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arun

2013-Mar-22 13:46 UTC

[R] Distance calculation

Hi Elisa,

I hope this is what you wanted.


dat1<-read.csv("peaks.csv",sep=",")
#Subset
dat2<-dat1[1:5,]
res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i)
do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))),
function(x) {x1<-rbind(dat2[i,],x);
abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2.t.[2]))}))))
res2<-do.call(cbind,lapply(seq_len(ncol(res1)),function(i)
c(c(tail(res1[seq(1,i,1),i],-1),0),res1[-c(1:i),i])))
row.names(res2)<-1:nrow(res2)
?res2
#????? [,1]???? [,2]???? [,3]???? [,4]???? [,5]
#1?? 0.0000?? 0.0000?? 0.0000 379.1364?? 0.0000
#2?? 0.0000?? 0.0000?? 0.0000 312.8267?? 0.0000
#3?? 0.0000?? 0.0000?? 0.0000 379.6576?? 0.0000
#4 379.1364 312.8267 379.6576?? 0.0000 324.4063
#5?? 0.0000?? 0.0000?? 0.0000 324.4063?? 0.0000

resWhole<-do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x) {x1<-rbind(dat1[i,],x);
abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2.t.[2]))}))))
res2Whole<-do.call(cbind,lapply(seq_len(ncol(resWhole)),function(i)
c(c(tail(resWhole[seq(1,i,1),i],-1),0),resWhole[-c(1:i),i])))
row.names(res2Whole)<-1:nrow(res2Whole)
dim(res2Whole)
#[1] 124 124
res2Whole[1:5,1:5]
#????? [,1]???? [,2]???? [,3]???? [,4]???? [,5]
#1?? 0.0000?? 0.0000?? 0.0000 379.1364?? 0.0000
#2?? 0.0000?? 0.0000?? 0.0000 312.8267?? 0.0000
#3?? 0.0000?? 0.0000?? 0.0000 379.6576?? 0.0000
#4 379.1364 312.8267 379.6576?? 0.0000 324.4063
#5?? 0.0000?? 0.0000?? 0.0000 324.4063?? 0.0000

A.K.



________________________________
From: eliza botto <eliza_botto at hotmail.com>
To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> 
Sent: Friday, March 22, 2013 8:26 AM
Subject: 



Dear Arun,
I hope you are fine.?
?the attached text file has my recent question and excel file contains the data.


thanks in advance

Elisa

arun

2013-Mar-26 05:53 UTC

head link

[R] Distance calculation

Hi Elisa,

Based on the formula you gave, this is what I got:

dat1<- read.csv("rate.csv",sep=",")

res<- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x)
{x1<-rbind(dat1[i,],x);colnames(x1)<-gsub("[.]","",colnames(x1));
?if({indx<- colSums(x1[,2:5]==0);indx[1]==0 & indx[2]==0 & indx[3]==1
& indx[4]==2}) #3 peaks 2 peak comparison
?{x2<- x1[order(x1$Peak3t,x1$Npeak3t),];
?with(x2,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))})
}
else if({indx[1]==0 & indx[2]==0 & indx[3]==1 & indx[4]==1}) #4
peaks 2 peak comparison. Peak3 Peak4 value compared with Peak1
{x4<- x1[order(x1$Peak3t,x1$Peak4t,x1$Npeak3t,x1$Npeak4t),];
with(x4,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))+abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2]))
})
}
else if({indx[1]==0 & indx[2]==1 & indx[3]==2 & indx[4]==2}) #2
peaks 1 peak comparison
{x5<- x1[order(x1$Peak2t,x1$Npeak2t),];
with(x5,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))
})

}

else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==2 }) #3
peak 1 peak comparison
{x6<- x1[order(x1$Peak2t,x1$Peak3t,x1$Npeak2t,x1$Npeak3t),];
with(x6,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2])) +
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2])) })
}
else if({indx[1]==0 & indx[2]==0 & indx[3]==0 & indx[4]==1}) # 4
peak 3 peak comparison
{x7<- x1[order(x1$Peak4t,x1$Npeak4t),];
with(x7,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+
abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak3v[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+
abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2])) })

