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Mao Jianfeng

2009-Jun-09 01:56 UTC

[R] how to substitute missing values (NAs) by the group means

Dear Ruser's

I ask for helps on how to substitute missing values (NAs) by mean of the
group it is belonging to.

my dummy dataframe is:
> df       group traits
1  BSPy01-10     NA
2  BSPy01-10    7.3
3  BSPy01-10    7.3
4  BSPy01-11    5.3
5  BSPy01-11    5.4
6  BSPy01-11    5.6
7  BSPy01-11     NA
8  BSPy01-11     NA
9  BSPy01-11    4.8
10 BSPy01-12    8.1
11 BSPy01-12    6.0
12 BSPy01-12    6.0
13 BSPy01-13    6.1


I want to substitute each "NA" by the group mean of which the
"NA" is
belonging to. For example, substitute the first record of traits "NA"
by the
mean of "BSPy01-10".

I have ever tried to solve this problem by using doBy package. But, I
failed. I ask for the right solutions by using doBy package or not.

The commands used and the output I got are as followed:

library(doBy)
df<-orderBy(~group,data=df)   # succeeded
f1<-function(x){m<-mean(x, na.ram=TRUE); x[is.na(x)]<-m; x} # succeeded
datatraits<-lapplyBy(traits~group,data=df, FUN=f1(traits)) # failed
errors: mean(x, na.ram = TRUE), can not find 'traits'.

Thanks in advance.

Sincerely,

Mao J-F

	[[alternative HTML version deleted]]

Henrique Dallazuanna

2009-Jun-09 02:24 UTC

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[R] how to substitute missing values (NAs) by the group means

Try this:

d$traits[is.na(d$traits)] <- ave(d$traits,
                                           d$group,
                                            FUN=function(x)mean(x,
na.rm = T))[is.na(d$traits)]


On 6/8/09, Mao Jianfeng <jianfeng.mao at gmail.com>
wrote:> Dear Ruser's
>
> I ask for helps on how to substitute missing values (NAs) by mean of the
> group it is belonging to.
>
> my dummy dataframe is:
>
>> df
>        group traits
> 1  BSPy01-10     NA
> 2  BSPy01-10    7.3
> 3  BSPy01-10    7.3
> 4  BSPy01-11    5.3
> 5  BSPy01-11    5.4
> 6  BSPy01-11    5.6
> 7  BSPy01-11     NA
> 8  BSPy01-11     NA
> 9  BSPy01-11    4.8
> 10 BSPy01-12    8.1
> 11 BSPy01-12    6.0
> 12 BSPy01-12    6.0
> 13 BSPy01-13    6.1
>
>
> I want to substitute each "NA" by the group mean of which the
"NA" is
> belonging to. For example, substitute the first record of traits
"NA" by the
> mean of "BSPy01-10".
>
> I have ever tried to solve this problem by using doBy package. But, I
> failed. I ask for the right solutions by using doBy package or not.
>
> The commands used and the output I got are as followed:
>
> library(doBy)
> df<-orderBy(~group,data=df)   # succeeded
> f1<-function(x){m<-mean(x, na.ram=TRUE); x[is.na(x)]<-m; x} #
succeeded
> datatraits<-lapplyBy(traits~group,data=df, FUN=f1(traits)) # failed
> errors: mean(x, na.ram = TRUE), can not find 'traits'.
>
> Thanks in advance.
>
> Sincerely,
>
> Mao J-F
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

-- 
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

David Winsemius

2009-Jun-09 02:39 UTC

head link

[R] how to substitute missing values (NAs) by the group means

On Jun 8, 2009, at 9:56 PM, Mao Jianfeng wrote:
> Dear Ruser's
>
> I ask for helps on how to substitute missing values (NAs) by mean of  
> the
> group it is belonging to.
>
> my dummy dataframe is:
>
>> df
>       group traits
> 1  BSPy01-10     NA
> 2  BSPy01-10    7.3
> 3  BSPy01-10    7.3
> 4  BSPy01-11    5.3
> 5  BSPy01-11    5.4
> 6  BSPy01-11    5.6
> 7  BSPy01-11     NA
> 8  BSPy01-11     NA
> 9  BSPy01-11    4.8
> 10 BSPy01-12    8.1
> 11 BSPy01-12    6.0
> 12 BSPy01-12    6.0
> 13 BSPy01-13    6.1
>
>
> I want to substitute each "NA" by the group mean of which the
"NA" is
> belonging to. For example, substitute the first record of traits  
> "NA" by the
> mean of "BSPy01-10".
>
> I have ever tried to solve this problem by using doBy package. But, I
> failed. I ask for the right solutions by using doBy package or not.
This should replace any NA by the mean with the group, or the non-NA  
value:

as.numeric(apply(df, 1, function (x) ifelse( is.na(x[2]),
                                              tapply(df$traits, df 
$group, mean, na.rm=TRUE)[x[1]] ,
                                              x[2] )
             )                               )

  [1] 7.300 7.300 7.300 5.300 5.400 5.600 5.275 5.275 4.800 8.100  
6.000 6.000 6.100

Whether that is the "right solution" depends on your artistic
standards.

