I'm trying to do the compositional analysis of habitat use with the compana() function in the adehabitatHS package (I use adehabitat because I can't install adehabitatHS).
Compana() needs two matrices: one of habitat use and one of avaiable habitat.
When I try to run the function it doesn't work (it never stops), so I have to abort the RStudio session.
I read that one problem could be the 0-values in some habitat types for some animals in the 'avaiable' matrix, whereas other animals have positive values for the same habitat. As done by other people, I replaced 0-values with small values (0,001), ran compana and it worked BUT the lambda values returned me NaN.
The problem is similar to the one found here
adehabitatHS compana test returns lambda = NaN?
They said they resolved using as 'used' habitat matrix the counts (integers) and not the proportions.
I tried also this approach, but never changed (it freezes when there are 0-values in the available matrix, or returns NaN value for Lambda if I replace 0- values wit small values).
I checked all matrices and they are ok, so I'm getting crazy.
I have 6 animals and 21 habitat types.
Can you resolve this BIG problem?
PARTIALLY SOLVED: Asking to some researchers, they told me that the number of habitats shouldn't be higher than the number of animals.
In fact I merged some habitats in order to have six animals per six habitats and now the function works when I replace 0-values in the 'avaiable' matrix with small values (e.d. 0.001).
Unfortunately this is not what I wanted, because I needed to find values (rankings, Log-ratios, etc..) for each habitat type (originally they were 21).
Related
I'm actually working on tuna tag-recapture data. I want to balance my sampling between two groups of individuals, the ones that where tagged in the reference area (Treated group) and the ones that where tagged outside this area (Control group). To do this, I used the MatchIt package.
I have 3 covariates: length (by 5 cm bins), month of tagging (January to December) and structure on which the tuna was tagged.
So there is the model: treatment ~ length + month + structure
This last variable, is a categorical variable with 5 levels coded as A to E. The level A is almost only represented in the Treated group (6000 individuals with structure = A, vs on 300 individuals with structure = A in control group).
I first used the nearest neighbour method, but the improvement in balance was not satisfying. So I ran exact and Coarsened Exact Matching methods.
I though that Exact methods should match pairs with the same values for each covariates. But in the output matched data, there are still more than 3000 individuals with structure = A in the treated group.
Do you guys have one explanation ? I red a lot but I didn't find answers.
Thanks
Exact and coarsened exact matching do not perform 1:1 matching. They find all members in the control group that exactly match each member in the treated group. Subclasses are formed based on each combination of the predictor values, and any subclass that has both treated and control units is retained, and others dropped. There is no pairing that takes place. Your results indicate that you have many control units that have identical (or near-identical in the case of CEM) values of the covariates as some treated units.
I am using R for the first time. I am trying to use the Chao1 function to estimate the diversity of my dataset. I have 20 columns, one for each species, and 8 rows (nine if you include the header), one for each plot. Each cell has a number, which is the number of individuals of that species found in that plot. For example, in my Excel file, cell A2 has the value "8", which means that 8 individuals of Species1 were found in the first plot.
I have downloaded the Fossil and Vegan packages, where I believe the Chao1 function is located. They are active in my library. I have imported my dataset as "speciesabund". I am now trying to run Chao1. According to the description (https://artax.karlin.mff.cuni.cz/r-help/library/fossil/html/chao1.html) I'm supposed to type
chao1(x, taxa.row = TRUE)
I assumed "x" was meant to represent my dataset, so I tried
chao1(speciesabund, taxa.row = TRUE)
instead. It did not work and returned me "Error: Unsupported use of matrix or array for column indexing." I assume this means that I need to do something more to my data before trying to use the Chao function, is that correct? If so, how do I do this?
Thank you so much for your help! I am using this for the first time, so I'm sorry if my question is dumb.
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 am looking for the best way to compare 2 or more images.
The images I have are now in matrix format, so basically I am comparing matrices.
They aren't square (but this isn't a problem).
This is an example of what I have with only two matrices:
#Original data
M1<-cbind(c(0,0,20,40,50,35),c(0,0,5,20,90,80),c(0,0,10,25,85,0),c(58,70,20,50,0,5))
#Data to be compared with M1
M2<-cbind(c(0,5,25,25,60,15),c(0,30,15,10,116,67),c(0,2,9,20,90,1),c(69,50,22,30,0,2))
I can check for the differences and the correlation, but I also want to be able to say for example, if:
high values in M2 occur in the same positions that M1
high values in M2 occur close to the positions in M1
high values in M2 occur far away
Same thing for low values.
