library(haven)
library(survey)
library(dplyr)
nhanesDemo <- read_xpt(url("https://wwwn.cdc.gov/Nchs/Nhanes/2015-2016/DEMO_I.XPT"))
# Rename variables into something more readable
nhanesDemo$fpl <- nhanesDemo$INDFMPIR
nhanesDemo$age <- nhanesDemo$RIDAGEYR
nhanesDemo$gender <- nhanesDemo$RIAGENDR
nhanesDemo$persWeight <- nhanesDemo$WTINT2YR
nhanesDemo$psu <- nhanesDemo$SDMVPSU
nhanesDemo$strata <- nhanesDemo$SDMVSTRA
# Select the necessary columns
nhanesAnalysis <- nhanesDemo %>%
select(fpl, age, gender, persWeight, psu, strata)
# Set up the design
nhanesDesign <- svydesign(id = ~psu,
strata = ~strata,
weights = ~persWeight,
nest = TRUE,
data = nhanesAnalysis)
# Select those between the agest of 18 and 79
ageDesign <- subset(nhanesDesign, age > 17 & age < 80 & !is.na(fpl))
quantile_results <- svyquantile(~fpl, ageDesign, quantiles=c(0.1, 0.5, 0.9))
print(quantile_results)
The default rounding of svyquantile appears to be two digits past the decimal place. How can I change this? I couldn't find anything in the documentation.
svyquantile does no rounding.
In this example, the two digit precision is the precision of the data: fpl is given to only two decimal places and by default svyquantile returns the left quantile, which is always one of the observed values. In fact, most of the distinct values of fpl occur multiple times: there are 20 observations equal to the 10th percentile, 29 equal to the median, and 1220 equal to the 90th percentile, so the quantile will be equal to one of the observed values in this example no matter what you specify for the qrule argument.
If you make fpl noisier, you'll get more digits
> ageDesign<-update(ageDesign, fpl_noisy=fpl+runif(nrow(ageDesign),0,0.005))
> svyquantile(~fpl_noisy, ageDesign, quantiles=c(0.1, 0.5, 0.9))
$fpl_noisy
quantile ci.2.5 ci.97.5 se
0.1 0.8027744 0.7128426 0.8841695 0.04019022
0.5 2.9711470 2.5921659 3.3747105 0.18357099
0.9 5.0031355 5.0027002 5.0035307 0.00019482
attr(,"hasci")
[1] TRUE
attr(,"class")
[1] "newsvyquantile"
Related
I have a data set that is split into 3 profiles
Profile 1 = 0.478 (95% confidence interval: 0.4, 0.56)
Profile 2 = 0.415 (95% confidence interval: 0.34, 0.49)
Profile 3 = 0.107 (95% confidence interval: 0.06, 0.15)
Profile 1 + Profile 2 + Profile 3 = 1
I want to create a stochastic model that selects a value for each profile from each proportion's confidence interval. I want to keep that these add up to one. I have been using
pro1_prop<- rpert (1, 0.4, 0.478, 0.56)
pro2_prop<- rpert (1, 0.34, 0.415, 0.49)
pro3_prop<- 1- (pro1_prop + pro2_prop)
But this does not seem robust enough. Also on some iterations, (pro1_prop + pro2_prop) >1 which results in a negative value for pro3_prop. Is there a better way of doing this? Thank you!
It is straightforward to sample from the posterior distributions of the proportions using Bayesian methods. I'll assume a multinomial model, where each observation is one of the three profiles.
Say the counts data for the three profiles are 76, 66, and 17.
Using a Dirichlet prior distribution, Dir(1/2, 1/2, 1/2), the posterior is also Dirichlet-distributed: Dir(76.5, 66.5, 17.5), which can be sampled using normalized random gamma variates.