If you accept that solution, you would execute:

df$traits <-  <the above expression>


Another approach only replacing the NA's, rather than the whole column:

df[is.na(df$traits), "traits"] <- tapply(df$traits, df$group, mean,
na.rm=TRUE)[ df[is.na(df$traits),"group"] ]

>
>
> The commands used and the output I got are as followed:
>
> library(doBy)
> df<-orderBy(~group,data=df)   # succeeded
> f1<-function(x){m<-mean(x, na.ram=TRUE); x[is.na(x)]<-m; x} #  
> succeeded
> datatraits<-lapplyBy(traits~group,data=df, FUN=f1(traits)) # failed
> errors: mean(x, na.ram = TRUE), can not find 'traits'.-- 

David Winsemius, MD
Heritage Laboratories
West Hartford, CT

Jorge Ivan Velez

2009-Jun-09 02:49 UTC

head link

[R] how to substitute missing values (NAs) by the group means

Dear Mao,
Here is another way:

yourdata$traits2 <- with(yourdata, do.call(c, tapply(traits, group,
function(y){
                                                         ym <-
mean(y,na.rm=TRUE)
                                                         y[is.na(y)]<- ym
                                                         y
                                                       }
                                              )))

HTH,

Jorge


On Mon, Jun 8, 2009 at 9:56 PM, Mao Jianfeng <jianfeng.mao@gmail.com>
wrote:
> Dear Ruser's
>
> I ask for helps on how to substitute missing values (NAs) by mean of the
> group it is belonging to.
>
> my dummy dataframe is:
>
> > df
>       group traits
> 1  BSPy01-10     NA
> 2  BSPy01-10    7.3
> 3  BSPy01-10    7.3
> 4  BSPy01-11    5.3
> 5  BSPy01-11    5.4
> 6  BSPy01-11    5.6
> 7  BSPy01-11     NA
> 8  BSPy01-11     NA
> 9  BSPy01-11    4.8
> 10 BSPy01-12    8.1
> 11 BSPy01-12    6.0
> 12 BSPy01-12    6.0
> 13 BSPy01-13    6.1
>
>
> I want to substitute each "NA" by the group mean of which the
"NA" is
> belonging to. For example, substitute the first record of traits
"NA" by
> the
> mean of "BSPy01-10".
>
> I have ever tried to solve this problem by using doBy package. But, I
> failed. I ask for the right solutions by using doBy package or not.
>
> The commands used and the output I got are as followed:
>
> library(doBy)
> df<-orderBy(~group,data=df)   # succeeded
> f1<-function(x){m<-mean(x, na.ram=TRUE); x[is.na(x)]<-m; x} #
succeeded
> datatraits<-lapplyBy(traits~group,data=df, FUN=f1(traits)) # failed
> errors: mean(x, na.ram = TRUE), can not find 'traits'.
>
> Thanks in advance.
>
> Sincerely,
>
> Mao J-F
>
>        [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
	[[alternative HTML version deleted]]

hadley wickham

2009-Jun-09 03:10 UTC

head link

[R] how to substitute missing values (NAs) by the group means

On Mon, Jun 8, 2009 at 8:56 PM, Mao Jianfeng<jianfeng.mao at gmail.com>
wrote:> Dear Ruser's
>
> I ask for helps on how to substitute missing values (NAs) by mean of the
> group it is belonging to.
>
> my dummy dataframe is:
>
>> df
> ? ? ? group traits
> 1 ?BSPy01-10 ? ? NA
> 2 ?BSPy01-10 ? ?7.3
> 3 ?BSPy01-10 ? ?7.3
> 4 ?BSPy01-11 ? ?5.3
> 5 ?BSPy01-11 ? ?5.4
> 6 ?BSPy01-11 ? ?5.6
> 7 ?BSPy01-11 ? ? NA
> 8 ?BSPy01-11 ? ? NA
> 9 ?BSPy01-11 ? ?4.8
> 10 BSPy01-12 ? ?8.1
> 11 BSPy01-12 ? ?6.0
> 12 BSPy01-12 ? ?6.0
> 13 BSPy01-13 ? ?6.1
>
>
> I want to substitute each "NA" by the group mean of which the
"NA" is
> belonging to. For example, substitute the first record of traits
"NA" by the
> mean of "BSPy01-10".
Here's yet another way, using the plyr package, http://had.co.nz/

library(plyr)
impute.mean <- function(x) replace(x, is.na(x), mean(x, na.rm = TRUE))
ddply(df, ~ group, transform, traits = impute.mean(traits))

Or if you wanted to make it a little more generic

impute <- function(x, fun) {
  missing <- is.na(x)
  replace(x, missing, fun(x[!missing]))
}
ddply(df, ~ group, transform, traits = impute(traits, mean))
ddply(df, ~ group, transform, traits = impute(traits, median))
ddply(df, ~ group, transform, traits = impute(traits, min))

Hadley


-- 
http://had.co.nz/

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