[R] Information Frequency problem calculation

[David Carlson]

If you want a symmetric table, try this:

> a<- read.table(text="Name1 Name2 category
+ mauro francesco E234
+ luca  giuseppe  E5578
+ luca  franco  E5569
+ maria luca E4556",sep="",header=TRUE,stringsAsFactors=FALSE)
> names <- sort(unique(c(a$Name1, a$Name2)))
> a$Name1 <- factor(a$Name1, names)
> a$Name2 <- factor(a$Name2, names)
> xtabs(~Name1+Name2, a)
           Name2
Name1       francesco franco giuseppe luca maria mauro
  francesco         0      0        0    0     0     0
  franco            0      0        0    0     0     0
  giuseppe          0      0        0    0     0     0
  luca              0      1        1    0     0     0
  maria             0      0        0    1     0     0
  mauro             1      0        0    0     0     0


-------------------------------------
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77840-4352





-----Original Message-----
From: r-help-bounces at r-project.org
[mailto:r-help-bounces at r-project.org] On Behalf Of arun
Sent: Monday, October 7, 2013 11:45 AM
To: jarod_v6 at libero.it
Cc: R help
Subject: 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",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.

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______________________________________________
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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.