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Nico Met

2013-Oct-08 14:47 UTC

[R] matrix of mean values

Dear all, I have a data set where I want to make a matrix of the mean
values using 3rd column (value). Where rows are unique “loc” and columns
would be “variables”. If there is one observation or no observation just
use “NA”.


> dput(d1)
structure(list(loc = structure(c(6L, 9L, 9L, 9L, 9L, 4L, 4L,

4L, 4L, 4L, 4L, 13L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 7L,

7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 10L,

10L, 10L, 10L, 10L, 11L, 11L, 12L, 12L, 12L, 12L, 12L, 12L, 1L,

3L, 3L, 3L, 2L), .Label = c("AX09_126", "AX09_628",
"AX09_924",

"IALA08_3299", "JA08_246", "JA08_3299",
"JA08_408", "JE08_246",

"JE08_3299", "JE08_408", "NO08_246",
"NO08_408", "PP08_3299"), class "factor"),

    variables = structure(c(5L, 6L, 6L, 6L, 6L, 6L, 5L, 5L, 5L,

    7L, 7L, 7L, 7L, 4L, 4L, 4L, 4L, 4L, 2L, 2L, 2L, 2L, 3L, 3L,

    3L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 1L, 5L, 5L, 5L, 6L, 6L, 7L,

    7L, 7L, 7L, 7L, 7L, 7L, 3L, 7L, 4L, 4L, 4L, 7L, 7L, 7L, 7L

    ), .Label = c("i", "r", "s", "t",
"x", "y", "z"), class = "factor"),

    value = structure(c(1L, 21L, 2L, 1L, 30L, 32L, 4L, 32L, 5L,

    20L, 5L, 1L, 24L, 12L, 26L, 24L, 12L, 15L, 26L, 14L, 15L,

    23L, 13L, 11L, 23L, 25L, 23L, 6L, 7L, 3L, 8L, 19L, 3L, 19L,

    28L, 16L, 18L, 18L, 17L, 31L, 29L, 10L, 27L, 22L, 9L, 22L,

    22L, 9L, 9L, 35L, 33L, 33L, 33L, 34L), .Label = c("0",
"0,3333",

    "142,1", "1506,00006", "1520,66664",
"170,36", "18,6185",

    "19,2933", "20,2125", "20,7423",
"225,94", "23,2245", "24,3526",

    "24,3685", "24,4593", "242,84",
"25,7422", "26,34", "26,5989",

    "30,33336284", "36,33336608", "39,913",
"41,4933", "43,9104",

    "47,1241", "49,3467", "49,5715",
"50,0037", "50,7974", "53,6667",

    "56,2778", "76", "76,5", "77,6667",
"82"), class = "factor")), .Names c("loc",

"variables", "value"), class = "data.frame",
row.names = c(NA,

-54L))



Thanks



Nico

	[[alternative HTML version deleted]]

arun

2013-Oct-08 14:54 UTC

head link

[R] matrix of mean values

Try:

d1$value<-as.numeric(gsub(",","",as.character(d1$value)))
library(reshape2)
res <- dcast(d1,loc~variables,value.var="value",mean)
A.K.


----- Original Message -----
From: Nico Met <nicomet80 at gmail.com>
To: R help <r-help at r-project.org>
Cc: 
Sent: Tuesday, October 8, 2013 10:47 AM
Subject: [R] matrix of mean values

Dear all, I have a data set where I want to make a matrix of the mean
values using 3rd column (value). Where rows are unique ?loc? and columns
would be ?variables?. If there is one observation or no observation just
use ?NA?.


> dput(d1)
structure(list(loc = structure(c(6L, 9L, 9L, 9L, 9L, 4L, 4L,

4L, 4L, 4L, 4L, 13L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 7L,

7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 10L,

10L, 10L, 10L, 10L, 11L, 11L, 12L, 12L, 12L, 12L, 12L, 12L, 1L,

3L, 3L, 3L, 2L), .Label = c("AX09_126", "AX09_628",
"AX09_924",

"IALA08_3299", "JA08_246", "JA08_3299",
"JA08_408", "JE08_246",

"JE08_3299", "JE08_408", "NO08_246",
"NO08_408", "PP08_3299"), class "factor"),

? ? variables = structure(c(5L, 6L, 6L, 6L, 6L, 6L, 5L, 5L, 5L,

? ? 7L, 7L, 7L, 7L, 4L, 4L, 4L, 4L, 4L, 2L, 2L, 2L, 2L, 3L, 3L,

? ? 3L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 1L, 5L, 5L, 5L, 6L, 6L, 7L,

