I have a vector that has a kurtosis of 2.95 (which is pretty high, Leptokurtic). Following is a sample of that data:
x = c(6.819, 8.948, 0, 67.556, -40.785, -18.951, -29.151, 1.008,
0, 18.034, -6.631, 6.294, 0.643, -28.921, 0, -2.133, -44.348,
-87.488, 7.063, 0, -74.428, -16.361, 50.963, -32.431, -82.233,
-26.953, -48.475, 64.043, 0, 1.576, -2.728, -5.9, -63.059, -1.061,
-15.018, -58.119, -32.092, 5.329, -19.968, 38.822, 66.897, 0,
-2.579, 82.696, 42.745, 79.677, 2.522, -11.475, 1.019, 2.719,
-3.634, -7.975, 0, 1.873, 21.732, -10.217, -24.002, -76.049,
35.045, 27.22, -71.366, 16.293, -48.762, 65.481, 66.615, -19.616,
6.016, 59.722, 88.235, 10.1, 0, -4.598, 5.446, 56.909, 0, -24.827,
0, 6.487, 0, 63.315, 28.397, 9.433, 19.085, 0, 6.591, -22.643,
32.235, -12.535, -1.787, 56.157, 68.819, 0, -21.936, 38.695,
-79.006, 24.888, -5.187, 10.368, -68.191, 0, -22.171, -78.783,
-14.119, 54.084, -13.597, 26.669, 0, -18.402, 80.309, -12.652,
1.801, -69.946, -87.67, -19.586, 38.085, -21.031, -36.957, 1.357,
0.17, 47.407, -59.598, 66.125, 10.97, 6.33, -38.837, 1.868, 38.169,
-46.662, -32.255, 25.816, 14.432, -18.57, -0.456, -0.638, 31.07,
72.794, 52.957, 13.858, -18.885, 0, -13.488, 11.689, 1.618, 19.373,
-57.526, 0, -0.655, 36.308, 50.231, 0.048, -80.157, 0, -64.805,
-70.864, 0.813, 52.143, -4.989, 42.166, 7.397, 87.437, -17.897,
-0.877, 68.363, 47.315, -2.181, 2.699, 36.278, 0, -2.924, 71.56,
74.406, -46.071, 56.158, 1.44, 0, 0, 0, -3.233, 37.084, -85.189,
0, -16.137, -84.499, -12.67, -14.117, 0, 23.757, -58.299, -34.956,
0.402, 0, -67.585, -14.314, -73.426, 23.158, 1.782, 0, 4.399,
18.871, -6.929)
Is there a way to normalize this data?
Since this data range between -90 to 90, the normalized data should be in a similar range and should not change vastly, i.e. the range should not be changed to -1 to 1 or -20 to 20 etc...
I have tried using atan(X), 1/x, log(x), and many other transformational techniques but they all tend to increase the skewness. Is there a way to normalize this data without skewing it?
I am sure there must be an easy solution to this.
It may not be what you want but you can almost always perfectly normalize a distribution (if there are no ties) using a normal scores transformation:
xq <- qnorm(rank(x)/(length(x)+1), mean=mean(x), sd=sd(x))
plot(sort(x),sort(xq))
hist(xq)
qqnorm(xq)
The new range is (-99.2, 99.6) (the old range was +/- 88).
If you need to change the range you could do it as follows:
newmin + (newmax-newmin)*scale(xq, center=min(qx), scale=diff(range(xq)))
but as suggested in the comments this may not actually be the right approach to solve your broader problem.
Related
I trained a BERT based encoder decoder model (EncoderDecoderModel) named ed_model with HuggingFace's transformers module.
