I'm doing PRC using the vegan-package but run into trouble when I attempt to perform an Anova on the results. I get the following error-message:
Error in doShuffleSet(spln[[i]], nset = nset, control) :
number of items to replace is not a multiple of replacement length
The problem originates in the shuffleSet-function of the permute-package. I created a reproducible example below. The weird thing is that the shuffle-function does not cause trouble, but the shuffleSet-function does.
In my experiment 3 treatments were given to 4 animals. The animals received the treatments in different orders. On every day, 5 samples were collected over time.
I would like to permute my observations within animals and not between them. Therefore I use AnimalID as a block.
I would like to permute days (in my actual experiments animals received the same treatment multiple times) but keep the measurements within a day intact. Hence I chose to permute Days freely and have no permutations within Days.
require(permute)
TreatmentLevels=3
Animals=4
TimeSteps=5
AnimalID=rep(letters[1:Animals],each=TreatmentLevels*TimeSteps)
Time=rep(1:TimeSteps,Animals=TreatmentLevels)
#treatments were given in different order per animal.
Day=rep(c(1,2,3,2,3,1,3,2,1,2,3,1),each=TimeSteps)
Treatment=rep(rep(LETTERS[1:TreatmentLevels],each=TimeSteps),Animals)
dataset=as.data.frame(cbind(AnimalID,Treatment,Day,Time))
ctrl=how(blocks = dataset$AnimalID,plots = Plots(strata=dataset$Day,type = "free"),
within=Within(type="none"), nperm = 999)
#this works
shuffle(60,control=ctrl)
#this giveas an error
shuffleSet(60,nset=1,control=ctrl)
shuffleSet(60,nset=10,control=ctrl)
The problem seems to be in the block. Because this works
dataset$AnimalDay=factor(paste0(dataset$AnimalID,dataset$Day))
ctrl=how(plots = Plots(strata=dataset$AnimalDay,type = "free"),
within=Within(type="none"), nperm = 999)
#this works
shuffle(60,control=ctrl)
shuffleSet(60,nset=1,control=ctrl)
shuffleSet(60,nset=10,control=ctrl)
The key problem seems to be nset = 1: the permutation is generated and shuffleSet works, but printing the result fails because one set is dropped to a vector and print expects a matrix. You can get the permutation, you can use the permutation, but you cannot print it.
We got to fix this.
Related
I am just getting started with R so I am sorry if I say things that dont make sense.
I am trying to make a for loop which does the following,
l_dtest[[1]]<-vector()
l_dtest[[2]]<-vector()
l_dtest[[3]]<-vector()
l_dtest[[4]]<-vector()
l_dtest[[5]]<-vector()
all the way up till any number which will be assigned as n. for example, if n was chosen to be 100 then it would repeat this all the way to > l_dtest[[100]]<-vector().
I have tried multiple different attempts at doing this and here is one of them.
n<-4
p<-(1:n)
l_dtest<-list()
for(i in p){
print((l_dtest[i]<-vector())<-i)
}
Again I am VERY new to R so I don't know what I am doing or what is wrong with this loop.
The detailed background for why I need to do this is that I need to write an R function that receives as input the size of the population "n", runs a simulation of the model below with that population size, and returns the number of generations it took to reach a MRCA (most recent common ancestor).
Here is the model,
We assume the population size is constant at n. Generations are discrete and non-overlapping. The genealogy is formed by this random process: in each
generation, each individual chooses two parents at random from the previous generation. The choices are made randomly and equally likely over the n possibilities and each individual chooses twice. All choices are made independently. Thus, for example, it is possible that, when an individual chooses his two parents, he chooses the same individual twice, so that in
fact he ends up with just one parent; this happens with probability 1/n.
I don't understand the specific step at the begining of this post or why I need to do it but my teacher said I do. I don't know if this helps but the next step is choosing parents for the first person and then combining the lists from the step I posted with a previous step. It looks like this,
sample(1:5, 2, replace=T)
#[1] 1 2
l_dtemp[[1]]<-union(l_dtemp[[1]], l_d[[1]]) #To my understanding, l_dtem[[1]] is now receiving the listdescandants from l_d[[1]] bcs the ladder chose l_dtemp[[1]] as first parent
l_dtemp[[2]]<-union(l_dtemp[[2]], l_d[[1]]) #Same as ^^ but for l_d[[1]]'s 2nd choice which is l_dtemp[[2]]
sample(1:5, 2, replace=T)
#[1] 1 3
l_dtemp[[1]]<-union(l_dtemp[[1]], l_d[[2]])
l_dtemp[[3]]<-union(l_dtemp[[3]], l_d[[2]])
I have mirrored some code to perform an analysis, and everything is working correctly (I believe). However, I am trying to understand a few lines of code related to splitting the data up into 40% testing and 60% training sets.