? ? 7L, 7L, 7L, 7L, 7L, 7L, 3L, 7L, 4L, 4L, 4L, 7L, 7L, 7L, 7L

? ? ), .Label = c("i", "r", "s", "t",
"x", "y", "z"), class = "factor"),

? ? value = structure(c(1L, 21L, 2L, 1L, 30L, 32L, 4L, 32L, 5L,

? ? 20L, 5L, 1L, 24L, 12L, 26L, 24L, 12L, 15L, 26L, 14L, 15L,

? ? 23L, 13L, 11L, 23L, 25L, 23L, 6L, 7L, 3L, 8L, 19L, 3L, 19L,

? ? 28L, 16L, 18L, 18L, 17L, 31L, 29L, 10L, 27L, 22L, 9L, 22L,

? ? 22L, 9L, 9L, 35L, 33L, 33L, 33L, 34L), .Label = c("0",
"0,3333",

? ? "142,1", "1506,00006", "1520,66664",
"170,36", "18,6185",

? ? "19,2933", "20,2125", "20,7423",
"225,94", "23,2245", "24,3526",

? ? "24,3685", "24,4593", "242,84",
"25,7422", "26,34", "26,5989",

? ? "30,33336284", "36,33336608", "39,913",
"41,4933", "43,9104",

? ? "47,1241", "49,3467", "49,5715",
"50,0037", "50,7974", "53,6667",

? ? "56,2778", "76", "76,5", "77,6667",
"82"), class = "factor")), .Names c("loc",

"variables", "value"), class = "data.frame",
row.names = c(NA,

-54L))



Thanks



Nico

??? [[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.

arun

2013-Oct-08 14:56 UTC

head link

[R] matrix of mean values

To make it matrix:


row.names(res)<- res[,1]
?res1<- as.matrix(res[,-1])
A.K.



----- Original Message -----
From: arun <smartpink111 at yahoo.com>
To: Nico Met <nicomet80 at gmail.com>
Cc: R help <r-help at r-project.org>
Sent: Tuesday, October 8, 2013 10:54 AM
Subject: Re: [R] matrix of mean values

Try:

d1$value<-as.numeric(gsub(",","",as.character(d1$value)))
library(reshape2)
res <- dcast(d1,loc~variables,value.var="value",mean)
A.K.



----- Original Message -----
From: Nico Met <nicomet80 at gmail.com>
To: R help <r-help at r-project.org>
Cc: 
Sent: Tuesday, October 8, 2013 10:47 AM
Subject: [R] matrix of mean values

Dear all, I have a data set where I want to make a matrix of the mean
values using 3rd column (value). Where rows are unique ?loc? and columns
would be ?variables?. If there is one observation or no observation just
use ?NA?.


