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Replace NA in column with value in adjacent column

This question is related to a post with a similar title (replace NA in an R vector with adjacent values). I would like to scan a column in a data frame and replace NA's with the value in the adjacent cell. In the aforementioned post, the solution was to replace the NA not with the value from the adjacent vector (e.g. the adjacent element in the data matrix) but was a conditional replace for a fixed value. Below is a reproducible example of my problem:

UNIT <- c(NA,NA, 200, 200, 200, 200, 200, 300, 300, 300,300) STATUS <-c('ACTIVE','INACTIVE','ACTIVE','ACTIVE','INACTIVE','ACTIVE','INACTIVE','ACTIVE','ACTIVE',                     'ACTIVE','INACTIVE')  TERMINATED <- c('1999-07-06' , '2008-12-05' , '2000-08-18' , '2000-08-18' ,'2000-08-18' ,'2008-08-18',                         '2008-08-18','2006-09-19','2006-09-19' ,'2006-09-19' ,'1999-03-15')  START <- c('2007-04-23','2008-12-06','2004-06-01','2007-02-01','2008-04-19','2010-11-29','2010-12-30',                    '2007-10-29','2008-02-05','2008-06-30','2009-02-07') STOP <- c('2008-12-05','4712-12-31','2007-01-31','2008-04-18','2010-11-28','2010-12-29','4712-12-31',                   '2008-02-04','2008-06-29','2009-02-06','4712-12-31') #creating dataframe TEST <- data.frame(UNIT,STATUS,TERMINATED,START,STOP); TEST                       UNIT   STATUS TERMINATED      START       STOP 1    NA   ACTIVE 1999-07-06 2007-04-23 2008-12-05 2    NA INACTIVE 2008-12-05 2008-12-06 4712-12-31 3   200   ACTIVE 2000-08-18 2004-06-01 2007-01-31 4   200   ACTIVE 2000-08-18 2007-02-01 2008-04-18 5   200 INACTIVE 2000-08-18 2008-04-19 2010-11-28 6   200   ACTIVE 2008-08-18 2010-11-29 2010-12-29 7   200 INACTIVE 2008-08-18 2010-12-30 4712-12-31 8   300   ACTIVE 2006-09-19 2007-10-29 2008-02-04 9   300   ACTIVE 2006-09-19 2008-02-05 2008-06-29 10  300   ACTIVE 2006-09-19 2008-06-30 2009-02-06 11  300 INACTIVE 1999-03-15 2009-02-07 4712-12-31  #using the syntax for a conditional replace and hoping it works :/           TEST$UNIT[is.na(TEST$UNIT)] <- TEST$STATUS; TEST      UNIT   STATUS TERMINATED      START       STOP 1     1   ACTIVE 1999-07-06 2007-04-23 2008-12-05 2     2 INACTIVE 2008-12-05 2008-12-06 4712-12-31 3   200   ACTIVE 2000-08-18 2004-06-01 2007-01-31 4   200   ACTIVE 2000-08-18 2007-02-01 2008-04-18 5   200 INACTIVE 2000-08-18 2008-04-19 2010-11-28 6   200   ACTIVE 2008-08-18 2010-11-29 2010-12-29 7   200 INACTIVE 2008-08-18 2010-12-30 4712-12-31 8   300   ACTIVE 2006-09-19 2007-10-29 2008-02-04 9   300   ACTIVE 2006-09-19 2008-02-05 2008-06-29 10  300   ACTIVE 2006-09-19 2008-06-30 2009-02-06 11  300 INACTIVE 1999-03-15 2009-02-07 4712-12-31 

The outcome should be:

      UNIT   STATUS TERMINATED      START       STOP 1   ACTIVE   ACTIVE 1999-07-06 2007-04-23 2008-12-05 2 INACTIVE INACTIVE 2008-12-05 2008-12-06 4712-12-31 3      200   ACTIVE 2000-08-18 2004-06-01 2007-01-31 4      200   ACTIVE 2000-08-18 2007-02-01 2008-04-18 5      200 INACTIVE 2000-08-18 2008-04-19 2010-11-28 6      200   ACTIVE 2008-08-18 2010-11-29 2010-12-29 7      200 INACTIVE 2008-08-18 2010-12-30 4712-12-31 8      300   ACTIVE 2006-09-19 2007-10-29 2008-02-04 9      300   ACTIVE 2006-09-19 2008-02-05 2008-06-29 10     300   ACTIVE 2006-09-19 2008-06-30 2009-02-06 11     300 INACTIVE 1999-03-15 2009-02-07 4712-12-31 
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hubert_farnsworth Avatar asked Mar 26 '13 05:03

hubert_farnsworth


People also ask

How do I change NA values in a column?

You can replace NA values with zero(0) on numeric columns of R data frame by using is.na() , replace() , imputeTS::replace() , dplyr::coalesce() , dplyr::mutate_at() , dplyr::mutate_if() , and tidyr::replace_na() functions.

How do you check if there is na in a column?

In R, the easiest way to find columns that contain missing values is by combining the power of the functions is.na() and colSums(). First, you check and count the number of NA's per column. Then, you use a function such as names() or colnames() to return the names of the columns with at least one missing value.


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2 Answers

It didn't work because status was a factor. When you mix factor with numeric then numeric is the least restrictive. By forcing status to be character you get the results you're after and the column is now a character vector:

TEST$UNIT[is.na(TEST$UNIT)] <- as.character(TEST$STATUS[is.na(TEST$UNIT)])  ##        UNIT   STATUS TERMINATED      START       STOP ## 1    ACTIVE   ACTIVE 1999-07-06 2007-04-23 2008-12-05 ## 2  INACTIVE INACTIVE 2008-12-05 2008-12-06 4712-12-31 ## 3       200   ACTIVE 2000-08-18 2004-06-01 2007-01-31 ## 4       200   ACTIVE 2000-08-18 2007-02-01 2008-04-18 ## 5       200 INACTIVE 2000-08-18 2008-04-19 2010-11-28 ## 6       200   ACTIVE 2008-08-18 2010-11-29 2010-12-29 ## 7       200 INACTIVE 2008-08-18 2010-12-30 4712-12-31 ## 8       300   ACTIVE 2006-09-19 2007-10-29 2008-02-04 ## 9       300   ACTIVE 2006-09-19 2008-02-05 2008-06-29 ## 10      300   ACTIVE 2006-09-19 2008-06-30 2009-02-06 ## 11      300 INACTIVE 1999-03-15 2009-02-07 4712-12-31 
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Tyler Rinker Avatar answered Sep 18 '22 08:09

Tyler Rinker


You have to do

TEST$UNIT[is.na(TEST$UNIT)] <- TEST$STATUS[is.na(TEST$UNIT)] 

so that the value will be replaced with the adjacent value. Otherwise there is a mismatch between the number of values to be replaced and the values to replace them with. This would result in the values being replaced in row order. It works in this case because the two values being replaced are the first two.

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Brennan Avatar answered Sep 22 '22 08:09

Brennan