}
else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==1}) #4 peak
1 peak comparison
{x8<-
x1[order(x1$Peak2t,x1$Peak3t,x1$Peak4t,x1$Npeak2t,x1$Npeak3t,x1$Npeak4t),];
with(x8,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))+abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2]))})

}
else ({ #cases where peaks are similar
with(x1,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+abs((Peak4v[1]-Peak4v[2])*(Peak4t[1]-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak3v[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+abs((Npeak4v[1]-Npeak4v[2])*(Npeak4t[1]-Npeak4t[2]))
})
})

}))))
res2<-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)<-1:nrow(res2)
res2[1:5,1:5]
#????? [,1]???? [,2]???? [,3]???? [,4]??? [,5]
#1?? 0.0000?? 0.0000?? 0.0000 857.6834?? 0.000
#2?? 0.0000?? 0.0000?? 0.0000 611.1167?? 0.000
#3?? 0.0000?? 0.0000?? 0.0000 854.3765?? 0.000
#4 857.6834 611.1167 854.3765?? 0.0000 579.756
#5?? 0.0000?? 0.0000?? 0.0000 579.7560?? 0.000
dim(res2)
#[1] 124 124



I also validated each of the cases by calculating the values:
For example:? stations 1 and 4, your previous email indicated

""
but the distance between 1 and 4 is not 379.1364, actually its 1495.01

""

Based on my calculation:

dat1[c(1,4),]
#? St. Peak1.t. Peak2.t. Peak3.t. Peak4.t. Npeak1.t. Npeak2.t. Npeak3.t.
#1?? 1??????? 5?????? 10??????? 0??????? 0???????? 7??????? 13???????? 0
#4?? 4??????? 4??????? 8?????? 10??????? 0???????? 6???????? 9??????? 14
?# Npeak4.t. Peak1.v. Peak2.v. Peak3.v. Peak4.v. Npeak1.v. Npeak2.v. Npeak3.v.
#1???????? 0 56.28785 17.43170? 0.00000??????? 0? 49.66025 81.759761??? 0.0000
#4???????? 0 27.91471 12.06137 11.52018??????? 0? 10.29134? 5.062441?? 17.8527
?# Npeak4.v.
#1???????? 0
#4???????? 0
?abs((56.28785-27.91471)*(5-4))+abs((17.43170-12.06137)*(10-8))+abs((56.28785-11.52018)*((5+12)-10))+abs((49.66025-10.29134)*(7-6))+abs((81.759761-5.062441)*(13-9))+abs((49.66025-17.8527)*((7+12)-14))
#[1] 857.6834 #which is the same value I got in the distance matrix.

If the formula is different, please let me know.
A.K.







________________________________
From: eliza botto <eliza_botto at hotmail.com>
To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> 
Sent: Monday, March 25, 2013 3:50 AM
Subject: RE: Distance calculation



Dear Arun,
I have a slight problem with this coding for the calculation of distance matrix.
The text files contains that problem where as data is in excel file.
kindly help me out.. :(

Elisa?

