Access data.table columns through vector indexes?

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

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

asked Nov 21 '18 at 10:39

Andrea Rivolta

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

add a comment |

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

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

asked Nov 21 '18 at 10:39

Andrea Rivolta

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

add a comment |

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

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

asked Nov 21 '18 at 10:39

Andrea Rivolta

133

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

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

asked Nov 21 '18 at 10:39

Andrea Rivolta

133

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

asked Nov 21 '18 at 10:39

Andrea Rivolta

133

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

edited Nov 21 '18 at 13:38

Henrik

41.5k994109

asked Nov 21 '18 at 10:39

Andrea Rivolta

133

asked Nov 21 '18 at 10:39

Andrea Rivolta

133

asked Nov 21 '18 at 10:39

Andrea Rivolta

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

add a comment |

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

add a comment |

2 Answers
2

active

oldest

votes

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).

edited Nov 21 '18 at 12:37

answered Nov 21 '18 at 10:47

jogo

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

add a comment |

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.

edited Nov 21 '18 at 12:58

answered Nov 21 '18 at 12:33

rookie

863

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

– Andrea Rivolta
Nov 21 '18 at 13:32

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

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).

edited Nov 21 '18 at 12:37

answered Nov 21 '18 at 10:47

jogo

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

add a comment |

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).

edited Nov 21 '18 at 12:37

answered Nov 21 '18 at 10:47

jogo

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

add a comment |

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).

edited Nov 21 '18 at 12:37

answered Nov 21 '18 at 10:47

jogo

10k92135

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).

edited Nov 21 '18 at 12:37

answered Nov 21 '18 at 10:47

jogo

10k92135

edited Nov 21 '18 at 12:37

answered Nov 21 '18 at 10:47

jogo

10k92135

answered Nov 21 '18 at 10:47

jogo

10k92135

answered Nov 21 '18 at 10:47

jogo

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

add a comment |

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

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

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

add a comment |

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.

edited Nov 21 '18 at 12:58

answered Nov 21 '18 at 12:33

rookie

863

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

– Andrea Rivolta
Nov 21 '18 at 13:32

add a comment |

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.

edited Nov 21 '18 at 12:58

answered Nov 21 '18 at 12:33

rookie

863

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

– Andrea Rivolta
Nov 21 '18 at 13:32

add a comment |

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.

edited Nov 21 '18 at 12:58

answered Nov 21 '18 at 12:33

rookie

863

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.

edited Nov 21 '18 at 12:58

answered Nov 21 '18 at 12:33

rookie

863

edited Nov 21 '18 at 12:58

answered Nov 21 '18 at 12:33

rookie

863

answered Nov 21 '18 at 12:33

rookie

863

answered Nov 21 '18 at 12:33

rookie

863

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

– Andrea Rivolta
Nov 21 '18 at 13:32

add a comment |

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

add a comment |

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