x <- c(76, 66, 17) # observations
# take 1M samples of the proportions from the posterior distribution
theta <- matrix(rgamma(3e6, rep(x + 1/2, each = 1e6)), ncol = 3)
theta <- theta/rowSums(theta)
head(theta)
#> [,1] [,2] [,3]
#> [1,] 0.5372362 0.3666786 0.09608526
#> [2,] 0.4008362 0.4365053 0.16265852
#> [3,] 0.5073144 0.3686412 0.12404435
#> [4,] 0.4752601 0.4367119 0.08802793
#> [5,] 0.4428575 0.4520680 0.10507456
#> [6,] 0.4494075 0.4178494 0.13274311
# compare the Bayesian credible intervals with the frequentist confidence intervals
cbind(
t(mapply(function(i) quantile(theta[,i], c(0.025, 0.975)), seq_along(x))),
t(mapply(function(y) setNames(prop.test(y, sum(x))$conf.int, c("2.5%", "97.5%")), x))
)
#> 2.5% 97.5% 2.5% 97.5%
#> [1,] 0.39994839 0.5537903 0.39873573 0.5583192
#> [2,] 0.33939396 0.4910900 0.33840295 0.4959541
#> [3,] 0.06581214 0.1614677 0.06535702 0.1682029
If samples within the individual 95% CIs are needed, simply reject samples that fall outside the desired interval.
TL;DR: Sample all three values (for example from a pert distribution, as you did) and norm those values afterwards so they add up to one.
Sampling all three values independently from each other and then dividing by their sum so that the normed values add up to one seems to be the easiest option as it is quite hard to sample from the set of legal values directly.
Legal values:
The downside of my approach is that the normed values are not necessarily legal (i.e. in the range of the confidence intervals) any more. However, for these values using a pert distribution, this only happens about 0.5% of the time.
Code:
library(plotly)
library(freedom)
library(data.table)
# define lower (L) and upper (U) bounds and expected values (E)
prof1L <- 0.4
prof1E <- 0.478
prof1U <- 0.56
prof2L <- 0.34
prof2E <- 0.415
prof2U <- 0.49
prof3L <- 0.06
prof3E <- 0.107
prof3U <- 0.15
dt <- as.data.table(expand.grid(
Profile1 = seq(prof1L, prof1U, by = 0.002),
Profile2 = seq(prof2L, prof2U, by = 0.002),
Profile3 = seq(prof3L, prof3U, by = 0.002)
))
# color based on how far the points are away from the center
dt[, color := abs(Profile1 - prof1E) + abs(Profile2 - prof2E) + abs(Profile3 - prof3E)]
# only keep those points that (almost) add up to one
dt <- dt[abs(Profile1 + Profile2 + Profile3 - 1) < 0.01]
# plot the legal values
fig <- plot_ly(dt, x = ~Profile1, y = ~Profile2, z = ~Profile3, color = ~color, colors = c('#BF382A', '#0C4B8E')) %>%
add_markers()
fig
# try to simulate the legal values:
# first sample without considering the condition that the profiles need to add up to 1
nSample <- 100000
dtSample <- data.table(
Profile1Sample = rpert(nSample, prof1L, prof1U, prof1E),
Profile2Sample = rpert(nSample, prof2L, prof2U, prof2E),
Profile3Sample = rpert(nSample, prof3L, prof3U, prof3E)
)
# we want to norm the samples by dividing by their sum
dtSample[, SampleSums := Profile1Sample + Profile2Sample + Profile3Sample]
dtSample[, Profile1SampleNormed := Profile1Sample / SampleSums]
dtSample[, Profile2SampleNormed := Profile2Sample / SampleSums]
dtSample[, Profile3SampleNormed := Profile3Sample / SampleSums]
# now get rid of the cases where the normed values are not legal any more
# (e.g. Profile 1 = 0.56, Profile 2 = 0.38, Profile 3 = 0.06 => dividing by their sum
# will make Profile 3 have an illegal value)
dtSample <- dtSample[
prof1L <= Profile1SampleNormed & Profile1SampleNormed <= prof1U &
prof2L <= Profile2SampleNormed & Profile2SampleNormed <= prof2U &
prof3L <= Profile3SampleNormed & Profile3SampleNormed <= prof3U
]
# see if the sampled values follow the desired distribution
hist(dtSample$Profile1SampleNormed)
hist(dtSample$Profile2SampleNormed)
hist(dtSample$Profile3SampleNormed)
Histogram of normed sampled values for Profile 1:
Ok, some thoughts on the matter.
Lets think about Dirichlet distribution, as one providing RV summed up to 1.
