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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