This is my first time with strucchange so bear with me. The problem I'm having seems to be that strucchange doesn't recognize my time series correctly but I can't figure out why and haven't found an answer on the boards that deals with this. Here's a reproducible example:
require(strucchange)
# time series
nmreprosuccess <- c(0,0.50,NA,0.,NA,0.5,NA,0.50,0.375,0.53,0.846,0.44,1.0,0.285,
0.75,1,0.4,0.916,1,0.769,0.357)
dat.ts <- ts(nmreprosuccess, frequency=1, start=c(1996,1))
str(dat.ts)
Time-Series [1:21] from 1996 to 2016: 0 0.5 NA 0 NA 0.5 NA 0.5 0.375 0.53 ...
To me this means that the time series looks OK to work with.
# obtain breakpoints
bp.NMSuccess <- breakpoints(dat.ts~1)
summary(bp.NMSuccess)
Which gives:
Optimal (m+1)-segment partition:
Call:
breakpoints.formula(formula = dat.ts ~ 1)
Breakpoints at observation number:
m = 1 6
m = 2 3 7
m = 3 3 14 16
m = 4 3 7 14 16
m = 5 3 7 10 14 16
m = 6 3 7 10 12 14 16
m = 7 3 5 7 10 12 14 16
Corresponding to breakdates:
m = 1 0.333333333333333
m = 2 0.166666666666667 0.388888888888889
m = 3 0.166666666666667
m = 4 0.166666666666667 0.388888888888889
m = 5 0.166666666666667 0.388888888888889 0.555555555555556
m = 6 0.166666666666667 0.388888888888889 0.555555555555556 0.666666666666667
m = 7 0.166666666666667 0.277777777777778 0.388888888888889 0.555555555555556 0.666666666666667
m = 1
m = 2
m = 3 0.777777777777778 0.888888888888889
m = 4 0.777777777777778 0.888888888888889
m = 5 0.777777777777778 0.888888888888889
m = 6 0.777777777777778 0.888888888888889
m = 7 0.777777777777778 0.888888888888889
Fit:
m 0 1 2 3 4 5 6 7
RSS 1.6986 1.1253 0.9733 0.8984 0.7984 0.7581 0.7248 0.7226
BIC 14.3728 12.7421 15.9099 20.2490 23.9062 28.7555 33.7276 39.4522
Here's where I start having the problem. Instead of reporting the actual breakdates it reports numbers which then makes it impossible to plot the break lines onto a graph because they're not at the breakdate (2002) but at 0.333.
plot.ts(dat.ts, main="Natural Mating")
lines(fitted(bp.NMSuccess, breaks = 1), col = 4, lwd = 1.5)
Nothing shows up for me in this graph (I think because it's so small for the scale of the graph).
In addition, when I try fixes that may possibly work around this problem,
fm1 <- lm(dat.ts ~ breakfactor(bp.NMSuccess, breaks = 1))
I get:
Error in model.frame.default(formula = dat.ts ~ breakfactor(bp.NMSuccess, :
variable lengths differ (found for 'breakfactor(bp.NMSuccess, breaks = 1)')
I get errors because of the NA values in the data so the length of dat.ts is 21 and the length of breakfactor(bp.NMSuccess, breaks = 1) 18 (missing the 3 NAs).
Any suggestions?
The problem occurs because breakpoints() currently can only (a) cope with NAs by omitting them, and (b) cope with times/date through the ts class. This creates the conflict because when you omit internal NAs from a ts it loses its ts property and hence breakpoints() cannot infer the correct times.
The "obvious" way around this would be to use a time series class that can cope with this, namely zoo. However, I just never got round to fully integrate zoo support into breakpoints() because it would likely break some of the current behavior.