> dput(d1)
structure(list(loc = structure(c(6L, 9L, 9L, 9L, 9L, 4L, 4L,

4L, 4L, 4L, 4L, 13L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 7L,

7L, 7L, 7L, 7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 10L,

10L, 10L, 10L, 10L, 11L, 11L, 12L, 12L, 12L, 12L, 12L, 12L, 1L,

3L, 3L, 3L, 2L), .Label = c("AX09_126", "AX09_628",
"AX09_924",

"IALA08_3299", "JA08_246", "JA08_3299",
"JA08_408", "JE08_246",

"JE08_3299", "JE08_408", "NO08_246",
"NO08_408", "PP08_3299"), class "factor"),

? ? variables = structure(c(5L, 6L, 6L, 6L, 6L, 6L, 5L, 5L, 5L,

? ? 7L, 7L, 7L, 7L, 4L, 4L, 4L, 4L, 4L, 2L, 2L, 2L, 2L, 3L, 3L,

? ? 3L, 3L, 3L, 3L, 1L, 1L, 1L, 2L, 1L, 5L, 5L, 5L, 6L, 6L, 7L,

? ? 7L, 7L, 7L, 7L, 7L, 7L, 3L, 7L, 4L, 4L, 4L, 7L, 7L, 7L, 7L

? ? ), .Label = c("i", "r", "s", "t",
"x", "y", "z"), class = "factor"),

? ? value = structure(c(1L, 21L, 2L, 1L, 30L, 32L, 4L, 32L, 5L,

? ? 20L, 5L, 1L, 24L, 12L, 26L, 24L, 12L, 15L, 26L, 14L, 15L,

? ? 23L, 13L, 11L, 23L, 25L, 23L, 6L, 7L, 3L, 8L, 19L, 3L, 19L,

? ? 28L, 16L, 18L, 18L, 17L, 31L, 29L, 10L, 27L, 22L, 9L, 22L,

? ? 22L, 9L, 9L, 35L, 33L, 33L, 33L, 34L), .Label = c("0",
"0,3333",

? ? "142,1", "1506,00006", "1520,66664",
"170,36", "18,6185",

? ? "19,2933", "20,2125", "20,7423",
"225,94", "23,2245", "24,3526",

? ? "24,3685", "24,4593", "242,84",
"25,7422", "26,34", "26,5989",

? ? "30,33336284", "36,33336608", "39,913",
"41,4933", "43,9104",

? ? "47,1241", "49,3467", "49,5715",
"50,0037", "50,7974", "53,6667",

? ? "56,2778", "76", "76,5", "77,6667",
"82"), class = "factor")), .Names c("loc",

"variables", "value"), class = "data.frame",
row.names = c(NA,

-54L))



Thanks



Nico

??? [[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.

arun

2013-Oct-15 14:38 UTC

head link

[R] matrix of mean values

Hi Nico,
Try:

setAs("character", "num.commas", 
??????? function(from) as.numeric(gsub(",", "", from) ) )
d1 <-
read.csv("cont.txt",header=TRUE,sep="\t",colClasses=c("numeric","character","num.commas"))
colnames(d1)[-3] <- c("loc","variables")
library(reshape2)
?res <- dcast(d1,loc~variables,value.var="value",mean)

row.names(res) <- res[,1]
?res1 <- as.matrix(res[,-1])
?dim(res1)
#[1] 129 190

A.K.

On Tuesday, October 15, 2013 8:07 AM, Nico Met <nicomet80 at gmail.com>
wrote:

Hi Arun, many thanks. Here I attached the data.

Regards

Nico

arun

2013-Oct-23 12:58 UTC

head link

[R] matrix of mean values

Hi Nico,

Try:
dat<-read.table("pollShed.txt", header = T, sep =
"\t",dec=",")
?as.vector(sapply(split(dat,dat$name1),nrow))
?#[1]? 86? 10? 43?? 3? 20?? 7? 25? 12? 18?? 3?? 7? 12? 12? 14?? 3? 14? 25 175?
59
#[20]? 49? 16?? 3? 16? 14? 25? 14? 19?? 3? 18?? 7?? 4?? 3? 45? 14? 16?? 3? 12
238
#[39]? 12?? 3? 25?? 9? 20? 12? 12? 15? 20?? 7 170? 16? 12? 45? 14? 25?? 3? 12?
14
#[58] 397?? 1?? 3? 10 542? 14? 14? 12? 24? 12?? 3? 14? 20?? 3?? 4?? 3? 40? ####
list element 59 produce the error
lst1 <- split(dat,dat$name1)

t.test(lst1[[59]][,2],dat[,2])
#Error in t.test.default(lst1[[59]][, 2], dat[, 2]) : 
#? not enough 'x' observations

?res <- sapply(lst1[sapply(lst1,nrow)>1],function(x)
t.test(x[,2],dat[,2])$p.value)
?head(res)
#????? -99999?????? 108946?????? 111364?????? 140400?????? 149067?????? 206262 
#4.508590e-01 9.142790e-05 4.911925e-09 7.124666e-01 1.376976e-03 3.161232e-02 



A.K.









On Wednesday, October 23, 2013 8:27 AM, Nico Met <nicomet80 at gmail.com>
wrote:

Many thanks Arun for the suggestion.?

However, I tried following and do not know why it shows error. Data attached.



?dat<-read.table("pollShed.txt", header = T, sep =
"\t",dec=",") # works fine?

?mean1<-sapply(split(dat,dat$name1),function(x)
t.test(x[,2],dat[,2])$statistic)?

# Error in t.test.default(x[, 2], dat[, 2]) : not enough 'x'
observations


?pval<-sapply(split(dat,dat$name1),function(x) t.test(x[,2],dat[,2])$p.value)

# Error in t.test.default(x[, 2], dat[, 2]) : not enough 'x'
observations

Thanks in advance

R help - Oct 2013 - matrix of mean values

[R] matrix of mean values

[R] matrix of mean values

[R] matrix of mean values

[R] matrix of mean values

[R] matrix of mean values