We're talking about Dir(a1, a2, a3), and have to find needed ai.
From the expression for E[Xi]=ai/Sum(i, ai), it is obvious we could get three ratios solving equations
a1/Sum(i, ai) = 0.478
a2/Sum(i, ai) = 0.415
a3/Sum(i, ai) = 0.107
Note, that we have only solved for RATIOS. In other words, if in the expression for E[Xi]=ai/Sum(i, ai) we multiply ai by the same value, mean will stay the same. So we have freedom to choose multiplier m, and what will change is the variance/std.dev. Large multiplier means smaller variance, tighter sampled values around the means
So we could choose m freely to satisfy three 95% CI conditions, three equations for variance but only one df. So it is not possible in general.
One cold play with numbers and the code
I essentially have two columns (vectors) with speed and accel in a data.frame as such:
speed acceleration
1 3.2694444 2.6539535522
2 3.3388889 2.5096979141
3 3.3888889 2.2722134590
4 3.4388889 1.9815256596
5 3.5000000 1.6777544022
6 3.5555556 1.3933215141
7 3.6055556 1.1439051628
8 3.6527778 0.9334115982
9 3.6722222 0.7561602592
I need to find for each value speed on the x axis (speed), what is the top 10% max values from the y axis (acceleration). This also needs to be in a specific interval. For example speed 3.2-3.4, 3.4-3.6, and so on. Can you please show me how a for loop would look like in this situation?
As #alistaire already pointed out, you have provided a very limited amount of data. So we first have to simulate I a bit more data based on which we can test our code.
set.seed(1)
# your data
speed <- c(3.2694444, 3.3388889, 3.3388889, 3.4388889, 3.5,
3.5555556, 3.6055556, 3.6527778, 3.6722222)
acceleration <- c(2.6539535522, 2.5096979141, 2.2722134590,
1.9815256596, 1.6777544022, 1.3933215141,
1.1439051628, 0.9334115982, 0.7561602592)
df <- data.frame(speed, acceleration)
# expand data.frame and add a little bit of noise to all values
# to make them 'unique'
df <- as.data.frame(do.call(
rbind,
replicate(15L, apply(df, 2, \(x) (x + runif(length(x), -1e-1, 1e-1) )),
simplify = FALSE)
))
The function create_intervals, as the name suggests, creates user-defined intervals. The rest of the code does the 'heavy lifting' and stores the desired result in out.
If you would like to have intervals of speed with equal widths, simply specify the number of groups (n_groups) you would like to have and leave the rest of the arguments (i.e. lwr, upr, and interval_span) unspecified.
# Cut speed into user-defined intervals
create_intervals <- \(n_groups = NULL, lwr = NULL, upr = NULL, interval_span = NULL) {
if (!is.null(lwr) & !is.null(upr) & !is.null(interval_span) & is.null(n_groups)) {
speed_low <- subset(df, speed < lwr, select = speed)
first_interval <- with(speed_low, c(min(speed), lwr))
middle_intervals <- seq(lwr + interval_span, upr - interval_span, interval_span)
speed_upp <- subset(df, speed > upr, select = speed)
last_interval <- with(speed_upp, c(upr, max(speed)))
intervals <- c(first_interval, middle_intervals, last_interval)
} else {
step <- with(df, c(max(speed) - min(speed))/n_groups)
intervals <- array(0L, dim = n_groups)
for(i in seq_len(n_groups)) {
intervals[i] <- min(df$speed) + i * step
}
}
return(intervals)
}
# three intervals with equal width
my_intervals <- create_intervals(n_groups = 3L)
# Compute values of speed when acceleration is greater then
# or equal to the 90th percentile
out <- lapply(1:(length(my_intervals)-1L), \(i) {
x <- subset(df, speed >= my_intervals[i] & speed <= my_intervals[i+1L])
x[x$acceleration >= quantile(x$acceleration, 0.9), ]
})
# function to round values to two decimal places
r <- \(x) format(round(x, 2), nsmall = 2L)
# assign names to each element of out
for(i in seq_along(out)) {
names(out)[i] <- paste0(r(my_intervals[i]), '-', r(my_intervals[i+1L]))
}
Output 1
> out
$`3.38-3.57`
speed acceleration
11 3.394378 2.583636
21 3.383631 2.267659
57 3.434123 2.300234
83 3.394886 2.580924
101 3.395459 2.460971
$`3.57-3.76`
speed acceleration
6 3.635234 1.447290
41 3.572868 1.618293
51 3.615017 1.420020
95 3.575412 1.763215
We could also compute the desired values of speed based on intervals that make more 'sense' than just equally spaced speed intervals, e.g. [min(speed), 3.3), [3.3, 3.45), [3.45, 3.6), and [3.6, max(speed)).
