Efficient solution to (recursively) replace NAs with the mean of lags, by group











up vote
0
down vote

favorite












I need to replace NAs with the mean of previous three values, by group.
Once an NA is replaced, it will serve as input for computing the mean corresponding to the next NA (if next NA is within the next three months).



Here it is an example:



id   date   value
1 2017-04-01 40
1 2017-05-01 40
1 2017-06-01 10
1 2017-07-01 NA
1 2017-08-01 NA
2 2014-01-01 27
2 2014-02-01 13


Data:



dt <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L), date = structure(c(17257, 17287, 17318, 17348, 17379, 16071, 16102), class = "Date"), value = c(40, 40, 10, NA, NA, 27, 13)), row.names = c(1L, 2L, 3L, 4L, 5L, 8L, 9L), class = "data.frame")



The output should look like:



id   date   value
1 2017-04-01 40.00
1 2017-05-01 40.00
1 2017-06-01 10.00
1 2017-07-01 30.00
1 2017-08-01 26.66
2 2014-01-01 27.00
2 2014-02-01 13.00


where 26.66 = (30 + 10 + 40)/3



What is an efficient way to do this (i.e. to avoid for loops)?










share|improve this question




















  • 1




    See this question.
    – Rui Barradas
    Nov 22 at 16:05















up vote
0
down vote

favorite












I need to replace NAs with the mean of previous three values, by group.
Once an NA is replaced, it will serve as input for computing the mean corresponding to the next NA (if next NA is within the next three months).



Here it is an example:



id   date   value
1 2017-04-01 40
1 2017-05-01 40
1 2017-06-01 10
1 2017-07-01 NA
1 2017-08-01 NA
2 2014-01-01 27
2 2014-02-01 13


Data:



dt <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L), date = structure(c(17257, 17287, 17318, 17348, 17379, 16071, 16102), class = "Date"), value = c(40, 40, 10, NA, NA, 27, 13)), row.names = c(1L, 2L, 3L, 4L, 5L, 8L, 9L), class = "data.frame")



The output should look like:



id   date   value
1 2017-04-01 40.00
1 2017-05-01 40.00
1 2017-06-01 10.00
1 2017-07-01 30.00
1 2017-08-01 26.66
2 2014-01-01 27.00
2 2014-02-01 13.00


where 26.66 = (30 + 10 + 40)/3



What is an efficient way to do this (i.e. to avoid for loops)?










share|improve this question




















  • 1




    See this question.
    – Rui Barradas
    Nov 22 at 16:05













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to replace NAs with the mean of previous three values, by group.
Once an NA is replaced, it will serve as input for computing the mean corresponding to the next NA (if next NA is within the next three months).



Here it is an example:



id   date   value
1 2017-04-01 40
1 2017-05-01 40
1 2017-06-01 10
1 2017-07-01 NA
1 2017-08-01 NA
2 2014-01-01 27
2 2014-02-01 13


Data:



dt <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L), date = structure(c(17257, 17287, 17318, 17348, 17379, 16071, 16102), class = "Date"), value = c(40, 40, 10, NA, NA, 27, 13)), row.names = c(1L, 2L, 3L, 4L, 5L, 8L, 9L), class = "data.frame")



The output should look like:



id   date   value
1 2017-04-01 40.00
1 2017-05-01 40.00
1 2017-06-01 10.00
1 2017-07-01 30.00
1 2017-08-01 26.66
2 2014-01-01 27.00
2 2014-02-01 13.00


where 26.66 = (30 + 10 + 40)/3



What is an efficient way to do this (i.e. to avoid for loops)?










share|improve this question















I need to replace NAs with the mean of previous three values, by group.
Once an NA is replaced, it will serve as input for computing the mean corresponding to the next NA (if next NA is within the next three months).



Here it is an example:



id   date   value
1 2017-04-01 40
1 2017-05-01 40
1 2017-06-01 10
1 2017-07-01 NA
1 2017-08-01 NA
2 2014-01-01 27
2 2014-02-01 13


Data:



dt <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L), date = structure(c(17257, 17287, 17318, 17348, 17379, 16071, 16102), class = "Date"), value = c(40, 40, 10, NA, NA, 27, 13)), row.names = c(1L, 2L, 3L, 4L, 5L, 8L, 9L), class = "data.frame")



The output should look like:



id   date   value
1 2017-04-01 40.00
1 2017-05-01 40.00
1 2017-06-01 10.00
1 2017-07-01 30.00
1 2017-08-01 26.66
2 2014-01-01 27.00
2 2014-02-01 13.00


where 26.66 = (30 + 10 + 40)/3



What is an efficient way to do this (i.e. to avoid for loops)?







r dplyr apply






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 at 19:00

























asked Nov 22 at 15:56









Luminita

1918




1918








  • 1




    See this question.
    – Rui Barradas
    Nov 22 at 16:05














  • 1




    See this question.
    – Rui Barradas
    Nov 22 at 16:05








1




1




See this question.
– Rui Barradas
Nov 22 at 16:05




See this question.
– Rui Barradas
Nov 22 at 16:05












2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Define a roll function which takes 3 or less previous values as a list and the current value and returns as a list the previous 2 values with the current value if the current value is not NA and the prevous 2 values with the mean if the current value is NA. Use that with Reduce and pick off the last value of each list in the result. Then apply all that to each group using ave.



