I'm using the Sim.DiffProc package in R to simulate a Stratonovich stochastic integral. Using the following code I can simulate 5 paths of the stochastic integral from t=0 to t=5:
fun=expression(w)
strat=st.int(fun, type="str", M=5, lower=0, upper=5)
How can I get the values of the stochastic integral in t=5 given that the st.int() function doesn't give the values in the various t as output?
I'm not sure what you mean by t=5. The $X matrix is a of times series:
> str(strat)
List of 8
$ X : mts [1:1001, 1:5] 0.0187 0.0177 0.0506 0.0357 0.0357 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:5] "X1" "X2" "X3" "X4" ...
..- attr(*, "tsp")= num [1:3] 0 5 200
..- attr(*, "class")= chr [1:3] "mts" "ts" "matrix"
$ fun : symbol w
$ type : chr "str"
$ subdivisions: int 1000
$ M : num 5
$ Dt : num 0.005
$ t0 : num 0
$ T : num 5
- attr(*, "class")= chr "st.int"
If it is the fifth row of the values matrix is what you mean, it would be:
> (strat$X[5 , ])
X1 X2 X3 X4 X5
0.0031517578 0.0161278426 0.0003616453 0.0097594992 0.0012617410
Related
I had a large dataset that contains more than 300,000 rows/observations and 22 variables. I used the CLARA method for the clustering and plotted the results using fviz_cluster. Using the silhouette method, I got 10 as my number of clusters and from there I applied it to my CLARA algorithm.
clara.res <- clara(df, 10, samples = 50,trace = 1,sampsize = 1000, pamLike = TRUE)
str(clara.res)
List of 10
$ sample : chr [1:1000] "100046" "100303" "10052" "100727" ...
$ medoids : num [1:10, 1:22] 0.925 0.125 0.701 0 0 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:10] "193751" "137853" "229261" "257462" ...
.. ..$ : chr [1:22] "COD" "DMW" "HER" "SPR" ...
$ i.med : int [1:10] 104171 42062 143627 174961 300065 13836 192832 207079 185241 228575
$ clustering: Named int [1:302251] 1 1 1 2 3 4 5 3 3 3 ...
..- attr(*, "names")= chr [1:302251] "1" "10" "100" "1000" ...
$ objective : num 0.37
$ clusinfo : num [1:10, 1:4] 71811 40181 46271 10155 31309 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:4] "size" "max_diss" "av_diss" "isolation"
$ diss : 'dissimilarity' num [1:499500] 1.392 2.192 0.937 2.157 1.643 ...
..- attr(*, "Size")= int 1000
..- attr(*, "Metric")= chr "euclidean"
..- attr(*, "Labels")= chr [1:1000] "100046" "100303" "10052" "100727" ...
$ call : language clara(x = df, k = 10, samples = 50, sampsize = 1000, trace = 1, pamLike = TRUE)
$ silinfo :List of 3
..$ widths : num [1:1000, 1:3] 1 1 1 1 1 1 1 1 1 1 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:1000] "83395" "181310" "34452" "42991" ...
.. .. ..$ : chr [1:3] "cluster" "neighbor" "sil_width"
..$ clus.avg.widths: num [1:10] 0.645 0.408 0.487 0.513 0.839 ...
..$ avg.width : num 0.612
$ data : num [1:302251, 1:22] 1 1 1 0.366 0.35 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : chr [1:302251] "1" "10" "100" "1000" ...
.. ..$ : chr [1:22] "COD" "DMW" "HER" "SPR" ...
- attr(*, "class")= chr [1:2] "clara" "partition"
For the plot:
fviz_cluster(clara.res,
palette = c(
"#004c6d",
"#00a1c1",
"#ffc334",
"#78ab63",
"#00ffff",
"#00cfe3",
"#6efa75",
"#cc0089",
"#ff9509",
"#ffb6de"
), # color palette
ellipse.type = "t",geom = "point",show.clust.cent = TRUE,repel = TRUE,pointsize = 0.5,
ggtheme = theme_classic()
)+ xlim(-7, 3) + ylim (-5, 4) + labs(title = "Plot of clusters")
The result:
I reckoned that this cluster plot is based on PCA and have been trying to figure out which variables in my original data were chosen as Dim1 and Dim2 or what these x and y-axis represent. Can somebody help me how to find out these Dim1 and Dim2 and eigenvalues/variance of the whole Dim that exist without running PCA separately?
I saw there are some other functions/packages for PCA such as get_eigenvalue in factoextra and FactomineR, but it seemed that will require me to use the PCA algorithm from the beginning? How can I integrate it directly with my CLARA results?