I used the BertTokenizer named as input_tokenizer
I tokenized the input with:
txt = "Some wonderful sentence to encode"
inputs = input_tokenizer(txt, return_tensors="pt").to(device)
print(inputs)
The output clearly shows that a input_ids is the return dict
{'input_ids': tensor([[ 101, 5660, 7975, 2127, 2053, 2936, 5061, 102]], device='cuda:0'), 'token_type_ids': tensor([[0, 0, 0, 0, 0, 0, 0, 0]], device='cuda:0'), 'attention_mask': tensor([[1, 1, 1, 1, 1, 1, 1, 1]], device='cuda:0')}
But when I try to predict, I get this error:
ed_model.forward(**inputs)
ValueError: You have to specify either input_ids or inputs_embeds
Any ideas ?
Well, apparently this is a known issue, for example: This issue of T5
The problem is that there's probably a renaming procedure in the code, since we use a encoder-decoder architecture we have 2 types of input ids.
The solution is to explicitly specify the type of input id
ed_model.forward(decoder_input_ids=inputs['input_ids'],**inputs)
I wish it was documented somewhere, but now you know :-)
I have a for-loop it looks like that:
for (ID in rownames(countDF)) {
avector <- as.vector(as.numeric(countDF2[rownames(countDF2)==ID,]))
nbfit <- fitdistr(avector,'negative binomial')
}
So I want to calculate the fitdistr function for each of IDs. But the problem is that for some of the IDs the function doesn't work and throws an error. Here it is:
Error in stats::optim(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, :
non-finite finite-difference value [2]
I want to skip these IDs somehow and continue with the others.
I've found a function try, but I don't understand how is it working.
I've tried it like this:
nbfir <- try(fitdistr(avector,'negative binomial'))
But the loop still breaks down with the error.
What should I do to fix it ?
You could use tryCatch and do nothing on catching an error.
for (ID in rownames(countDF)) {
avector <- as.vector(as.numeric(countDF2[rownames(countDF2)==ID,]))
tryCatch(
nbfit <- fitdistr(avector,'negative binomial'),
error = function(e) {})
}
I created the following function to determine the lag of two variables.
However, this function takes only two parameters, and I would like to run it over my whole dataset:
datSel <- structure(list(stat.resProp.Dwell.4 = c(0.000887705, 0.007954085,
-0.025859667, 0.024097552, 0.114052787, 0.023329207, 0.042143181,
-0.092587287, -0.004050228, -0.001624696, 0.020121403, -0.100502922,
0.057354185, 0.025463388, 0.037409854, 0.001561281, -0.028482938,
-0.004827041, 0.014411779, -0.029034298, 0.021053409, -0.067963182,
0.032070259, -0.038091783, 0.039751534, 0.027802281, -0.027802281,
-0.013355791, 0.009201236, -0.073403679, 0.021277398, -0.033901552,
0.012624153, -0.065733979, 0.032017801, -0.072042665, 0.041936911,
0.002861232, 0.017933468, -0.01698154, 0.006638242, -0.08375153,
-0.007220248, 0.0255507, 0.019980685, 0.013752673, 0.026000502,
-0.021134312, -0.019608471, 0.0166916, -0.021654389, 0.066402455,
0.024828862, -0.083302632, 0.042518482, -0.052439198, 0.037186281,
-0.056311172, -0.012270093), stat.lohn = c(0, -0.007558004, -0.015289567,
0, 0, -0.009609384, -0.019500305, 0, 0, -0.012458015, -0.025391532,
-0.000983501, 0, -0.00165265, -0.003313516, 0.000204576, 0, -0.004898564,
-0.009869709, 0, 0, -0.010574012, -0.021489482, 0, 0, -0.011534651,