roll <- function(prev, cur) {
prev <- unlist(prev)
list(tail(prev, 2), if (is.na(cur)) mean(prev) else cur)
}

reduce_roll <- function(x) {
sapply(Reduce(roll, init = x[1], x[-1], acc = TRUE), tail, 1)
}

transform(dt, value = ave(value, id, FUN = reduce_roll))


giving:



  id       date    value
1 1 2017-04-01 40
2 1 2017-05-01 40
3 1 2017-06-01 10
4 1 2017-07-01 30
5 1 2017-08-01 26.66667
8 2 2014-01-01 27
9 2 2014-02-01 13





share|improve this answer























  • Neat answer, thanks. Also, I corrected the cut off data (it was a copy-paste mistake which I didn't notice).
    – Luminita
    Nov 22 at 19:04


















up vote
1
down vote













The following uses base R only and does what you need.



sp <- split(dt, dt$id)
sp <- lapply(sp, function(DF){
for(i in which(is.na(DF$value))){
tmp <- DF[seq_len(i - 1), ]
DF$value[i] <- mean(tail(tmp$value, 3))
}
DF
})

result <- do.call(rbind, sp)
row.names(result) <- NULL

result
# id date value
#1 1 2017-01-04 40.00000
#2 1 2017-01-05 40.00000
#3 1 2017-01-06 10.00000
#4 1 2017-01-07 30.00000
#5 1 2017-01-08 26.66667
#6 2 2014-01-01 27.00000
#7 2 2014-01-02 13.00000





share|improve this answer





















    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53434559%2fefficient-solution-to-recursively-replace-nas-with-the-mean-of-lags-by-group%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Define a roll function which takes 3 or less previous values as a list and the current value and returns as a list the previous 2 values with the current value if the current value is not NA and the prevous 2 values with the mean if the current value is NA. Use that with Reduce and pick off the last value of each list in the result. Then apply all that to each group using ave.



    roll <- function(prev, cur) {
    prev <- unlist(prev)
    list(tail(prev, 2), if (is.na(cur)) mean(prev) else cur)
    }

    reduce_roll <- function(x) {
    sapply(Reduce(roll, init = x[1], x[-1], acc = TRUE), tail, 1)
    }

    transform(dt, value = ave(value, id, FUN = reduce_roll))


    giving:



      id       date    value
    1 1 2017-04-01 40
    2 1 2017-05-01 40
    3 1 2017-06-01 10
    4 1 2017-07-01 30
    5 1 2017-08-01 26.66667
    8 2 2014-01-01 27
    9 2 2014-02-01 13





    share|improve this answer























    • Neat answer, thanks. Also, I corrected the cut off data (it was a copy-paste mistake which I didn't notice).
      – Luminita
      Nov 22 at 19:04















    up vote
    1
    down vote



    accepted










    Define a roll function which takes 3 or less previous values as a list and the current value and returns as a list the previous 2 values with the current value if the current value is not NA and the prevous 2 values with the mean if the current value is NA. Use that with Reduce and pick off the last value of each list in the result. Then apply all that to each group using ave.



    roll <- function(prev, cur) {
    prev <- unlist(prev)
    list(tail(prev, 2), if (is.na(cur)) mean(prev) else cur)
    }

    reduce_roll <- function(x) {
    sapply(Reduce(roll, init = x[1], x[-1], acc = TRUE), tail, 1)
    }

    transform(dt, value = ave(value, id, FUN = reduce_roll))


    giving:



      id       date    value
    1 1 2017-04-01 40
    2 1 2017-05-01 40
    3 1 2017-06-01 10
    4 1 2017-07-01 30
    5 1 2017-08-01 26.66667
    8 2 2014-01-01 27
    9 2 2014-02-01 13





    share|improve this answer























    • Neat answer, thanks. Also, I corrected the cut off data (it was a copy-paste mistake which I didn't notice).
      – Luminita
      Nov 22 at 19:04













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Define a roll function which takes 3 or less previous values as a list and the current value and returns as a list the previous 2 values with the current value if the current value is not NA and the prevous 2 values with the mean if the current value is NA. Use that with Reduce and pick off the last value of each list in the result. Then apply all that to each group using ave.



    roll <- function(prev, cur) {
    prev <- unlist(prev)
    list(tail(prev, 2), if (is.na(cur)) mean(prev) else cur)
    }

    reduce_roll <- function(x) {
    sapply(Reduce(roll, init = x[1], x[-1], acc = TRUE), tail, 1)
    }

    transform(dt, value = ave(value, id, FUN = reduce_roll))


    giving:



      id       date    value
    1 1 2017-04-01 40
    2 1 2017-05-01 40
    3 1 2017-06-01 10
    4 1 2017-07-01 30
    5 1 2017-08-01 26.66667
    8 2 2014-01-01 27
    9 2 2014-02-01 13





    share|improve this answer














    Define a roll function which takes 3 or less previous values as a list and the current value and returns as a list the previous 2 values with the current value if the current value is not NA and the prevous 2 values with the mean if the current value is NA. Use that with Reduce and pick off the last value of each list in the result. Then apply all that to each group using ave.