> Date: Fri, 22 Mar 2013 06:46:18 -0700
> From: smartpink111 at yahoo.com
> Subject: Re: Distance calculation
> To: eliza_botto at hotmail.com
> CC: r-help at r-project.org
> 
> Hi Elisa,
> 
> I hope this is what you wanted.
> 
> 
> dat1<-read.csv("peaks.csv",sep=",")
> #Subset
> dat2<-dat1[1:5,]
> res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i)
do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))),
function(x) {x1<-rbind(dat2[i,],x);
abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2.t.[2]))}))))
> res2<-do.call(cbind,lapply(seq_len(ncol(res1)),function(i)
c(c(tail(res1[seq(1,i,1),i],-1),0),res1[-c(1:i),i])))
> row.names(res2)<-1:nrow(res2)
> ?res2
> #????? [,1]???? [,2]???? [,3]???? [,4]???? [,5]
> #1?? 0.0000?? 0.0000?? 0.0000 379.1364?? 0.0000
> #2?? 0.0000?? 0.0000?? 0.0000 312.8267?? 0.0000
> #3?? 0.0000?? 0.0000?? 0.0000 379.6576?? 0.0000
> #4 379.1364 312.8267 379.6576?? 0.0000 324.4063
> #5?? 0.0000?? 0.0000?? 0.0000 324.4063?? 0.0000
> 
> resWhole<-do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x) {x1<-rbind(dat1[i,],x);
abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2.t.[2]))}))))
> res2Whole<-do.call(cbind,lapply(seq_len(ncol(resWhole)),function(i)
c(c(tail(resWhole[seq(1,i,1),i],-1),0),resWhole[-c(1:i),i])))
> row.names(res2Whole)<-1:nrow(res2Whole)
> dim(res2Whole)
> #[1] 124 124
> res2Whole[1:5,1:5]
> #????? [,1]???? [,2]???? [,3]???? [,4]???? [,5]
> #1?? 0.0000?? 0.0000?? 0.0000 379.1364?? 0.0000
> #2?? 0.0000?? 0.0000?? 0.0000 312.8267?? 0.0000
> #3?? 0.0000?? 0.0000?? 0.0000 379.6576?? 0.0000
> #4 379.1364 312.8267 379.6576?? 0.0000 324.4063
> #5?? 0.0000?? 0.0000?? 0.0000 324.4063?? 0.0000
> 
> A.K.
> 
> 
> 
> ________________________________
> From: eliza botto <eliza_botto at hotmail.com>
> To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com>
> Sent: Friday, March 22, 2013 8:26 AM
> Subject: 
> 
> 
> 
> Dear Arun,
> I hope you are fine.?
> ?the attached text file has my recent question and excel file contains the
data.
> 
> 
> thanks in advance
> 
> Elisa

arun

2013-Mar-26 16:19 UTC

head link

[R] Distance calculation

HI,
I am not sure about the error.? I modified the script and added 12 (hopefully
all the locations) and didn't had any problems.

dat1<- read.csv("rate.csv",sep=",")

res<- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x)
{x1<-rbind(dat1[i,],x);colnames(x1)<-gsub("[.]","",colnames(x1));
?if({indx<- colSums(x1[,2:5]==0);indx[1]==0 & indx[2]==0 & indx[3]==1
& indx[4]==2}) #3 peaks 2 peak comparison
?{x2<- x1[order(x1$Peak3t,x1$Npeak3t),];
?with(x2,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))})
}
else if({indx[1]==0 & indx[2]==0 & indx[3]==1 & indx[4]==1}) #4
peaks 2 peak comparison. Peak3 Peak4 value compared with Peak1
{x4<- x1[order(x1$Peak3t,x1$Peak4t,x1$Npeak3t,x1$Npeak4t),];
with(x4,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+abs((Peak2v[1]-Peak4v[2])*((Peak2t[1]+12+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))+abs((Npeak2v[1]-Npeak4v[2])*((Npeak2t[1]+12+12)-Npeak4t[2]))
})
}
else if({indx[1]==0 & indx[2]==1 & indx[3]==2 & indx[4]==2}) #2
peaks 1 peak comparison
{x5<- x1[order(x1$Peak2t,x1$Npeak2t),];
with(x5,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))
})

}

else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==2 }) #3
peak 1 peak comparison
{x6<- x1[order(x1$Peak2t,x1$Peak3t,x1$Npeak2t,x1$Npeak3t),];
with(x6,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12+12)-Peak3t[2])) +
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12+12)-Npeak3t[2])) })
}
else if({indx[1]==0 & indx[2]==0 & indx[3]==0 & indx[4]==1}) # 4
peak 3 peak comparison
{x7<- x1[order(x1$Peak4t,x1$Npeak4t),];
with(x7,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+
abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak3v[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+
abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2])) })