To cut a long story short: Your best choice at the moment is to do the book-keeping about the times yourself and not expect breakpoints() to do it for you. The additional work is not so huge. First, we create a time series with the response and the time vector and omit the NAs:
d <- na.omit(data.frame(success = nmreprosuccess, time = 1996:2016))
d
## success time
## 1 0.000 1996
## 2 0.500 1997
## 4 0.000 1999
## 6 0.500 2001
## 8 0.500 2003
## 9 0.375 2004
## 10 0.530 2005
## 11 0.846 2006
## 12 0.440 2007
## 13 1.000 2008
## 14 0.285 2009
## 15 0.750 2010
## 16 1.000 2011
## 17 0.400 2012
## 18 0.916 2013
## 19 1.000 2014
## 20 0.769 2015
## 21 0.357 2016
Then we can estimate the breakpoint(s) and afterwards transform from the "number" of observations back to the time scale. Note that I'm setting the minimal segment size h explicitly here because the default of 15% is probably somewhat small for this short series. 4 is still small but possibly enough for estimating of a constant mean.
bp <- breakpoints(success ~ 1, data = d, h = 4)
bp
## Optimal 2-segment partition:
##
## Call:
## breakpoints.formula(formula = success ~ 1, h = 4, data = d)
##
## Breakpoints at observation number:
## 6
##
## Corresponding to breakdates:
## 0.3333333
We ignore the break "date" at 1/3 of the observations but simply map back to the original time scale:
d$time[bp$breakpoints]
## [1] 2004
To re-estimate the model with nicely formatted factor levels, we could do:
lab <- c(
paste(d$time[c(1, bp$breakpoints)], collapse = "-"),
paste(d$time[c(bp$breakpoints + 1, nrow(d))], collapse = "-")
)
d$seg <- breakfactor(bp, labels = lab)
lm(success ~ 0 + seg, data = d)
## Call:
## lm(formula = success ~ 0 + seg, data = d)
##
## Coefficients:
## seg1996-2004 seg2005-2016
## 0.3125 0.6911
Or for visualization:
plot(success ~ time, data = d, type = "b")
lines(fitted(bp) ~ time, data = d, col = 4, lwd = 2)
abline(v = d$time[bp$breakpoints], lty = 2)
One final remark: For such short time series where just a simple shift in the mean is needed, one could also consider conditional inference (aka permutation tests) rather than the asymptotic inference employed in strucchange. The coin package provides the maxstat_test() function exactly for this purpose (= short series where a single shift in the mean is tested).
library("coin")
maxstat_test(success ~ time, data = d, dist = approximate(99999))
## Approximative Generalized Maximally Selected Statistics
##
## data: success by time
## maxT = 2.3953, p-value = 0.09382
## alternative hypothesis: two.sided
## sample estimates:
## "best" cutpoint: <= 2004
This finds the same breakpoint and provides a permutation test p-value. If however, one has more data and needs multiple breakpoints and/or further regression coefficients, then strucchange would be needed.
Related
So I have a data set that is based on HR data training which asks tech and common questions.
The rows represent an employee and the columns represent the score they got on each question. The columns also include demographic data. I only want to see the row total of the tech and common questions though and not include the demographic data.
techs<-grep("^T",rownames(dat))
commons<-grep("^C",rownames(dat))
I used this to try to group the columns together but when I do:
total<-rowsum(commons,techs)
and try to put it in a linear regression:
Mod1Train<-lm(total~.,data=dat[Train,])
it says that there are different variable lengths.
I'm a super newbie to R, so sorry in advance if I'm really off.
in the future it would be ever so helpful if you provided a sample of your data. It's hard for us to help when we're guessing about that. Please see this link https://stackoverflow.com/help/minimal-reproducible-example.
Having said that LOL and realizing you're new I'll take a guess...