To my current understanding, the code randomly assigns each row into group 1 or 2. Subsequently, all the the rows assigned to 1 are pulled into the training set, and the 2's into the testing.
Later, I realized that sampling with replacement is not want I wanted for my data analysis. Although in this case I am unsure of what is actually being replaced. Currently, I do not believe it is the actual data itself being replaced, rather the "1" and "2" place holders. I am looking to understand exactly how these lines of code work. Based on my results, it seems as it is working accomplishing what I want. I need to confirm whether or not the data itself is being replaced.
To test the lines in question, I created a dataframe with 10 unique values (1 through 10).
If the data values themselves were being sampled with replacement, I would expect to see some duplicates in "training1" or "testing2". I ran these lines of code 10 times with 10 different set.seed numbers and the data values were never duplicated. To me, this suggest the data itself is not being replaced.
If I set replace= FALSE I get this error:
Error in sample.int(x, size, replace, prob) :
cannot take a sample larger than the population when 'replace = FALSE'
set.seed(8)
test <-sample(2, nrow(df), replace = TRUE, prob = c(.6,.4))
training1 <- df[test==1,]
testing2 <- df[test==2,]
Id like to split up my data into 60-40 training and testing. Although I am not sure that this is actually happening. I think the prob function is not doing what I think it should be doing. I've noticed the prob function does not actually split the data exactly into 60percent and 40percent. In the case of the n=10 example, it can result in 7 training 2 testing, or even 6 training 4 testing. With my actual larger dataset with ~n=2000+, it averages out to be pretty close to 60/40 (i.e., 60.3/39.7).
The way you are sampling is bound to result in a undesired/ random split size unless number of observations are huge, formally known as law of large numbers. To make a more deterministic split, decide on the size/ number of observation for the train data and use it to sample from nrow(df):
set.seed(8)
# for a 60/40 train/test split
train_indx = sample(x = 1:nrow(df),
size = 0.6*nrow(df),
replace = FALSE)
train_df <- df[train_indx,]
test_df <- df[-train_indx,]
I recommend splitting the code based on Mankind_008's answer. Since I ran quite a bit of analysis based on the original code, I spent a few hours looking into what it does exactly.
The original code:
test <-sample(2, nrow(df), replace = TRUE, prob = c(.6,.4))
Answer From ( https://www.datacamp.com/community/tutorials/machine-learning-in-r ):
"Note that the replace argument is set to TRUE: this means that you assign a 1 or a 2 to a certain row and then reset the vector of 2 to its original state. This means that, for the next rows in your data set, you can either assign a 1 or a 2, each time again. The probability of choosing a 1 or a 2 should not be proportional to the weights amongst the remaining items, so you specify probability weights. Note also that, even though you don’t see it in the DataCamp Light chunk, the seed has still been set to 1234."
One of my main concerns that the data values themselves were being replaced. Rather it seems it allows the 1 and 2 placeholders to be assigned over again based on the probabilities.
Lets say I have this data. My objective is to extraxt combinations of sequences.
I have one constraint, the time between two events may not be more than 5, lets call this maxGap.
User <- c(rep(1,3)) # One users
Event <- c("C","B","C") # Say this is random events could be anything from LETTERS[1:4]
Time <- c(c(1,12,13)) # This is a timeline
df <- data.frame(User=User,
Event=Event,
Time=Time)
If want to use these sequences as binary explanatory variables for analysis.
Given this dataframe the result should be like this.
res.df <- data.frame(User=1,
C=1,
B=1,
CB=0,
BC=1,
CBC=0)
(CB) and (CBC) will be 0 since the maxGap > 5.
I was trying to write a function for this using many for-loops, but it becomes very complex if the sequence becomes larger and the different number of evets also becomes larger. And also if the number of different User grows to 100 000.
Is it possible of doing this in TraMineR with the help of seqeconstraint?
