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_v6 at libero.it" <jarod_v6 at libero.it>
To: smartpink111 at 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: smartpink111 at yahoo.com
>Data: 08/10/2013 15.12
>A: "jarod_v6 at libero.it"<jarod_v6 at 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_v6 at libero.it" <jarod_v6 at libero.it>
>To: smartpink111 at 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: smartpink111 at yahoo.com
>>Data: 07/10/2013 18.45
>>A: "jarod_v6 at libero.it"<jarod_v6 at libero.it>
>>Cc: "R help"<r-help at 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_v6 at libero.it" <jarod_v6 at libero.it>
>>To: r-help at 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 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.
>>
>>
>