This can be accomplished by leaving n_groups unspecified and instead specify lwr, upr, and an interval_span that makes sense. For instance, it makes sense to have a interval span of 0.15 when the lower limit is 3.3 and the upper limit is 3.6.
# custom boundaries based on a lower limit and upper limit
my_intervals <- create_intervals(lwr = 3.3, upr = 3.6, interval_span = 0.15)
Output 2
> out
$`3.18-3.30`
speed acceleration
37 3.238781 2.696456
82 3.258691 2.722076
$`3.30-3.45`
speed acceleration
11 3.394378 2.583636
19 3.328292 2.711825
73 3.315306 2.644580
83 3.394886 2.580924
$`3.45-3.60`
speed acceleration
4 3.520530 2.018930
40 3.517329 2.032943
58 3.485247 2.079893
67 3.458031 2.078545
$`3.60-3.76`
speed acceleration
6 3.635234 1.447290
34 3.688131 1.218969
51 3.615017 1.420020
78 3.628465 1.348873
Note: use function(x) instead of \(x) if you use a version of R <4.1.0
I am trying to identify the highest correlation values among the independent variables (e.g.> = 0.8 | <= -0.8) and then exclude the independent variable that has the lowest correlation value with the dependent variable to avoid multicollinearity in linear models.
First I would like to identify the highest correlations between the independent variables and then exclude those that have the lowest correlations with the dependent variable that is in the first row and column called EC1
The dataset if you want to run it
cor_26_EC<-rcorr(x=as.matrix(data_26_EM[c(4,8:length(data_26_EM))]),type="pearson")
cor_test<-(cor_26_EC$r)
> head(cor_test)
EC1 DEM slope aspect northernness plan_curv prof_curv convergence twi
EC1 1.0000000 -0.68580505 0.36444948 -0.17735481 0.17735481 -0.14541592 -0.21159663 -0.10027208 -0.10220409
DEM -0.6858051 1.00000000 -0.47325220 0.06090698 -0.06090698 0.28021257 0.34739247 0.24297883 -0.02919072
slope 0.3644495 -0.47325220 1.00000000 -0.02321129 0.02321129 0.04219001 0.01703231 0.03937512 -0.56400210
aspect -0.1773548 0.06090698 -0.02321129 1.00000000 -1.00000000 -0.01574986 -0.01260762 0.04838931 0.02877949
northernness 0.1773548 -0.06090698 0.02321129 -1.00000000 1.00000000 0.01574986 0.01260762 -0.04838931 -0.02877949
plan_curv -0.1454159 0.28021257 0.04219001 -0.01574986 0.01574986 1.00000000 0.59109001 0.73023077 -0.51818538
(it continues...)
If I understand correctly, you want something along the lines of this:
library("dplyr")
library("reshape2")
x <- read.delim("~/Documents/stack/EM_26.txt")
c <- cor(x)
## set diagonal to NA because we don't want autocorrelation
diag(c) <- NA
## reshape from matrix to long table
mdf <- melt(c)
## Choose max correlation for each variable
## You could also filter here by a threshold value, eg filter(abs(value) > 0.8)
max_cors <- mdf %>% group_by(Var1) %>% filter(value == max(value, na.rm = TRUE))
## For each max correlation, drop the one which has lower correlation with EC1
drop_variables <- apply(max_cors,
1,
function(row) {
row[which.min(c(c["EC1", row[[1]]], c["EC1", row[[2]]]))]
}
)
unique(drop_variables)
#> [1] "MS1" "z" "EC05"
Update:
the following code should be reproducible
someFrameA = data.frame(label="A", amount=rnorm(10000, 100, 20))
someFrameB = data.frame(label="B", amount=rnorm(1000, 50000, 20))
wholeFrame = rbind(someFrameA, someFrameB)
fit <- e1071::naiveBayes(label ~ amount, wholeFrame)
wholeFrame$predicted = predict(fit, wholeFrame)
nrow(subset(wholeFrame, predicted != label))
In my case, this gave 243 misclassifications.
