}
Vul <- distinct(d1[,c(1,2)])
dim(VC) <- c(length(unlist(str_split(as.character(d1[2,]$Y_vals), pattern =
","))),length(distinct(d1[,c(1,2)])$Name)) ## (rows, cols)
VC
VC_t <- t(VC)
Vulnerability <- matrix(apply(VC_t, 1, function(x) paste(x, collapse = ',')
;,")))
print(unique(VC ))
}
}
}
Vul <- distinct(d1[,c(65,4,3)])
dim(VC) <- c(length(unlist(str_split(as.character(d1[1,]$Y_vals), pattern =
","))),length(distinct(d1[,c(65,4,3)])$Name)) ## (rows, cols)
V
'Greece', NA)
Any help much appreciated!
Best,
ioanna
A<- data.frame( name1 = c('fields', 'fields', 'fields'),
name2= c('category', 'asset', 'country'),
value
in some cases there will be one country reported and in some
others multiple. How can i optimise the code?
Best,
From: Rainer M Krug
Sent: 16 January 2020 14:47
To: Ioanna Ioannou
Subject: Re: [R] How to save multiple values of a variable in a json file in R
Ch
n from
the json file to an R data.frame? See below for the json file.
Best,
ioanna
{
"pk": 670,
"model": "vulnerability.generalinformation",
"fields": {
"category": "Structure class",
"article_title"
aforementioned two. In
fact, the third is identical to the first. Could you please optimize?
Thank you very much again,
Best,
ioanna
-Original Message-
From: Jim Lemon [mailto:drjimle...@gmail.com]
Sent: Friday, December 20, 2019 9:04 PM
To: Ioannou, Ioanna
Cc: r-help mailing list
Subject
check the code I sent
last and based on that give your solution?
Many thanks.
Get Outlook for Android<https://aka.ms/ghei36>
From: Jim Lemon
Sent: Friday, December 20, 2019 11:40:28 AM
To: Ioannou, Ioanna
Cc: r-help mailing list
Subject: Re: [R] How to cr
mp; DS2_rows,calc_vars] -
D[calc_rows & DS3_rows,calc_vars]) +
0.43 * (D[calc_rows & DS3_rows,calc_vars] -
D[calc_rows & DS4_rows,calc_vars]) +
1.0* D[calc_rows & DS4_rows,calc_vars]
}
}
}
-----Original Mes
Hello everyone,
Could you please let me know how to create a new data.frame with the output of
the 2 unique loops. Essentially i want a data.frame with the IM, Taxonomy and
VC . MInd you VC is a vector with 33 elements.
Any ideas?
best,
ioanna
D<- data.frame(Ref.No = c(1622, 1623, 1624, 1
')[10:13])
So the question is how can I do that in an automated way for all possible
combinations and store the results in new data.frame which would look like
this:
Ref.No. Region IM.type TaxonomyIM_1IM_2IM_3IM_4VC_1
VC_2VC_3VC_4
1622South America
[10:13])
So the question is how can I do that in an automated way for all possible
combinations and store the results in new data.frame which would look like
this:
Ref.No. Region IM.type TaxonomyIM_1IM_2IM_3IM_4VC_1
VC_2VC_3VC_4
1622South America PGA
.16, 0.16, 0.16, 0.16, 0.16, 0.16),
IM_1 = c(0.24, 0.24, 0.24, 0.24, 0.24, 0.24, 0.24, 0.24),
Prob.of.exceedance_1 = c(0,0,0,0,0,0,0,0),
Prob.of.exceedance_2 = c(0,0,0,0,0,0,0,0),
Prob.of.exceedance_3 =
c(0.26,0.001,0.00
27;ER+ETR_H1')[10:13])
So the question is how can I do that in an automated way for all possible
combinations and store the results in new data.frame which would look like
this:
Ref.No. Region IM.type TaxonomyIM_1 IM_2 IM_3 IM_4 VC_1 VC_2
VC_3 VC_4
1622 South Americ
+ETR_H1')[10:13])
So the question is how can I do that in an automated way for all possible
combinations and store the results in new data.frame which would look like
this:
Ref.No. Region IM.type TaxonomyIM_1 IM_2 IM_3 IM_4 VC_1 VC_2
VC_3 VC_4
1622 South Americ
Hello everyone,
Thank you for this. Nonetheless it is not exactly want i need.
I need mydata[[1]] to provide the values for all 3 variables (Y, X1 and X2) of
the first imputation only. As it stands it returns the whole database.
Any ideas?
Best,
ioanna
.
Can you help me please?
Best,
ioanna
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27;D','B'),
b= c('N','N','Y','N','Y','N','N') )
I want to add a column c which will change 'C' to 'D' if column b is 'Y'.
error. Any idea how to fix
the problem?
Any help much appreciated,
Best,
Ioanna
new<-read(Sample.csv)
new$Use<-factor(new$Use)
MissingData <- missing_data.frame(new)
MissingData <- change(MissingData, y = "DS", what = "type", to =
"ordered-catego
error. Any idea how to fix
the problem?
Any help much appreciated,
Best,
Ioanna
new<-read(Sample.csv)
new$Use<-factor(new$Use)
MissingData <- missing_data.frame(new)
MissingData <- change(MissingData, y = "DS", what = "type", to =
"ordered-catego
for illustration purposes:
d <- data.frame(x = runif(N))
d$y <- d$x^2 - d$x + 1 + (1+d$x)*rnorm(N, sd = 0.1)
Any help much appreciated.
