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Home/ Questions/Simple solutions for the r number of items to replace is not a multiple of replacement length error
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Sacha Papin
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Sacha Papin
Asked: May 11, 20222022-05-11T18:26:33+00:00 2022-05-11T18:26:33+00:00In: r

Simple solutions for the r number of items to replace is not a multiple of replacement length error

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I get an error

number of items to replace is not a multiple of replacement length

when I try to run the following code:

combi$DT[is.na(combi$DT) & ! is.na(combi$OD) ] <- combi$OD
id DT OD
 67 2010-12-12 2010-12-12
 68 NA NA
 69 NA 2010-12-12
 70 NA NA

How to fix r number of items to replace is not a multiple of replacement length. Please give me some good ideas.

dataframe
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    lyytutoria Expert
    2022-05-27T09:32:59+00:00Added an answer on May 27, 2022 at 9:32 am

    The cause: The replacement length is not multiplied by the number of items that need to be replaced. The number of items you can replace is determined by the number is.na(combi$DT) & !is.na(combi$OD) rows that are less than combi rows (and the replacement length).

    Solution: The easiest way to fix this error is to simply use ifelse:

    1 combi$DT <- ifelse(is.na(combi$DT) & !is.na(combi$OD), combi$OD, combi$DT)

    N.B. the and !is.na(combi$OD) is redundant in the event that the two are NA the substitute would be NA. You can use only

    1 combi$DT <- ifelse(is.na(combi$DT), combi$OD, combi$DT)
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  2. Sacha Moulin
    2022-05-25T19:23:40+00:00Added an answer on May 25, 2022 at 7:23 pm

    This warning occurs because you’re trying to assign all combi$OD at the combi$DT NA locations. If you have 100 rows with 2 variables each with 5 NAs, you tell it to replace the 5 NAs in variable1 by 100 values. This is why the warning. This is a better alternative.

    combi$DT[is.na(combi$DT) & !is.na(combi$OD)] <- combi$OD[is.na(combi$DT) & !is.na(combi$OD)]
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