Note these two rows:
(row num, label, amount, prediction)
10252 B 50024.81895 A
2955 A 100.55977 A
10678 B 50010.26213 B
While the input is only different by 12.6, the classification changes. It's curious that the posterior probabilities for rows like this are so close:
> predict(fit, wholeFrame[10683, ], type="raw")
A B
[1,] 0.5332296 0.4667704
Original Question:
I am trying to classify some bank transactions using the transaction amount. I had many other text based features in my original model, but noticed something fishy when using just the numeric one.
> head(trainingSet)
category amount
1 check 688.00
2 non-businesstransaction 2.50
3 non-businesstransaction 36.00
4 non-businesstransaction 243.22
5 payroll 302.22
6 non-businesstransaction 16.18
fit <- e1071::naiveBayes(category ~ amount, data=trainingSet)
fit
Naive Bayes Classifier for Discrete Predictors
Call:
naiveBayes.default(x = X, y = Y, laplace = laplace)
A-priori probabilities:
Y
bankfee check creditcardpayment e-commercedeposit insurance
0.029798103 0.189613233 0.054001459 0.018973486 0.008270494
intrabanktransfer loanpayment mcapayment non-businesstransaction nsf
0.045001216 0.015689613 0.011432741 0.563853077 0.023351982
other payroll taxpayment utilitypayment
0.003405497 0.014838239 0.005716371 0.016054488
Conditional probabilities:
amount
Y [,1] [,2]
bankfee 103.58490 533.67098
check 803.44668 2172.12515
creditcardpayment 819.27502 2683.43571
e-commercedeposit 42.15026 59.24806
insurance 302.16500 727.52321
intrabanktransfer 1795.54065 11080.73658
loanpayment 308.43233 387.71165
mcapayment 356.62755 508.02412
non-businesstransaction 162.41626 951.65934
nsf 44.92198 78.70680
other 9374.81071 18074.36629
payroll 1192.79639 2155.32633
taxpayment 1170.74340 1164.08019
utilitypayment 362.13409 1064.16875
According to the e1071 docs, the first column for "conditional probabilities" is the mean of the numeric variable, and the other is the standard deviation. These means and stdevs are correct, as are the apriori probabilities.
So, it's troubling that this row:
> thatRow
category amount
40 other 11268.53
receives these posteriors:
> predict(fit, newdata=thatRow, type="raw")
bankfee check creditcardpayment e-commercedeposit insurance intrabanktransfer loanpayment mcapayment
[1,] 4.634535e-96 7.28883e-06 9.401975e-05 0.4358822 4.778703e-51 0.02582751 1.103762e-174 1.358662e-101
non-businesstransaction nsf other payroll taxpayment utilitypayment
[1,] 1.446923e-29 0.5364704 0.001717378 1.133719e-06 2.059156e-18 2.149142e-24
Note that "nsf" has about 300X the score than "other" does. Since this transaction has an amount of 11.2k dollars, if it were to follow that "nsf" distribution, it would be over 100 standard deviations from the mean. Meanwhile, since "other" transactions have a sample mean of about 9k dollars with a large standard deviation, I would think that this transaction is much more probable as an "other". While "nsf" is more likely wrt the prior probabilities, they aren't so different as to outweigh that tail observation, and there are plenty of other viable candidates besides "other" as well.
I was assuming that this package just looked at the normal(mew=samplemean, stdev=samplestdev) pdf and used that value to multiply, but is that not the case? I can't quite figure out how to see the source.
Datatypes seem to be fine too:
> class(trainingSet$amount)
[1] "numeric"
> class(trainingSet$category)
[1] "factor"
The "naive bayes classifier for discrete predictors" in the printout is maybe odd, since this is a continuous predictor, but I assume this package can handle continuous predictors.