-0.023476287, 0, 0, -0.00814845, -0.016498838, 0, 0, -0.0099856,
-0.020275409, -0.002818337, 0, -0.007212389, -0.014582736, 0,
0, -0.004121565, -0.008294445, 0, 0, -0.010766386, -0.021886884,
0, 0, -0.010179741, -0.02067574, 0, 0, -0.011797067, -0.024020039,
-0.002017983, -0.007343864, -0.007398196, -0.014962644), stat.resProp.Dwell.1 = c(0.012777325,
-0.002991775, -0.057819571, -0.00796817, -0.019386714, 0, 0.009740337,
0.005638356, -0.035148694, 0, 0.027084134, -0.160377856, 0.101169235,
-0.043007944, 0.043007944, -0.002580647, -0.015625318, 0.023347364,
0.007662873, -0.09607383, -0.024575906, 0.056733018, -0.000904568,
-0.058703392, 0.011450507, 0.007561473, 0.037879817, -0.032246,
0.042169401, -0.001796946, -0.024580209, -0.148788737, 0.082097362,
-0.000985707, -0.00098668, 0.003940892, -0.049380309, 0.005151995,
0.027371197, -0.025317808, 0.019299736, -0.047382704, -0.010604553,
0.082827084, -0.04516573, 0.003075348, 0.007139245, 0.022111454,
-0.004982571, -0.038701368, 0.018519048, -0.049096021, 0.061254226,
-0.020346582, 0.023363175, -0.00402415, -0.014213437, 0.023245109,
0.027587957), stat.carReg = c(0.022775414, 0.008073857, 0.002624717,
0.169431097, -0.144595366, 0.066716837, -0.086971929, 0.037928208,
0.071752161, -0.046824102, 0.106085873, 0.049965928, -0.057984255,
-0.091650262, 0.090732857, -0.082282389, 0.053376121, -0.044203971,
-0.022855425, 0.025856271, 0.000136493, 0.05579193, -0.293966656,
0.013645739, 0.059732986, 0.187020956, -0.145234848, 0.11041385,
-0.126539687, -0.000949877, 0.031473389, 0.020267816, -0.02180532,
-0.07175183, 0.147500145, -0.040559138, 0.008394819, 0.049045337,
-0.043050615, 0.094358754, -0.058408438, -0.005018402, -0.061717889,
0.100150837, -0.071100417, -0.084393865, 0.002854733, 0.002141389,
-0.026538398, 0.013480513, -0.046002189, -0.030495611, 0.052899746,
0.012842017, 0.064086498, 0.020757573, -0.043441298, -0.009563043,
0.048033848)), .Names = c("stat.resProp.Dwell.4", "stat.lohn",
"stat.resProp.Dwell.1", "stat.carReg"), row.names = c(NA, -59L
), class = "data.frame")
The function and my function call is:
select.lags<-function(x,y,max.lag=8) {
y<-as.numeric(y)
y.lag<-embed(y,max.lag+1)[,-1,drop=FALSE]
x.lag<-embed(x,max.lag+1)[,-1,drop=FALSE]
t<-tail(seq_along(y),nrow(y.lag))
ms=lapply(1:max.lag,function(i) lm(y[t]~y.lag[,1:i]+x.lag[,1:i]))
pvals<-mapply(function(i) anova(ms[[i]],ms[[i-1]])[2,"Pr(>F)"],max.lag:2)
ind<-which(pvals<0.05)[1]
ftest<-ifelse(is.na(ind),1,max.lag-ind+1)
aic<-as.numeric(lapply(ms,AIC))
bic<-as.numeric(lapply(ms,BIC))
structure(list(ic=cbind(aic=aic,bic=bic),pvals=pvals,
selection=list(aic=which.min(aic),bic=which.min(bic),ftest=ftest)))
}
for (i in length(datSel) ) {
for (y in length(datSel) ) {
d1<-ts(datSel[i])
d2<-ts(datSel[y])
lag <- select.lags(d1,d2,5)
}
}
As output of lag I get:
> lag
$ic
aic bic
[1,] -115.3623 -109.56679
[2,] -114.3370 -106.60972
[3,] -116.2026 -106.54350
[4,] -114.7030 -103.11210
[5,] -112.7153 -99.19253
[6,] -110.8018 -95.34721
[7,] -110.0812 -92.69477
[8,] -110.1427 -90.82446
$pvals
[1] 0.1952302 0.3017934 0.7858944 0.9176337 0.5040079 0.0604511 0.3406657
$selection
$selection$aic
[1] 3
$selection$bic
[1] 1
$selection$ftest
[1] 1
As you can see I get only 8 results back, however, my data.frame has 20 variables.