By high values I mean maximum values, for example if the max value in M1 is in position (M1_maxvalue(x,y)), than I M2 max value should be a similar value observed in M1 as well as in the same or close position M1_maxvalue(x,y).
I can extract the positions, the variation of the positions of the maximum values, however I am looking for existent methods where I can base my comparisons.
What type of calculations can I use to do such type of analysis?
I can use both image processing packages as well as matrices algorithms.
Sounds like a job better handled with ImageJ or SAODS9 at http://hea-www.harvard.edu/RD/ds9/ .
IIRC those apps have built-in tools for spot and blob-finding, which may save you a lot of time and pain.
I want to run a nested logistic regression in R, but the examples I found online didn't help much. I read over an example from this website (Step by step procedure on how to run nested logistic regression in R) which is similar to my problem, but I found that it seems not resolved in the end (The questioner reported errors and I didn't see more answers).
So I have 9 predictors (continuous scores), and 1 categorical dependent variable (DV). The DV is called "effect", and it can be divided into 2 general categories: "negative (0)" and "positive (1)". I know how to run a simple binary logit regression (using the general grouping way, i.e., negative (0) and positive (1)), but this is not enough. "positive" can be further grouped into two types: "physical (1)" and "mental (2)". So I want to run a nested model which includes these 3 categories (negative (0), physical (1), and mental (2)), and reflects the nature that "physical" and "mental" are nested in "positive". Maybe R can compare these two models (general vs. detailed) together? So I created two new columns, one is called "effect general", in which the individual scores are "negative (0)" and "positive (1)"; the other is called "effect detailed", which contains 3 values - negative (0), physical (1), and mental (2). I ran a simple binary logit regression only using "effect general", but I don't know how to run a nested logit model for "effect detailed".
From the example I searched and other materials, the R package "mlogit" seems right, but I'm stuck with how to make it work for my data. I don't quite understand the examples in R-help, and this part in the example from this website I mentioned earlier (...shape='long', alt.var='town.list', nests=list(town.list)...) makes me very confused: I can see that my data shape should be 'wide', but I have no idea what "alt.var" and "nests" are...
I also looked at page 19 of the mlogit manual for examples of nested logit model calls. But I still cannot decide what I need in terms of options. (http://cran.r-project.org/web/packages/mlogit/mlogit.pdf)
Could someone provide me with detailed steps and notes on how to do it? I'm sure this example (if well discussed and resolved) is also going to help me and others a lot!
Thanks for your help!!!
I can help you with understanding the mlogit structure. When using the mlogit.data() command, specify choice = yourchoicevariable (and id.var = respondentid if you have a panel dataset, i.e. you have multiple responses from the same individual), along with the shape='wide' argument. The new data.frame created will be in long format, with a line for each choice situation, negative, physical, mental. So you will have 3 rows for which you only had one in the wide data format. Whatever your MN choice var is, it will now be a column of logical values, with TRUE for the row that the respondent chose. The row names will now have be in the format of observation#.level(choice variable) So in your case, if the first row of your dataset the person had a response of negative, you would see:
row.name | choice
1.negative | TRUE
1.physical | FALSE
1.mental | FALSE
Also not that the actual factor level for each choice is stored in an index called alt of the mlogit.data.frame which you can see by index(your.data.frame) and the observation number (i.e. the row number from your wide format data.frame) is stored in chid. Which is in essence what the row.name is telling you, i.e. chid.alt. Also note you DO NOT have to specify alt.var if your data is in wide format, only long format. The mlogit.data function does that for you as I have just described. Essentially, it takes unique(choice) when you specify your choice variable and creates the alt.var for you, so it is redundant if your data is in wide format.
You then specify the nests by adding to the mlogit() command a named list of the nests like this, assuming your factor levels are just '0','1','2':
mlogit(..., nests = c(negative = c('0'), positive = c('1','2')
or if the factor levels were 'negative','physical','mental' it would be the like this:
mlogit(..., nests = c(negative = c('negative'), positive = c('physical','mental')
Also note a nest of one still MUST be specified with a c() argument per the package documentation. The resulting model will then have the iv estimate between nests if you specify the un.nest.el=T argument, or nest specific estimates if un.nest.el=F
You may find Kenneth Train's Examples useful