Also, my Dim1 only consists of 12.3% and Dim2 8.8%, does it mean that these variables are not representative enough or? considering that I would have 22 dimensions in total (from my 22 variables), I think it's alright, no? I am not sure how these percentages of Dim1 and Dim2 affect my cluster results. I was thinking to do the screeplot from my CLARA results but I also can't figure it out.
I'd appreciate any insights.
I need a faster way of doing linear regression than the lm() method. I found that lm.fit() is quite a bit faster but I'm wondering how to use the results. For example using this code:
x = 1:5
y = 5:1
regr = lm.fit(as.matrix(x), y)
str(regr)
Outputs:
List of 8
$ coefficients : Named num 0.636
..- attr(*, "names")= chr "x1"
$ residuals : num [1:5] 4.364 2.727 1.091 -0.545 -2.182
$ effects : Named num [1:5] -4.719 1.69 -0.465 -2.619 -4.774
..- attr(*, "names")= chr [1:5] "x1" "" "" "" ...
$ rank : int 1
$ fitted.values: num [1:5] 0.636 1.273 1.909 2.545 3.182
$ assign : NULL
$ qr :List of 5
..$ qr : num [1:5, 1] -7.416 0.27 0.405 0.539 0.674
..$ qraux: num 1.13
..$ pivot: int 1
..$ tol : num 1e-07
..$ rank : int 1
..- attr(*, "class")= chr "qr"
$ df.residual : int 4
I'm expecting intercept = 6 and slope = -1 but the result above doesn't contain anyhing near that. Also, does lm.fit() output r squared?
lm.fit allows to do things much more manually, so, as #MrFlick commented, we must include the intercept manually as well using cbind(1, x) as the design matrix. The R^2 is not provided but we may easily compute it:
x <- 1:5
y <- 5:1 + rnorm(5)
regr <- lm.fit(cbind(1, x), y)
regr$coef
# x
# 5.2044349 -0.5535963
1 - var(regr$residuals) / var(y) # R^2
# [1] 0.3557227
1 - var(regr$residuals) / var(y) * (length(y) - 1) / regr$df.residual # Adj. R^2
# [1] 0.1409636
I have been following an online example for R Kohonen self-organising maps (SOM) which suggested that the data should be centred and scaled before computing the SOM.
However, I've noticed the object created seems to have attributes for centre and scale, in which case am I really applying a redundant step by centring and scaling first? Example script below
# Load package
require(kohonen)
# Set data
data(iris)
# Scale and centre
dt <- scale(iris[, 1:4],center=TRUE)
# Prepare SOM
set.seed(590507)
som1 <- som(dt,
somgrid(6,6, "hexagonal"),
rlen=500,
keep.data=TRUE)
str(som1)
The output from the last line of the script is:
List of 13
$ data :List of 1
..$ : num [1:150, 1:4] -0.898 -1.139 -1.381 -1.501 -1.018 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : NULL
.. .. ..$ : chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length"
"Petal.Width"
.. ..- attr(*, "scaled:center")= Named num [1:4] 5.84 3.06 3.76 1.2
.. .. ..- attr(*, "names")= chr [1:4] "Sepal.Length" "Sepal.Width"
"Petal.Length" "Petal.Width"
.. ..- attr(*, "scaled:scale")= Named num [1:4] 0.828 0.436 1.765 0.762
.. .. ..- attr(*, "names")= chr [1:4] "Sepal.Length" "Sepal.Width"
"Petal.Length" "Petal.Width"
$ unit.classif : num [1:150] 3 5 5 5 4 2 4 4 6 5 ...
$ distances : num [1:150] 0.0426 0.0663 0.0768 0.0744 0.1346 ...
$ grid :List of 6
..$ pts : num [1:36, 1:2] 1.5 2.5 3.5 4.5 5.5 6.5 1 2 3 4 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : NULL
.. .. ..$ : chr [1:2] "x" "y"
..$ xdim : num 6
..$ ydim : num 6
..$ topo : chr "hexagonal"
..$ neighbourhood.fct: Factor w/ 2 levels "bubble","gaussian": 1
..$ toroidal : logi FALSE
..- attr(*, "class")= chr "somgrid"
$ codes :List of 1
..$ : num [1:36, 1:4] -0.376 -0.683 -0.734 -1.158 -1.231 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:36] "V1" "V2" "V3" "V4" ...
.. .. ..$ : chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length"
"Petal.Width"
$ changes : num [1:500, 1] 0.0445 0.0413 0.0347 0.0373 0.0337 ...
