Dear Jarod,
You can call me Arun.
a<- read.table(text="NAME1 NAME2 RNA
mauro francesco E234
luca giuseppe E5578
luca franco E5569
maria luca E4556
maria mauro E4556
luisa mara E4755
mara luca E234
luigi veronica E234
mauro veronica E235",sep="",header=TRUE,stringsAsFactors=FALSE)
res <- do.call(rbind,lapply(unique(c(a[,1],a[,2])),function(x)
{data.frame(RESULTS= x,GROUP1= sum(a[,1] %in% x), GROUP2=sum(a[,2] %in%
x),stringsAsFactors=FALSE)}))
res
# RESULTS GROUP1 GROUP2
#1 mauro 2 1
#2 luca 2 2
#3 maria 2 0
#4 luisa 1 0
#5 mara 1 1
#6 luigi 1 0
#7 francesco 0 1
#8 giuseppe 0 1
#9 franco 0 1
#10 veronica 0 2
A.K.
----- Original Message -----
From: "jarod...@libero.it" <jarod...@libero.it>
To: smartpink...@yahoo.com
Cc:
Sent: Tuesday, October 8, 2013 10:07 AM
Subject: Re: [R] Information Frequency problem calculation
Dear Master,
thanks for your help!!
I have this long database of breeder divide in two list. eache list have
particular group rna identify (catcghory of birds)
GROUP1 GROUP2
NAME1 NAME2 RNA
mauro francesco E234
luca giuseppe E5578
luca franco E5569
maria luca E4556
maria mauro E4556"
luisa mara E4755
mara luca E234
luigi veronica E234
mauro veronica E235
What I want is to have a frequency of presence of all my data:
So :
RESULTS group1 group2
MAURO 2 1
I want know how many times one name it is found in group 1 or in group2 and
how many times have different RNA (identify of group).
Thanks for help!!
Jarod
>----Messaggio originale----
>Da: smartpink...@yahoo.com
>Data: 08/10/2013 15.12
>A: "jarod...@libero.it"<jarod...@libero.it>
>Ogg: Re: R: Re: [R] Information Frequency problem calculation
>
>Hi Jarod,
>
>Could you show your expected outcome as this is still confusing. As I
understand from your initial post, you are comparing two columns from your data
frame (ie. Name1 and Name2). My idea was to compare each entry of the first
column and compare it with the full set of second column to see where they are
the same.
>for example.
>a<- read.table(text="Name1 Name2 category
>mauro francesco E234
>luca giuseppe E5578
>luca franco E5569
>maria luca E4556
>maria mauro E4556",sep="",header=TRUE,stringsAsFactors=FALSE)
>
> lapply(seq_len(nrow(a)),function(i) a[i,1])
>[[1]]
>[1] "mauro"
>
>[[2]]
>[1] "luca"
>
>[[3]]
>[1] "luca"
>
>[[4]]
>[1] "maria"
>
>[[5]]
>[1] "maria"
>
>
>lapply(seq_len(nrow(a)),function(i) sum(a[,2]%in% a[i,1])) #here, it
calculates the sum of the number of matches for each entry of column 1.
>[[1]]
>[1] 1
>
>[[2]]
>[1] 1
>
>[[3]]
>[1] 1
>
>[[4]]
>[1] 0
>
>[[5]]
>[1] 0
>
>
>
> I think this is not you wanted. Your idea of "ex mauro it is
>presence on E234 and E4556 so I found 2 time 5 condition and 1 time fo
single
>condition" is a bit confusing.
>
>Did you meant that "E4556" is repeated twice, so it should be 2 times?. As I
pointed out, if the expected result table is also shown, it would be easier to
understand than words.
>
>
>A.K.
>
>
>
>
>
>
>----- Original Message -----
>From: "jarod...@libero.it" <jarod...@libero.it>
>To: smartpink...@yahoo.com
>Cc:
>Sent: Tuesday, October 8, 2013 3:57 AM
>Subject: R: Re: [R] Information Frequency problem calculation
>
>Thanks so much for your suggestion.. I have a little difficult to understand
>the comand .. could you help me?
>So I interested in calculate the frequence of some names ex mauro it is
>presence on E234 and E4556 so I found 2 time 5 condition and 1 time fo
single
>condition
>Tahanks for your help!!
>
>>----Messaggio originale----
>>Da: smartpink...@yahoo.com
>>Data: 07/10/2013 18.45
>>A: "jarod...@libero.it"<jarod...@libero.it>
>>Cc: "R help"<r-help@r-project.org>
>>Ogg: Re: [R] Information Frequency problem calculation
>>
>>Hi,
>>Not sure what your expected output would be:
>>a<- read.table(text="Name1 Name2 category
>>mauro francesco E234
>>luca giuseppe E5578
>>luca franco E5569
>>maria luca E4556"
>maria mauro E4556
>
>
>
>,sep="",header=TRUE,stringsAsFactors=FALSE)
>>sapply(seq_len(nrow(a)),function(i) sum(a[,2] %in% a[i,1]))
>>#[1] 0 1 1 0
>>
>>
>>A.K.
>>
>>
>>
>>
>>
>>----- Original Message -----
>>From: "jarod...@libero.it" <jarod...@libero.it>
>>To: r-help@r-project.org
>>Cc:
>>Sent: Monday, October 7, 2013 12:09 PM
>>Subject: [R] Information Frequency problem calculation
>>
>>Dear All,
>>
>>I Have a dataframe like that:
>>
>>Name1 Name2 category
>>
>>mauro francesco E234
>>luca giuseppe E5578
>>luca franco E5569
>>maria luca E4556
>>...
>>I would like to calculate the frequency of many time in my data I found in
>>the list name:
>>
>>a<-read.table("pippo.csv",header=T,sep="\t")
>>name1<-as.character(a[,1])
>>name2<-as.character(a[,2])
>>category<-as.character(a[,3])
>>
>>for(i in 1:lenght(name1)){
>>re <-which(name1 == name2)
>>
>>}
>>
>>So how can create a table of frequncy of many times I found a name in
column
>>one respect to the second ?
>>Thanks in advance for your help!
>>M.
>>
>>______________________________________________
>>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.
>>
>>
>
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