    roll <- function(prev, cur) {
    prev <- unlist(prev)
    list(tail(prev, 2), if (is.na(cur)) mean(prev) else cur)
    }

    reduce_roll <- function(x) {
    sapply(Reduce(roll, init = x[1], x[-1], acc = TRUE), tail, 1)
    }

    transform(dt, value = ave(value, id, FUN = reduce_roll))


    giving:



      id       date    value
    1 1 2017-04-01 40
    2 1 2017-05-01 40
    3 1 2017-06-01 10
    4 1 2017-07-01 30
    5 1 2017-08-01 26.66667
    8 2 2014-01-01 27
    9 2 2014-02-01 13






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 22 at 22:09

























    answered Nov 22 at 16:48









    G. Grothendieck

    144k9125230




    144k9125230












    • Neat answer, thanks. Also, I corrected the cut off data (it was a copy-paste mistake which I didn't notice).
      – Luminita
      Nov 22 at 19:04


















    • Neat answer, thanks. Also, I corrected the cut off data (it was a copy-paste mistake which I didn't notice).
      – Luminita
      Nov 22 at 19:04
















    Neat answer, thanks. Also, I corrected the cut off data (it was a copy-paste mistake which I didn't notice).
    – Luminita
    Nov 22 at 19:04




    Neat answer, thanks. Also, I corrected the cut off data (it was a copy-paste mistake which I didn't notice).
    – Luminita
    Nov 22 at 19:04












    up vote
    1
    down vote













    The following uses base R only and does what you need.



    sp <- split(dt, dt$id)
    sp <- lapply(sp, function(DF){
    for(i in which(is.na(DF$value))){
    tmp <- DF[seq_len(i - 1), ]
    DF$value[i] <- mean(tail(tmp$value, 3))
    }
    DF
    })

    result <- do.call(rbind, sp)
    row.names(result) <- NULL

    result
    # id date value
    #1 1 2017-01-04 40.00000
    #2 1 2017-01-05 40.00000
    #3 1 2017-01-06 10.00000
    #4 1 2017-01-07 30.00000
    #5 1 2017-01-08 26.66667
    #6 2 2014-01-01 27.00000
    #7 2 2014-01-02 13.00000





    share|improve this answer

























      up vote
      1
      down vote













      The following uses base R only and does what you need.



      sp <- split(dt, dt$id)
      sp <- lapply(sp, function(DF){
      for(i in which(is.na(DF$value))){
      tmp <- DF[seq_len(i - 1), ]
      DF$value[i] <- mean(tail(tmp$value, 3))
      }
      DF
      })

      result <- do.call(rbind, sp)
      row.names(result) <- NULL

      result
      # id date value
      #1 1 2017-01-04 40.00000
      #2 1 2017-01-05 40.00000
      #3 1 2017-01-06 10.00000
      #4 1 2017-01-07 30.00000
      #5 1 2017-01-08 26.66667
      #6 2 2014-01-01 27.00000
      #7 2 2014-01-02 13.00000





      share|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        The following uses base R only and does what you need.



        sp <- split(dt, dt$id)
        sp <- lapply(sp, function(DF){
        for(i in which(is.na(DF$value))){
        tmp <- DF[seq_len(i - 1), ]
        DF$value[i] <- mean(tail(tmp$value, 3))
        }
        DF
        })

        result <- do.call(rbind, sp)
        row.names(result) <- NULL

        result
        # id date value
        #1 1 2017-01-04 40.00000
        #2 1 2017-01-05 40.00000
        #3 1 2017-01-06 10.00000
        #4 1 2017-01-07 30.00000
        #5 1 2017-01-08 26.66667
        #6 2 2014-01-01 27.00000
        #7 2 2014-01-02 13.00000





        share|improve this answer












        The following uses base R only and does what you need.



        sp <- split(dt, dt$id)
        sp <- lapply(sp, function(DF){
        for(i in which(is.na(DF$value))){
        tmp <- DF[seq_len(i - 1), ]
        DF$value[i] <- mean(tail(tmp$value, 3))
        }
        DF
        })

        result <- do.call(rbind, sp)
        row.names(result) <- NULL

        result
        # id date value
        #1 1 2017-01-04 40.00000
        #2 1 2017-01-05 40.00000
        #3 1 2017-01-06 10.00000
        #4 1 2017-01-07 30.00000
        #5 1 2017-01-08 26.66667
        #6 2 2014-01-01 27.00000
        #7 2 2014-01-02 13.00000






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 22 at 16:16









        Rui Barradas

        15.4k31730




        15.4k31730






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53434559%2fefficient-solution-to-recursively-replace-nas-with-the-mean-of-lags-by-group%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Trompette piccolo

            Slow SSRS Report in dynamic grouping and multiple parameters

            Simon Yates (cyclisme)