but when i do the assignment it says: Error in [<-.data.frame(*tmp*`, I, value = 1) : 'value' is the wrong length

    – Andrea Rivolta
    Nov 21 '18 at 13:17








  • 1





    Ok, i solved it: the mapping introduced coordinates repetitions. Using unique(I) everything works fine. From 1600 to 2 seconds. Thank you very much

    – Andrea Rivolta
    Nov 21 '18 at 13:30



















  • Thanks, but i need to avoid for loops

    – Andrea Rivolta
    Nov 21 '18 at 10:51











  • In my second variant I hided the loop (and the hidden loop is in another place).

    – jogo
    Nov 21 '18 at 11:51






  • 1





    The second variant works very fast (less than a second). I had to make some adjustment because (i did not specify this aspect in the post to have it easier) because for each row i have multiple range of columns (length(indexes) > dim(df)[1]), but i have been able to manage this through indexes mapping. Now, if i View (or print) df[I] everything is ok, but when i do the assignment it says: Error in [<-.data.frame(*tmp*`, I, value = 1) : 'value' is the wrong length

    – Andrea Rivolta
    Nov 21 '18 at 13:17








  • 1





    Ok, i solved it: the mapping introduced coordinates repetitions. Using unique(I) everything works fine. From 1600 to 2 seconds. Thank you very much

    – Andrea Rivolta
    Nov 21 '18 at 13:30

















Thanks, but i need to avoid for loops

– Andrea Rivolta
Nov 21 '18 at 10:51





Thanks, but i need to avoid for loops

– Andrea Rivolta
Nov 21 '18 at 10:51













In my second variant I hided the loop (and the hidden loop is in another place).

– jogo
Nov 21 '18 at 11:51





In my second variant I hided the loop (and the hidden loop is in another place).

– jogo
Nov 21 '18 at 11:51




1




1





The second variant works very fast (less than a second). I had to make some adjustment because (i did not specify this aspect in the post to have it easier) because for each row i have multiple range of columns (length(indexes) > dim(df)[1]), but i have been able to manage this through indexes mapping. Now, if i View (or print) df[I] everything is ok, but when i do the assignment it says: Error in [<-.data.frame(*tmp*`, I, value = 1) : 'value' is the wrong length

– Andrea Rivolta
Nov 21 '18 at 13:17







The second variant works very fast (less than a second). I had to make some adjustment because (i did not specify this aspect in the post to have it easier) because for each row i have multiple range of columns (length(indexes) > dim(df)[1]), but i have been able to manage this through indexes mapping. Now, if i View (or print) df[I] everything is ok, but when i do the assignment it says: Error in [<-.data.frame(*tmp*`, I, value = 1) : 'value' is the wrong length

– Andrea Rivolta
Nov 21 '18 at 13:17






1




1





Ok, i solved it: the mapping introduced coordinates repetitions. Using unique(I) everything works fine. From 1600 to 2 seconds. Thank you very much

– Andrea Rivolta
Nov 21 '18 at 13:30





Ok, i solved it: the mapping introduced coordinates repetitions. Using unique(I) everything works fine. From 1600 to 2 seconds. Thank you very much

– Andrea Rivolta
Nov 21 '18 at 13:30













0














Try this, it avoids any looping or lapply and is vectorized. This takes advantage of the fact that a data.frame is really a list. 



impute <- function(lst, start, end){ lst[start:end] <- 1; lst }



fill <- function(df, start, end){

  cols <- names(df)

  lst <- as.list(as.data.frame(t(df)))

  res <- as.data.frame(t(Vectorize(impute)(lst, start, end)))

  names(res) <- names(df)

  row.names(res) <- row.names(df)

  res

}



res <- fill(df, indexesStart, indexesEnd)


Takes around 5 seconds to do 1 million rows on my MacBook Pro.






share|improve this answer


























  • This solution works as well, but jogo's way is faster! Thank you anyway :)

    – Andrea Rivolta
    Nov 21 '18 at 13:32
















0














Try this, it avoids any looping or lapply and is vectorized. This takes advantage of the fact that a data.frame is really a list. 



impute <- function(lst, start, end){ lst[start:end] <- 1; lst }



fill <- function(df, start, end){

  cols <- names(df)

  lst <- as.list(as.data.frame(t(df)))

  res <- as.data.frame(t(Vectorize(impute)(lst, start, end)))

  names(res) <- names(df)

  row.names(res) <- row.names(df)

  res

}



res <- fill(df, indexesStart, indexesEnd)


Takes around 5 seconds to do 1 million rows on my MacBook Pro.






share|improve this answer


























  • This solution works as well, but jogo's way is faster! Thank you anyway :)

    – Andrea Rivolta
    Nov 21 '18 at 13:32














0












0








0







Try this, it avoids any looping or lapply and is vectorized. This takes advantage of the fact that a data.frame is really a list. 



impute <- function(lst, start, end){ lst[start:end] <- 1; lst }



fill <- function(df, start, end){

  cols <- names(df)

  lst <- as.list(as.data.frame(t(df)))

  res <- as.data.frame(t(Vectorize(impute)(lst, start, end)))

  names(res) <- names(df)

  row.names(res) <- row.names(df)

  res

}



res <- fill(df, indexesStart, indexesEnd)


Takes around 5 seconds to do 1 million rows on my MacBook Pro.






share|improve this answer















Try this, it avoids any looping or lapply and is vectorized. This takes advantage of the fact that a data.frame is really a list. 



impute <- function(lst, start, end){ lst[start:end] <- 1; lst }



fill <- function(df, start, end){

  cols <- names(df)

  lst <- as.list(as.data.frame(t(df)))

  res <- as.data.frame(t(Vectorize(impute)(lst, start, end)))

  names(res) <- names(df)

  row.names(res) <- row.names(df)

  res

}



res <- fill(df, indexesStart, indexesEnd)


Takes around 5 seconds to do 1 million rows on my MacBook Pro.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 21 '18 at 12:58

























answered Nov 21 '18 at 12:33









rookierookie

863




863













  • This solution works as well, but jogo's way is faster! Thank you anyway :)

    – Andrea Rivolta
    Nov 21 '18 at 13:32



















  • This solution works as well, but jogo's way is faster! Thank you anyway :)

    – Andrea Rivolta
    Nov 21 '18 at 13:32

















This solution works as well, but jogo's way is faster! Thank you anyway :)

– Andrea Rivolta
Nov 21 '18 at 13:32





This solution works as well, but jogo's way is faster! Thank you anyway :)

– Andrea Rivolta
Nov 21 '18 at 13:32


















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