Let's make pretend data that I imagine is a smaller imaginary version of yours...
set.seed(2020)
emplid <- 1:10
gender <- sample(c("Male", "Female"), size = 10, replace = TRUE)
Tech1 <- sample(10:20, size = 10, replace = TRUE)
Tech2 <- sample(10:20, size = 10, replace = TRUE)
Tech3 <- sample(10:20, size = 10, replace = TRUE)
Common1 <- sample(10:20, size = 10, replace = TRUE)
Common2 <- sample(10:20, size = 10, replace = TRUE)
Common3 <- sample(10:20, size = 10, replace = TRUE)
Kathryn <- data.frame(emplid, gender, Tech1, Tech2, Tech3, Common1, Common2, Common3)
Kathryn
#> emplid gender Tech1 Tech2 Tech3 Common1 Common2 Common3
#> 1 1 Female 10 17 15 18 17 15
#> 2 2 Female 17 13 11 20 11 13
#> 3 3 Male 17 11 19 18 10 12
#> 4 4 Female 19 16 15 14 15 16
#> 5 5 Female 11 13 20 20 16 13
#> 6 6 Male 15 11 17 19 17 13
#> 7 7 Male 11 13 11 15 14 11
#> 8 8 Female 12 14 10 11 17 19
#> 9 9 Female 11 13 15 18 11 10
#> 10 10 Female 17 20 12 12 14 15
If you're new may want to invest some time learning the tidyverse which could make this simple like here Efficiently sum across multiple columns in R
Per your note in the comments, you have a pattern we can match for summing questions. You were close with your attempt at grep but we want the values back so we need value = TRUE which we'll store and make use of.
techqs <- grep(x = names(Kathryn), pattern = "^Tech", value = TRUE)
commonqs <- grep(x = names(Kathryn), pattern = "^Common", value = TRUE)
Kathryn$TechScores <- rowSums(Kathryn[,techqs])
Kathryn$CommonScores <- rowSums(Kathryn[,commonqs])
### Commented out how to do it manually.
# Kathryn$TechScores <- rowSums(Kathryn[,c("TQ1", "TQ2", "TQ3")])
# Kathryn$CommonScores <- rowSums(Kathryn[,c("CQ1", "CQ2", "CQ3")])
Kathryn$TotalScore <- Kathryn$TechScores + Kathryn$CommonScores
Now to regress which is where the statistical problem comes in. Are you really trying to predict the total score from the components??? That's not hard in r but it leads to silly answers.
Kathryn_model <- lm(formula = TotalScore ~ TechScores + CommonScores, data = Kathryn)
summary(Kathryn_model)
#> Warning in summary.lm(Kathryn_model): essentially perfect fit: summary may be
#> unreliable
#>
#> Call:
#> lm(formula = TotalScore ~ TechScores + CommonScores, data = Kathryn)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -3.165e-14 -1.905e-15 9.290e-16 8.590e-15 1.183e-14
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 8.089e-14 6.345e-14 1.275e+00 0.243
#> TechScores 1.000e+00 9.344e-16 1.070e+15 <2e-16 ***
#> CommonScores 1.000e+00 1.130e-15 8.853e+14 <2e-16 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 1.43e-14 on 7 degrees of freedom
#> Multiple R-squared: 1, Adjusted R-squared: 1
#> F-statistic: 9.875e+29 on 2 and 7 DF, p-value: < 2.2e-16
I don't understand your code and what you search for
rowsums don't make "a row total" but, quite on the contrary, adds rows between themselves. It returns a matrix, not a vector. Is that what you want ?
Otherwise, maybe you're looking for rowSums, which computes every rows totals of a matrix.
(by the way, if you need it, the matrix product is %*% in R)
Are you sure you have understood lm ?
In lm, there should be something like
lm(y~x,data=adataframe)
"adataframe" is the eventual dataframe/matrix where lm seeks both the response and the input variable,named "y" and "x" here. It is optional. If not found, y and x are seeked in the Global Env as if the columns names are not found in data, they are seeked in the Global environment. It is sometimes better however, to have such a matrix-like object, to avoid common errors.
So if you want to use lm, maybe you should first try to obtain 2 vectors, one for x and one for y, have them in a data.frame with 2 columns (x and y), and call the code above, if I have correctly understood
Note : if you want to remove the constant, use then
lm(y~x+0,data=adataframe)
This is my first time with strucchange so bear with me. The problem I'm having seems to be that strucchange doesn't recognize my time series correctly but I can't figure out why and haven't found an answer on the boards that deals with this. Here's a reproducible example:
require(strucchange)
# time series
nmreprosuccess <- c(0,0.50,NA,0.,NA,0.5,NA,0.50,0.375,0.53,0.846,0.44,1.0,0.285,
0.75,1,0.4,0.916,1,0.769,0.357)
dat.ts <- ts(nmreprosuccess, frequency=1, start=c(1996,1))
str(dat.ts)
Time-Series [1:21] from 1996 to 2016: 0 0.5 NA 0 NA 0.5 NA 0.5 0.375 0.53 ...