Best,
Ioanna
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.
Any ideas?
For example
N <- 250
xeval <- 0:100/100
## ex1
d <- data.frame(x = runif(N))
d$y <- d$x^2 - d$x + 1 + rnorm(N, sd = 0.1)
r <- locpol(y~x,d)
plot(r)
Best,
Ioanna
-
E-Mail: (Ted Harding
s(X1,Y3,lty=2)
Any ideas how?
Best
IOanna
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s(X1,Y3,lty=2)
Any ideas how?
Best
IOanna
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Hello all,
A simple question. When I use grf from the package 'geoR' , I adopt the
exponential model. For this model is the parameter phi in m or km?
Best
ioanna
[[alternative HTML version deleted]]
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Hello all,
A simple question. When I use grf from the package 'geoR' , I adopt the
exponential model. For this model is the parameter range in m or km?
Best
ioanna
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D,none))
Thanks in advance,
IOanna
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: 21 February 2014 00:19
To: r-help@r-project.org
Cc: ioanna ioannou
Subject: Re: [R] Data manipulation in a data.frame
Also,
rownames(which(t(!!A[,-1]),arr.ind=TRUE))
A.K.
On Thursday, Fe
hat:
A<-data.frame(A=c(10,100,1000,30,50,60,300),
B=c(0,1,1,1,0,0,0),
C=c(0,0,0,0,1,1,0),
D=c(1,0,0,0,0,0,1),
Variable=c(D,B,B,B,C,C,D))
How can I do it?
Best
IOanna
[[alternative HTML version d
Fantastic. Thanks very much! Is there an easy way to plot the points and
the 4 areas?
Best,
Ioanna
-Original Message-
From: David Carlson [mailto:dcarl...@tamu.edu]
Sent: 25 November 2013 15:21
To: 'IOANNA'; r-help@r-project.org
Subject: RE: [R] Aggregating spatial data
ing to these points.
My question is how to divide this area in 4 sub-areas of equal points each
and produce the counts of z1= '1', '2' , '3' in each quarter as well as mean
values of z2 for each quarter.
Best,
Ioanna
understand why as the code seems to run if I change the limits of
the grid:
sim2 <- grf(nslon*nslat, grid="reg", nx=nslon, ny=nslat,cov.pars=c(1,
range),
nsim=N, cov.model = "exponential",
xlims=c(0.02,5.00),ylims=c(0.02,5.00) )
Any ideas?
Best
{
ds[i]<-6
}
}
}
Best,
IOanna
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and provide commented, minimal, self-contained, reproducible code.
ks fine. What if my data were clustered requiring a bandwidth that
varies with x? How can I do that?
Thanks in advance,
Ioanna
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Hello all,
I have data in the form of a table:
X Y1Y2
0.1 3 2
0.2 2 1
And I would like to transform in the form:
X Y
0.1 Y1
0.1 Y1
0.1 Y1
0.1 Y2
0.1 Y2
0.2 Y1
0.2 Y1
0.2 Y2
Any ideas how?
Thanks in advance,
IOanna
Thanks again. I hope its clearer now.
Ioanna
-Original Message-
From: John Kane [mailto:jrkrid...@inbox.com]
Sent: 15 March 2013 12:51
To: IOANNA; r-help@r-project.org
Subject: RE: [R] Data manipulation
What zero values? And are they acutall zeros or are the NA's, that is,
m
Thanks a lot!
-Original Message-
From: John Kane [mailto:jrkrid...@inbox.com]
Sent: 15 March 2013 13:41
To: Blaser Nello; IOANNA; r-help@r-project.org
Subject: Re: [R] Data manipulation
Nice. That does look like it. IOANNA?
John Kane
Kingston ON Canada
> -Original Mess
achieve what I want?
Best regards,
Ioanna
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, 2 Afxentiou Ampelokipi
Thessaloniki Greece, 4 Afxentiou Ampelokipi Thessaloniki Greece, 55
Agathonos Ampelokipi Thessaloniki Greece)
For (i in 1:4){
Y<-geocode('X')
print Y[i]
}
Best wishes,
Ioanna
[[alternative HTML
residuals for the nonparametric
model?
Thanks, Ioanna
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ctor(class),
family=binomial("logit"))
This works. However, if I change the family to Gaussian:
ced.logr <- glm(ced.del ~ cat + follows + factor(class), family=gaussian)
I get the error:
Error: (subscript) logical subscript too long
I would like to use the probit function. Is this
ll the standard error of the parameters and the
function:
SEconc<-0.05033
I don't know how to do this. Any help?
Regards,
Ioanna
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5 Feb 2012 03:42:05 -0800
From: ml-node+s789695n4358800...@n4.nabble.com
To: ii54...@msn.com
Subject: Re: plotting confidence bands from predict.nls
On 05/02/2012 08:10, ioanna wrote:
> How do you use bootstrap to estimate the confidence as well as the prediction
> intervals i
How do you use bootstrap to estimate the confidence as well as the prediction
intervals in nonlinear regression ?
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Sent from the R help mailing list archive at Nabble.c
Hello,
Lets assume I have an ordinal response variable representing the
D<-c(D0,D1,D2,D3,D4) where D0 is no damage and D4 is collapse which I want
to correlate with a continuous predictor variable, wind speed at the
location of each building.
is there a function in R which I can use to es
Is it possible to apply a kernel smoothing regression whose estimator or
indeed the confidence intervals cannot take negative values or values
greater than 1?
Best regards,
Ioanna
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