$ alpha : num [1:2] 0.05 0.01
$ radius : Named num [1:2] 3.61 0
..- attr(*, "names")= chr [1:2] "66.66667%" ""
$ user.weights : num 1
$ distance.weights: num 1
$ whatmap : int 1
$ maxNA.fraction : int 0
$ dist.fcts : chr "sumofsquares"
- attr(*, "class")= chr "kohonen"
Note notice that in lines 7 and 10 of the output there are references to centre and scale. I would appreciate an explanation as to the process here.
Your step with scaling is not redundant because in source code there are no scaling, and attributes, that you see in 7 and 10 are attributes from train dataset.
To check this, just run and compare results of this chunk of code:
# Load package
require(kohonen)
# Set data
data(iris)
# Scale and centre
dt <- scale(iris[, 1:4],center=TRUE)
#compare train datasets
str(dt)
str(as.matrix(iris[, 1:4]))
# Prepare SOM
set.seed(590507)
som1 <- kohonen::som(dt,
kohonen::somgrid(6,6, "hexagonal"),
rlen=500,
keep.data=TRUE)
#without scaling
som2 <- kohonen::som(as.matrix(iris[, 1:4]),
kohonen::somgrid(6,6, "hexagonal"),
rlen=500,
keep.data=TRUE)
#compare results of som function
str(som1)
str(som2)
I would like to extract the p-values from the Anderson-Darling test (ad.test from package kSamples). The test result is a list of 12 containing a 2x3 matrix. The p value is part of the 2x3 matrix and is present in element 7.
When using the following code:
lapply(AD_result, "[[", 7)
I get the following subset of AD test results (first 2 of a total of 50 shown)
[[1]]
AD T.AD asympt. P-value
version 1: 1.72 0.94536 0.13169
version 2: 1.51 0.66740 0.17461
[[2]]
AD T.AD asympt. P-value
version 1: 12.299 14.624 6.9248e-07
version 2: 11.900 14.144 1.1146e-06
My question is how to extract only the p-value (e.g. from version 1) and put these 50 results into a vector
The output from str(AD_result) is:
List of 55
$ :List of 12
..$ test.name : chr "Anderson-Darling"
..$ k : int 2
..$ ns : int [1:2] 103 2905
..$ N : int 3008
..$ n.ties : int 2873
..$ sig : num 0.762
..$ ad : num [1:2, 1:3] 1.72 1.51 0.945 0.667 0.132 ...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:2] "version 1:" "version 2:"
.. .. ..$ : chr [1:3] "AD" "T.AD" " asympt. P-value"
..$ warning : logi FALSE
..$ null.dist1: NULL
..$ null.dist2: NULL
..$ method : chr "asymptotic"
..$ Nsim : num 1
..- attr(*, "class")= chr "kSamples"
You could try:
unlist(lapply(AD_result, function(x) x$ad[,3]))
I'm working from caracal's great example conducting a factor analysis on dichotomous data and I'm now struggling to extract the factors from the object produced by the psych package's fa.poly function.
Can anyone help me extract the factors from the fa.poly object (and look at the correlation)?
Please see caracal's example for the working example.
In this example you create an object with:
faPCdirect <- fa.poly(XdiNum, nfactors=2, rotate="varimax") # polychoric FA
so somewhere in faPCdirect there is what you want. I recommend using str() to inspect the structure of faPCdirect
> str(faPCdirect)
List of 5
$ fa :List of 34
..$ residual : num [1:6, 1:6] 4.79e-01 7.78e-02 -2.97e-0...
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:6] "X1" "X2" "X3" "X4" ...
.. .. ..$ : chr [1:6] "X1" "X2" "X3" "X4" ...
..$ dof : num 4
..$ fit
...skip stuff....
..$ BIC : num 4.11
..$ r.scores : num [1:2, 1:2] 1 0.0508 0.0508 1
.. ..- attr(*, "dimnames")=List of 2
.. .. ..$ : chr [1:2] "MR2" "MR1"
.. .. ..$ : chr [1:2] "MR2" "MR1"
..$ R2 : Named num [1:2] 0.709 0.989
.. ..- attr(*, "names")= chr [1:2] "MR2" "MR1"
..$ valid : num [1:2] 0.819 0.987
..$ score.cor : num [1:2, 1:2] 1 0.212 0.212 1
So this says that this object is a list of five, with the first element called fa and that contains an element called score.cor that is a 2x2 matrix. I think what you want is the off diagonal.
> faPCdirect$fa$score.cor
[,1] [,2]
[1,] 1.0000000 0.2117457
[2,] 0.2117457 1.0000000