[R] Handling "NA" in summation

[ce]

Sorry I misunderstood questions .
I think original poster wants result to be 0 , right ? 
Then 

dataframe$A[ is.na(dataframe$B) ] <- 0
dataframe$B[ is.na(dataframe$B) ] <- 0

is this case you would lose  dataframe$A data or 

dataframe$A + ( dataframe$B[ is.na(dataframe$B) ] <- -1*dataframe$A )



-----Original Message-----
From: "Sarah Goslee" [sarah.goslee at gmail.com]
Date: 09/06/2015 07:00 PM
CC: "r-help" <r-help at r-project.org>
Subject: Re: [R] Handling "NA" in summation

I'm not quite sure how you get zero from that situation. Do you expect
the answer to be 20?

How about:

> dataframe <- data.frame(A=20, B=NA)

> dataframe$A + dataframe$B

[1] NA

> ?sum

> sum(dataframe$A, dataframe$B, na.rm=TRUE)

[1] 20

Sarah

On Sun, Sep 6, 2015 at 6:48 PM, ce <zadig_1 at excite.com> wrote:
>
>
> I use something like :
>
> dataframe[ is.na(dataframe) ] <- 0
> dataframe[ is.nan(dataframe) ] <- 0
> dataframe[ is.infinite(dataframe) ] <- 0
>
> -----Original Message-----
> From: "Olu Ola via R-help" [r-help at r-project.org]
> Date: 09/06/2015 06:24 PM
> To: r-help at r-project.org
> Subject: [R] Handling "NA" in summation
>
> Hello,
> I am currently working with a dataframe which has some missing values represented by "NA". whenever, I add two columns in which at least one of the pair of an observation is "NA", the sum returns zero. That is for the same observation, if
>
> dataframe$A = 20
> dataframe$B = NA
>
> dataframe$A + dataframe$B  returns zero.
>
> I do not want to delete the observations with the NA's. How do I go about carrying out the necessary operations without deleting the observations with the NA's
>
> Thank you
>

-- 
Sarah Goslee
http://www.functionaldiversity.org

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