Any recommendation what I am doing wrong?
I appreciate your replies!
If you want to e.g. store the result of the AIC criterion:
lag.aic.store = matrix(NA, 4, 4)
for (i in 1:length(datSel) ) {
for (y in 1:length(datSel) ) {
d1<-ts(datSel[,i])
d2<-ts(datSel[,y])
lag <- select.lags(d1,d2,5)
lag.store.aic[i,y] = lag$selection$aic
}
}
You get 8 values in $ic because max.lag is 8, it has nothing to do with your number of variables.
Please also note that i added commas when indexing by variable for clarity and that you have to loop through 1:length(datSel) as otherwise you will only catch the last variable.
When I enter the following commands directly into the R console
library("xts")
mySeries <- xts(c(1.0, 2.0, 3.0, 5.0, 6.0), order.by=c(ISOdatetime(2001, 1, 1, 0, 0, 0), ISOdatetime(2001, 1, 2, 0, 0, 0), ISOdatetime(2001, 1, 3, 0, 0, 0), ISOdatetime(2001, 1, 4, 0, 0, 0), ISOdatetime(2001, 1, 5, 0, 0, 0)))
resultingSeries <- to.monthly(mySeries)
resultingSeries
I will get an output like this
mySeries.Open mySeries.High mySeries.Low mySeries.Close
Jan 2001 1 6 1 6
When I look into the attributes, I see the following output
attributes(resultingSeries)
$dim
[1] 1 4
$dimnames
$dimnames[[1]]
NULL
$dimnames[[2]]
[1] "mySeries.Open" "mySeries.High" "mySeries.Low" "mySeries.Close"
$index
[1] 978307200
attr(,"tclass")
[1] "yearmon"
$tclass
[1] "POSIXct" "POSIXt"
$tzone
[1] ""
$class
[1] "xts" "zoo"
$.indexCLASS
[1] "yearmon"
This is the same I get in Java. I'm wondering where the magic happens so that I see the nice output I get in R. I have no access to the event loop, since I'm using JRI like this (since, it's the recommended way and simplifies error handling):
REngine engine = REngine.engineForClass("org.rosuda.REngine.JRI.JRIEngine");
REXP result = engine.parseAndEval(...)
/edit
In Java I execute each command from above as follows:
REXP result = engine.parseAndEval("resultingSeries") // or any other command
What I get is
org.rosuda.REngine.REXPDouble#4ac66122+[12]
The payload being doubles: 1, 6, 1, 6
The attributes are the same as specified above.
Now R does some magic to display the output above. Is there a way I can get the same output without having to create it manually by myself? Where's the implementation stored, that R gets the above mentioned output?
Here is a piece of code that will work, here i extracted the first element of the field mySeries.Open from the object resultingSeries (which i converted to a data frame) which is equal to 1, notice that you can't pass all of the resultingSeries object strait into Java, you will need to break it down.
package stackoverflow;
import org.rosuda.JRI.REXP;
import org.rosuda.JRI.Rengine;
/**
*
* #author yschellekens
*/
public class StackOverflow {
public static void main(String[] args) throws Exception {
String[] Rargs = {"--vanilla"};
Rengine rengine = new Rengine( Rargs, false, null);
rengine.eval("library('xts')");
rengine.eval("mySeries <- xts(c(1.0, 2.0, 3.0, 5.0, 6.0), order.by=c(ISOdatetime(2001, 1, 1, 0, 0, 0), ISOdatetime(2001, 1, 2, 0, 0, 0), ISOdatetime(2001, 1, 3, 0, 0, 0), ISOdatetime(2001, 1, 4, 0, 0, 0), ISOdatetime(2001, 1, 5, 0, 0, 0)))");
rengine.eval("resultingSeries <- to.monthly(mySeries)");
rengine.eval("resultingSeries<-as.data.frame(resultingSeries)");
REXP result= rengine.eval("resultingSeries$mySeries.Open");
System.out.println("Greeting from R: "+result.asDouble());
}
}
And the Java output:
run:
Greeting from R: 1.0
I figured out the following workaround. The solution is far from perfect.