}
else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==1}) #4 peak
1 peak comparison
{x8<-
x1[order(x1$Peak2t,x1$Peak3t,x1$Peak4t,x1$Npeak2t,x1$Npeak3t,x1$Npeak4t),];
with(x8,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12+12)-Peak3t[2]))+abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12+12+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12+12)-Npeak3t[2]))+abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12+12+12)-Npeak4t[2]))})

}
else ({ #cases where peaks are similar
with(x1,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+abs((Peak4v[1]-Peak4v[2])*(Peak4t[1]-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak3v[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+abs((Npeak4v[1]-Npeak4v[2])*(Npeak4t[1]-Npeak4t[2]))
})
})

}))))
res2<-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)<-1:nrow(res2)

?res2[1:5,1:5]
#????? [,1]???? [,2]???? [,3]???? [,4]??? [,5]
#1?? 0.0000?? 0.0000?? 0.0000 857.6834?? 0.000
#2?? 0.0000?? 0.0000?? 0.0000 611.1167?? 0.000
#3?? 0.0000?? 0.0000?? 0.0000 854.3765?? 0.000
#4 857.6834 611.1167 854.3765?? 0.0000 579.756
#5?? 0.0000?? 0.0000?? 0.0000 579.7560?? 0.000


A.K.



________________________________
From: eliza botto <eliza_botto at hotmail.com>
To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> 
Sent: Tuesday, March 26, 2013 10:41 AM
Subject: RE: Distance calculation



Dear Arun,
I checked the command and did some addition to make them?purposeful. I didnt
change any, except adding 12 at specific location.

dat1<- read.csv("ratef.csv",sep=",")

res<- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x) {x1<-

rbind(dat1[i,],x);colnames(x1)<-gsub("[.]","",colnames(x1));
?if({indx<- colSums(x1[,2:5]==0);indx[1]==0 & indx[2]==0 & indx[3]==1
& indx[4]==2}) #3 peaks 2 peak comparison
?{x2<- x1[order(x1$Peak3t,x1$Npeak3t),];
?with(x2,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))})
}
else if({indx[1]==0 & indx[2]==0 & indx[3]==1 & indx[4]==1}) #4
peaks 2 peak comparison. Peak3 Peak4 value compared with Peak1
{x4<- x1[order(x1$Peak3t,x1$Peak4t,x1$Npeak3t,x1$Npeak4t),];
with(x4,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+abs

((Peak2v[1]-Peak4v[2])*((Peak2t[1]+12+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+ abs

((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))+abs((Npeak2v[1]-Npeak4v[2])*((Npeak2t[1]+12+12)-Npeak4t[2]))
})
}
else if({indx[1]==0 & indx[2]==1 & indx[3]==2 & indx[4]==2}) #2
peaks 1 peak comparison
{x5<- x1[order(x1$Peak2t,x1$Npeak2t),];
with(x5,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t

[2]))+abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2])) })

}

else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==2 }) #3
peak 1 peak comparison?
{x6<- x1[order(x1$Peak2t,x1$Peak3t,x1$Npeak2t,x1$Npeak3t),];
with(x6,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12+12)-Peak3t

[2])) + abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t

[1]+12+12)-Npeak3t[2])) })
}
else if({indx[1]==0 & indx[2]==0 & indx[3]==0 & indx[4]==1}) # 4
peak 3 peak comparison
{x7<- x1[order(x1$Peak4t,x1$Npeak4t),];
with(x7,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+
abs

((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs

((Npeak3v[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+
abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2])) })

}
else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==1}) #4 peak
1 peak comparison
{x8<-
x1[order(x1$Peak2t,x1$Peak3t,x1$Peak4t,x1$Npeak2t,x1$Npeak3t,x1$Npeak4t),];
with(x8,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12+12)-Peak3t

[2]))+abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12+12+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-

Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12+12)-Npeak3t[2]))+abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12+12+12)-Npeak4t[2]))})

}
else ({ #cases where peaks are similar
with(x1,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+abs

((Peak4v[1]-Peak4v[2])*(Peak4t[1]-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak3v

[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+abs((Npeak4v[1]-Npeak4v[2])*(Npeak4t[1]-Npeak4t[2]))
})
})

}))))
res2<-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)<-1:nrow(res2)?