I had similar results with the klaR package. Maybe I need to set the kernel option on that?
The threshold argument is a large part of this. The code in the package has a bit like this:
L <- sapply(1:nrow(newdata), function(i) {
ndata <- newdata[i, ]
L <- log(object$apriori) + apply(log(sapply(seq_along(attribs),
function(v) {
nd <- ndata[attribs[v]]
if (is.na(nd)) rep(1, length(object$apriori)) else {
prob <- if (isnumeric[attribs[v]]) {
msd <- object$tables[[v]]
msd[, 2][msd[, 2] <= eps] <- threshold
dnorm(nd, msd[, 1], msd[, 2])
} else object$tables[[v]][, nd]
prob[prob <= eps] <- threshold
prob
}
The threshold (and this is documented) will replace any probabilities less than eps. So, if the normal pdf for the continuous variable is 0.000000000, it will become .001 by default.
> wholeFrame$predicted = predict(fit, wholeFrame, threshold=0.001)
> nrow(subset(wholeFrame, predicted != label))
[1] 249
> wholeFrame$predicted = predict(fit, wholeFrame, threshold=0.0001)
> nrow(subset(wholeFrame, predicted != label))
[1] 17
> wholeFrame$predicted = predict(fit, wholeFrame, threshold=0.00001)
> nrow(subset(wholeFrame, predicted != label))
[1] 3
Now, I believe that the quantities returned by the sapply are incorrect, since when "debugging" it, I got something like .012 for what should have been dnorm(49990, 100, 20), and I think something gets left out / mixed up with the mean and standard deviation matrix, but in any case, setting the threshold will help with this.
.001*(10/11) > pdfB*(1/11) or A having higher posterior than B due to this situation means that pdfB has to be less than .01 by chance.
> dnorm(49977, 50000, 20)
[1] 0.01029681
> 2*pnorm(49977, 50000, 20)
[1] 0.2501439
And since there were 1000 observations in class B, we should expect about 250 misclassifications, which is pretty close to the original 243.
I want to perform winsorization in a dataframe like this:
event_date beta_before beta_after
2000-05-05 1.2911707054 1.3215648954
1999-03-30 0.5089734305 0.4269575657
2000-05-05 0.5414700258 0.5326762272
2000-02-09 1.5491034852 1.2839988507
1999-03-30 1.9380674599 1.6169735009
1999-03-30 1.3109909155 1.4468207148
2000-05-05 1.2576420753 1.3659492507
1999-03-30 1.4393018341 0.7417777965
2000-05-05 0.2624037804 0.3860641307
2000-05-05 0.5532216441 0.2618245169
2000-02-08 2.6642931822 2.3815576738
2000-02-09 2.3007578964 2.2626960407
2001-08-14 3.2681270302 2.1611010935
2000-02-08 2.2509121123 2.9481325199
2000-09-20 0.6624503316 0.947935581
2006-09-26 0.6431111805 0.8745333151
By winsorization I mean to find the max and min for beta_before for example. That value should be replaced by the second highest or second lowest value in the same column, without loosing the rest of the details in the observation. For example. In this case, in beta_before the max value is 3.2681270302 and should be replaced by 3.2681270302. The same process will be followed for the min and then for the beta_after variable. Therefore, only 2 values per column will be changes, the highest and the minimum, the rest will remain the same.
Any advice? I tried different approaches in plyr, but I ended up replacing the whole observation, which I don’t want to do. I would like to create 2 new variables, for example beta_before_winsorized and beta _after_winsorized
I thought winsorizing usually finds the value x% (typically 10%, 15%, or 20%) from the bottom of the ordered list, and replaces all the values below it with that value. Same with the top. Here you're just choosing the top and bottom value, but winsorizing usually involves specifying a percentage of values at the top and bottom to replace.