R offers a command to save its console output as characters vector.
capture.output( {command} )
We can access the output using
REXPString s = rengine.parseAndEval("capture.output( to.monthly(mySeries))")
String[] output = result.asStrings()
The variable output will contain all output lines
[0] mySeries.Open mySeries.High mySeries.Low mySeries.Close
[1]Jan 2001 1 6 1 6
Alternatively you coud use JRIEngine and attack yourself to the event loop, which it did not want in my case (due to the more complicated error handling).
This question already has answers here:
Put a fixed title in an interactive 3D plot using rgl package, R
(4 answers)
Closed 9 years ago.
Editing to provide more code and a runnable example.
If I open a 3D plot using the following code:
library(rgl)
GainTargets = seq(from=1, to=12, by=1)
PredBarTargets = seq(from=3, to=16, by=1)
data=structure(c(1176.18, 1379.34, 1280.67, 1149.02, 1034.01, 1028.74,
944.06, 807.54, 715.39, 691.08, 679.14, 644.02, 642.98, 577.61,
1004.22, 1238.13, 1216.16, 1104.92, 1032.46, 1038.11, 919.98,
855.47, 706.54, 668.51, 665.67, 689.57, 686, 623.82, 729.61,
1127.32, 1137.55, 1084.25, 955.15, 1005.86, 955.6, 881.66, 839.93,
767.45, 731.49, 727.19, 696.45, 634.52, 377.81, 917.08, 1087.57,
1036.17, 920.88, 993.75, 964.35, 848.94, 874.95, 780.58, 725.08,
695.12, 709.32, 663.93, 227, 787.53, 949.31, 972.59, 950.3, 946.22,
876.33, 881.49, 827.44, 767.66, 755.26, 731.71, 744.37, 692.86,
32.65, 404.91, 514.82, 652.8, 697.77, 859.85, 808.12, 794.15,
746.61, 719.72, 709.59, 662.36, 695.06, 687.76, 0, 262.09, 347.17,
442.35, 453.76, 684.06, 638.13, 664.69, 721.68, 688.32, 653.87,
665.21, 680.52, 685.13, 0, 38.77, 264.4, 432.8, 408.33, 457.38,
474.23, 453.76, 570.74, 591.96, 593.7, 636.75, 585.43, 577.72,
0, 0, 92.65, 255.08, 388.66, 432.73, 418.19, 425.02, 436.3, 452.17,
489.17, 499.27, 500.93, 495.26, 0, 0, 0, 90.28, 309.44, 324.99,
357.05, 376.11, 412.2, 337.4, 338.54, 370.11, 389.95, 501.99,
0, 0, 0, 0, 106.55, 190.84, 216.41, 322.11, 337.05, 313.99, 325.03,
356.98, 381.47, 373.77, 0, 0, 0, 0, 0, 198.84, 120.76, 204.33,
146.22, 289.93, 284.82, 337.22, 347.41, 337.29), .Dim = c(14L,
12L))
Points<-as.matrix(data)
open3d()
persp3d(x=PredBarTargets, y=GainTargets, z=Points,
col="green3", main="Total Points")
then I do get a title on the plot, but unfortunately the title moves around with mouse actions.
Does anyone know how to get the title text "Total Points" to be fixed at the top of the window such that it doesn't move? Also, if there's a way to associate the axis text with the axis that that has the numbers on it then that might be more readable.
Thanks
This cannot be done in rgl at the moment. You can only place text in 3d coordinates and have it rotate with the plot. Why not contact the maintainer of rgl and put this on a feature request list.
Note this Q is a duplicate of another question on SO.