But in the end i am getting the following error
Error in abs((Peak1v[1] - Peak1v[2]) * (Peak1t[1] - Peak1t[2])) + abs((Peak2v[1]
- ?:?
? non-numeric argument to binary operator

When i use your commands, i dont get this error.

Elisa
> Date: Mon, 25 Mar 2013 22:53:34 -0700
> From: smartpink111 at yahoo.com
> Subject: Re: Distance calculation
> To: eliza_botto at hotmail.com
> CC: r-help at r-project.org
> 
> Hi Elisa,
> 
> Based on the formula you gave, this is what I got:
> 
> dat1<- read.csv("rate.csv",sep=",")
> 
> res<- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x)
{x1<-rbind(dat1[i,],x);colnames(x1)<-gsub("[.]","",colnames(x1));
> ?if({indx<- colSums(x1[,2:5]==0);indx[1]==0 & indx[2]==0 &
indx[3]==1 & indx[4]==2}) #3 peaks 2 peak comparison
> ?{x2<- x1[order(x1$Peak3t,x1$Npeak3t),];
> ?with(x2,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+
> abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))})
> }
> else if({indx[1]==0 & indx[2]==0 & indx[3]==1 & indx[4]==1}) #4
peaks 2 peak comparison. Peak3 Peak4 value compared with Peak1
> {x4<- x1[order(x1$Peak3t,x1$Peak4t,x1$Npeak3t,x1$Npeak4t),];
> with(x4,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))+abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2]))
})
> }
> else if({indx[1]==0 & indx[2]==1 & indx[3]==2 & indx[4]==2}) #2
peaks 1 peak comparison
> {x5<- x1[order(x1$Peak2t,x1$Npeak2t),];
>
with(x5,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))
})
> 
> }
> 
> else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==2 })
#3 peak 1 peak comparison
> {x6<- x1[order(x1$Peak2t,x1$Peak3t,x1$Npeak2t,x1$Npeak3t),];
> with(x6,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2])) +
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2])) })
> }
> else if({indx[1]==0 & indx[2]==0 & indx[3]==0 & indx[4]==1}) #
4 peak 3 peak comparison
> {x7<- x1[order(x1$Peak4t,x1$Npeak4t),];
>
with(x7,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+
abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak3v[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+
abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2])) })
> 
> }
> else if({indx[1]==0 & indx[2]==1 & indx[3]==1 & indx[4]==1}) #4
peak 1 peak comparison
> {x8<-
x1[order(x1$Peak2t,x1$Peak3t,x1$Peak4t,x1$Npeak2t,x1$Npeak3t,x1$Npeak4t),];
> with(x8,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+
abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Peak1v[1]-Peak3v[2])*((Peak1t[1]+12)-Peak3t[2]))+abs((Peak1v[1]-Peak4v[2])*((Peak1t[1]+12)-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+
abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))+
abs((Npeak1v[1]-Npeak3v[2])*((Npeak1t[1]+12)-Npeak3t[2]))+abs((Npeak1v[1]-Npeak4v[2])*((Npeak1t[1]+12)-Npeak4t[2]))})
> 
> }
> else ({ #cases where peaks are similar
>
with(x1,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))+abs((Peak3v[1]-Peak3v[2])*(Peak3t[1]-Peak3t[2]))+abs((Peak4v[1]-Peak4v[2])*(Peak4t[1]-Peak4t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))+abs((Npeak3v[1]-Npeak3v[2])*(Npeak3t[1]-Npeak3t[2]))+abs((Npeak4v[1]-Npeak4v[2])*(Npeak4t[1]-Npeak4t[2]))
})
> })
> 
> }))))
> res2<-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
> row.names(res2)<-1:nrow(res2)
> res2[1:5,1:5]
> #????? [,1]???? [,2]???? [,3]???? [,4]??? [,5]
> #1?? 0.0000?? 0.0000?? 0.0000 857.6834?? 0.000
> #2?? 0.0000?? 0.0000?? 0.0000 611.1167?? 0.000
> #3?? 0.0000?? 0.0000?? 0.0000 854.3765?? 0.000
> #4 857.6834 611.1167 854.3765?? 0.0000 579.756
> #5?? 0.0000?? 0.0000?? 0.0000 579.7560?? 0.000
> dim(res2)
> #[1] 124 124
> 
> 
> 
> I also validated each of the cases by calculating the values:
> For example:? stations 1 and 4, your previous email indicated
> 
> ""
> but the distance between 1 and 4 is not 379.1364, actually its 1495.01
> 
> ""
> 
> Based on my calculation:
> 
> dat1[c(1,4),]
> #? St. Peak1.t. Peak2.t. Peak3.t. Peak4.t. Npeak1.t. Npeak2.t. Npeak3.t.
> #1?? 1??????? 5?????? 10??????? 0??????? 0???????? 7??????? 13???????? 0
> #4?? 4??????? 4??????? 8?????? 10??????? 0???????? 6???????? 9??????? 14
> ?# Npeak4.t. Peak1.v. Peak2.v. Peak3.v. Peak4.v. Npeak1.v. Npeak2.v.
Npeak3.v.
> #1???????? 0 56.28785 17.43170? 0.00000??????? 0? 49.66025 81.759761???
0.0000
> #4???????? 0 27.91471 12.06137 11.52018??????? 0? 10.29134? 5.062441??
17.8527
> ?# Npeak4.v.
> #1???????? 0
> #4???????? 0
>
?abs((56.28785-27.91471)*(5-4))+abs((17.43170-12.06137)*(10-8))+abs((56.28785-11.52018)*((5+12)-10))+abs((49.66025-10.29134)*(7-6))+abs((81.759761-5.062441)*(13-9))+abs((49.66025-17.8527)*((7+12)-14))
> #[1] 857.6834 #which is the same value I got in the distance matrix.
> 
> If the formula is different, please let me know.
> A.K.
> 
> 
> 
> 
> 
> 
> 
> ________________________________
> From: eliza botto <eliza_botto at hotmail.com>
> To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com>
> Sent: Monday, March 25, 2013 3:50 AM
> Subject: RE: Distance calculation
> 
> 
> 
> Dear Arun,
> I have a slight problem with this coding for the calculation of distance
matrix. The text files contains that problem where as data is in excel file.
> kindly help me out.. :(
> 
> Elisa?
> 
> 
> > Date: Fri, 22 Mar 2013 06:46:18 -0700
> > From: smartpink111 at yahoo.com
> > Subject: Re: Distance calculation
> > To: eliza_botto at hotmail.com
> > CC: r-help at r-project.org
> > 
> > Hi Elisa,
> > 
> > I hope this is what you wanted.
> > 
> > 
> > dat1<-read.csv("peaks.csv",sep=",")
> > #Subset
> > dat2<-dat1[1:5,]
> > res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i)
do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))),
function(x) {x1<-rbind(dat2[i,],x);
abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2.t.[2]))}))))
> > res2<-do.call(cbind,lapply(seq_len(ncol(res1)),function(i)
c(c(tail(res1[seq(1,i,1),i],-1),0),res1[-c(1:i),i])))
> > row.names(res2)<-1:nrow(res2)
> > ?res2
> > #????? [,1]???? [,2]???? [,3]???? [,4]???? [,5]
> > #1?? 0.0000?? 0.0000?? 0.0000 379.1364?? 0.0000
> > #2?? 0.0000?? 0.0000?? 0.0000 312.8267?? 0.0000
> > #3?? 0.0000?? 0.0000?? 0.0000 379.6576?? 0.0000
> > #4 379.1364 312.8267 379.6576?? 0.0000 324.4063
> > #5?? 0.0000?? 0.0000?? 0.0000 324.4063?? 0.0000
> > 
> > resWhole<-do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x) {x1<-rbind(dat1[i,],x);
abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2.t.[2]))}))))
> > res2Whole<-do.call(cbind,lapply(seq_len(ncol(resWhole)),function(i)
c(c(tail(resWhole[seq(1,i,1),i],-1),0),resWhole[-c(1:i),i])))
> > row.names(res2Whole)<-1:nrow(res2Whole)
> > dim(res2Whole)
> > #[1] 124 124
> > res2Whole[1:5,1:5]
> > #????? [,1]???? [,2]???? [,3]???? [,4]???? [,5]
> > #1?? 0.0000?? 0.0000?? 0.0000 379.1364?? 0.0000
> > #2?? 0.0000?? 0.0000?? 0.0000 312.8267?? 0.0000
> > #3?? 0.0000?? 0.0000?? 0.0000 379.6576?? 0.0000
> > #4 379.1364 312.8267 379.6576?? 0.0000 324.4063
> > #5?? 0.0000?? 0.0000?? 0.0000 324.4063?? 0.0000
> > 
> > A.K.
> > 
> > 
> > 
> > ________________________________
> > From: eliza botto <eliza_botto at hotmail.com>
> > To: "smartpink111 at yahoo.com" <smartpink111 at
yahoo.com>
> > Sent: Friday, March 22, 2013 8:26 AM
> > Subject: 
> > 
> > 
> > 
> > Dear Arun,
> > I hope you are fine.?
> > ?the attached text file has my recent question and excel file contains
the data.
> > 
> > 
> > thanks in advance
> > 
> > Elisa