Here is how you would do that with TraMineR
df.seqe <- seqecreate(id=df$User, timestamp=df$Time, event=df$Event)
constr <- seqeconstraint(maxGap=5)
subseq <- seqefsub(df.seqe, minSupport=0, constraint=constr)
(presence <- seqeapplysub(subseq, method="presence"))
which gives
(B) (B)-(C) (C)
1-(C)-11-(B)-1-(C) 1 1 1
presence is a table with a column for each subsequence that occurs at least once in the data set. So, if you have several individuals (event sequences), the table will have one row per individual and the columns will be the binary variable you are looking for. (See also TraMineR: Can I get the complete sequence if I give an event sub sequence? )
However, be aware that TraMineR works fine only with subsequences of length up to about 4 or 5. We suggest to set maxK=3 or 4 in seqefsub. The number of individuals should not be a problem, nor should the number of different possible events (the alphabet) as long as you restrict the maximal subsequence length you are looking for.
Hope this helps
I am working with a large federal dataset with thousands of observations and thousands of variables. Replicate weights are provided. I am using the "survey" package in R to apply these weights:
els.weighted=svrepdesign(data=els, repweights = ~els$F3F1PNLWT,
combined.weights = TRUE).
I am interested in some categorical descriptive characteristics of a subset of the population, such as family living arrangements. I want to get these sorted out into a contingency table that shows frequency. I would like to sort people based on four variables (none of which are binary, but all of which are numeric) This is what I would like to get:
.
The blank boxes are where the cross-tabulation/frequency counts would show. (I only put in 3 columns beneath F1COMP for brevity's sake, but it has 9 outcomes – indexed 1-9)
My current code: svyby(~F1FCOMP, ~F1RTRCC +BYS33C +F1A10 +byurban, els.weighted, svytotal)
This code does sort the data, but it sorts every single combination, by default. I want them pared down to represent only specific subpopulations of each variable. I tried:
svyby(~F1FCOMP, ~F1RTRCC==2 |F1RTRCC==3 +BYS33C==1 +F1A10==2 | F1A10==3 +byurban==3, els.weighted, svytotal)
But got stopped:
Error: unexpected '==' in "svyby(~F1FCOMP, ~F1RTRCC==2 |F1RTRCC==3 +BYS33C=="
Additionally, my current version of the code tells me how many cases occur for each combination, This is a picture of what my current output looks like. There are hundreds more rows, 1 for each combination, when I keep scrolling down.
This is a picture of what my current output looks like. There are hundreds more rows, 1 for each combination, when I keep scrolling down
.
You can see in that picture that I only get one number for F1FCOMP per row – the number of cases who fit the specified combination – a specific subpopulation. I want to know more about that subpopulation. That is, F1COMP has nine different outcomes (indexed 1-9), and I want to see how many of each subpopulation fits into each of the 9 outcomes of F1COMP.
I've been given this to do by the GENELAND tutorial to give population names to a dataset of populations of 60 individuals :
pop.mbrship1<-rep(c(1,2,3), each=60)
Nevertheless, my dataset comprises 10 populations of irregular sizes to which i would give the names of 1,2,3,4,5,6,7,8,9,10 and the distribution of my individuals (represented by one row each) would be :
1:24,25:39,40:58,59:79,80:103,104:126,127:147,148:171,172:191,192:214
I'd be tempted to use each population number as number of repeats which would make it
pop.mbrship1<-rep[c(1,2,3,4,5,6,7,8,9,10), each=c(24,15,19,21,24,23,21,24,20,23)]
Or try their distribution...
pop.mbrship1<-rep[c(1,2,3,4,5,6,7,8,9,10),
c(1:24,25:39,40:58,59:79,80:103,104:126,127:147,148:171,172:191,192:214)]
In both case, R gives me Error: unexpected '>' in ">"
I'm sure i'm really close to having it work but i've spent a shameful amount of time on this and i'd defenetly need a hand. Thanks a lot!
I'm looking at the geneland tutorial and I see that they have > at the beginning of the lines that you're copying/editing.
You are copying everything including the console pointer > all you need to copy/paste is :
# replicates each element 60 times
pop.mbrship1 <- rep(c(1,2,3),each=60)
# replicates each element, respectively
pop.mbrship2 <- rep(c(1,2,3),times=c(60,40,30))
Your answer is what Henrik said above, without a preceding>.
pop.mbrship1 <- rep(c(1,2,3,4,5,6,7,8,9,10), c(24,15,19,21,24,23,21,24,20,23))
# same as
pop.mbrship1 <- rep(c(...),times=c(...))