Here is a function that does the winsorzation you describe:
winsorize <- function(x) {
Min <- which.min(x)
Max <- which.max(x)
ord <- order(x)
x[Min] <- x[ord][2]
x[Max] <- x[ord][length(x)-1]
x
}
If you data are in a data frame dat, then we can windsoroize the data using your procedure via:
dat2 <- dat
dat2[, -1] <- sapply(dat[,-1], winsorize)
which results in:
R> dat2
event_date beta_before beta_after
1 2000-05-05 1.2911707 1.3215649
2 1999-03-30 0.5089734 0.4269576
3 2000-05-05 0.5414700 0.5326762
4 2000-02-09 1.5491035 1.2839989
5 1999-03-30 1.9380675 1.6169735
6 1999-03-30 1.3109909 1.4468207
7 2000-05-05 1.2576421 1.3659493
8 1999-03-30 1.4393018 0.7417778
9 2000-05-05 0.5089734 0.3860641
10 2000-05-05 0.5532216 0.3860641
11 2000-02-08 2.6642932 2.3815577
12 2000-02-09 2.3007579 2.2626960
13 2001-08-14 2.6642932 2.1611011
14 2000-02-08 2.2509121 2.3815577
15 2000-09-20 0.6624503 0.9479356
16 2006-09-26 0.6431112 0.8745333
I'm not sure where you got the value you suggest should replace the max in beta_before as the second highest is 2.6642932 in the snippet of data provided and that is what my function has used to replace with the maximum value with.
Note the function will only work if there is one minimum and maximum values respectively in each column owing to the way which.min() and which.max() are documented to work. If you have multiple entries taking the same max or min value then we would need something different:
winsorize2 <- function(x) {
Min <- which(x == min(x))
Max <- which(x == max(x))
ord <- order(x)
x[Min] <- x[ord][length(Min)+1]
x[Max] <- x[ord][length(x)-length(Max)]
x
}
should do it (latter is not tested).
Strictly speaking, "winsorization" is the act of replacing the most extreme data points with an acceptable percentile (as mentioned in some of the other answers). One fairly standard R function to do this is winsor from the psych package. Try:
dat$beta_before = psych::winsor(dat$beta_before, trim = 0.0625)
dat$beta_after = psych::winsor(dat$beta_after , trim = 0.0625)
I chose trim = to be 0.0625 (the 6.25th percentile and 93.75th percentile) because you only have 16 data points and you want to "rein in" the top and bottom ones: 1/16 = 0.0625
Note that this might make the extreme data equal to a percentile number which may or may not exist in your data set: the theoretical n-th percentile of the data.
The statar package works very well for this. Copying the relevant snippet from the readme file:
# winsorize (default based on 5 x interquartile range)
v <- c(1:4, 99)
winsorize(v)
winsorize(v, replace = NA)
winsorize(v, probs = c(0.01, 0.99))
winsorize(v, cutpoints = c(1, 50))
https://github.com/matthieugomez/statar
follow up from my previous point about actually replacing the to-be-trimmed values with value at trim position:
winsorized.sample<-function (x, trim = 0, na.rm = FALSE, ...)
{
if (!is.numeric(x) && !is.complex(x) && !is.logical(x)) {
warning("argument is not numeric or logical: returning NA")
return(NA_real_)
}
if (na.rm)
x <- x[!is.na(x)]
if (!is.numeric(trim) || length(trim) != 1L)
stop("'trim' must be numeric of length one")
n <- length(x)
if (trim > 0 && n) {
if (is.complex(x))
stop("trimmed sample is not defined for complex data")
if (any(is.na(x)))
return(NA_real_)
if (trim >= 0.5) {
warning("trim >= 0.5 is odd...trying it anyway")
}
lo <- floor(n * trim) + 1
hi <- n + 1 - lo
#this line would work for just trimming
# x <- sort.int(x, partial = unique(c(lo, hi)))[lo:hi]
#instead, we're going to replace what would be trimmed
#with value at trim position using the next 7 lines
idx<-seq(1,n)
myframe<-data.frame(idx,x)
myframe<-myframe[ order(x,idx),]
myframe$x[1:lo]<-x[lo]
myframe$x[hi:n]<-x[hi]
myframe<-myframe[ order(idx,x),]
x<-myframe$x
}
x
}
#test it
mydist<-c(1,20,1,5,2,40,5,2,6,1,5)
mydist2<-winsorized.sample(mydist, trim=.2)
mydist
mydist2
descStat(mydist)
descStat(mydist2)