arun

2013-Mar-27 14:39 UTC

head link

[R] Distance calculation

HI,
Try this:


dat1<- read.csv("DQP.csv",sep="\t")
res<- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i)
do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))),
function(x)
{x1<-rbind(dat1[i,],x);colnames(x1)<-gsub("[.]","",colnames(x1));
?if({indx<- colSums(x1[,2:3]==0);indx[1]==0 & indx[2]==1 }) #2 peaks 1
peak comparison
{x2<- x1[order(x1$Peak2t,x1$Npeak2t),];
with(x2,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2]))
})
}
else #cases where peaks are similar 
{with(x1,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2]))
+
abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))})
}

}))))

res2<-do.call(cbind,lapply(seq_len(ncol(res)),function(i)
c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)<-1:nrow(res2)

?dim(res2)
#[1] 124 124
?res2[1:5,1:5]
#??? [,1]?? [,2]? [,3]? [,4]?? [,5]
#1? 0.000? 1.512 7.031 3.662 13.030
#2? 1.512? 0.000 7.109 4.880 18.731
#3? 7.031? 7.109 0.000 0.056? 1.280
#4? 3.662? 4.880 0.056 0.000? 0.584
#5 13.030 18.731 1.280 0.584? 0.000


A.K.



________________________________
From: eliza botto <eliza_botto at hotmail.com>
To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> 
Sent: Wednesday, March 27, 2013 8:58 AM
Subject: RE: Distance calculation



Dear Arun,
I would like to ask a small question.
In the distance calculation procedure, if there are only 2 peaks and 1 peaks
stations and there are no 3 and 4 peaks stations, like the file i attached. How
to modify the script below to eliminate 3 peaks and 4 peaks scripts??
I hope you mind my hasty questioning
Thanks in advance
Elisa

R help - Mar 2013 - Distance calculation

[R] Distance calculation

[R] Distance calculation

[R] Distance calculation

[R] Distance calculation