To me this means that the time series looks OK to work with.
# obtain breakpoints
bp.NMSuccess <- breakpoints(dat.ts~1)
summary(bp.NMSuccess)
Which gives:
Optimal (m+1)-segment partition:
Call:
breakpoints.formula(formula = dat.ts ~ 1)
Breakpoints at observation number:
m = 1 6
m = 2 3 7
m = 3 3 14 16
m = 4 3 7 14 16
m = 5 3 7 10 14 16
m = 6 3 7 10 12 14 16
m = 7 3 5 7 10 12 14 16
Corresponding to breakdates:
m = 1 0.333333333333333
m = 2 0.166666666666667 0.388888888888889
m = 3 0.166666666666667
m = 4 0.166666666666667 0.388888888888889
m = 5 0.166666666666667 0.388888888888889 0.555555555555556
m = 6 0.166666666666667 0.388888888888889 0.555555555555556 0.666666666666667
m = 7 0.166666666666667 0.277777777777778 0.388888888888889 0.555555555555556 0.666666666666667
m = 1
m = 2
m = 3 0.777777777777778 0.888888888888889
m = 4 0.777777777777778 0.888888888888889
m = 5 0.777777777777778 0.888888888888889
m = 6 0.777777777777778 0.888888888888889
m = 7 0.777777777777778 0.888888888888889
Fit:
m 0 1 2 3 4 5 6 7
RSS 1.6986 1.1253 0.9733 0.8984 0.7984 0.7581 0.7248 0.7226
BIC 14.3728 12.7421 15.9099 20.2490 23.9062 28.7555 33.7276 39.4522
Here's where I start having the problem. Instead of reporting the actual breakdates it reports numbers which then makes it impossible to plot the break lines onto a graph because they're not at the breakdate (2002) but at 0.333.
plot.ts(dat.ts, main="Natural Mating")
lines(fitted(bp.NMSuccess, breaks = 1), col = 4, lwd = 1.5)
Nothing shows up for me in this graph (I think because it's so small for the scale of the graph).
In addition, when I try fixes that may possibly work around this problem,
fm1 <- lm(dat.ts ~ breakfactor(bp.NMSuccess, breaks = 1))
I get:
Error in model.frame.default(formula = dat.ts ~ breakfactor(bp.NMSuccess, :
variable lengths differ (found for 'breakfactor(bp.NMSuccess, breaks = 1)')
I get errors because of the NA values in the data so the length of dat.ts is 21 and the length of breakfactor(bp.NMSuccess, breaks = 1) 18 (missing the 3 NAs).
Any suggestions?
The problem occurs because breakpoints() currently can only (a) cope with NAs by omitting them, and (b) cope with times/date through the ts class. This creates the conflict because when you omit internal NAs from a ts it loses its ts property and hence breakpoints() cannot infer the correct times.
The "obvious" way around this would be to use a time series class that can cope with this, namely zoo. However, I just never got round to fully integrate zoo support into breakpoints() because it would likely break some of the current behavior.
To cut a long story short: Your best choice at the moment is to do the book-keeping about the times yourself and not expect breakpoints() to do it for you. The additional work is not so huge. First, we create a time series with the response and the time vector and omit the NAs:
d <- na.omit(data.frame(success = nmreprosuccess, time = 1996:2016))
d
## success time
## 1 0.000 1996
## 2 0.500 1997
## 4 0.000 1999
## 6 0.500 2001
## 8 0.500 2003
## 9 0.375 2004
## 10 0.530 2005
## 11 0.846 2006
## 12 0.440 2007
## 13 1.000 2008
## 14 0.285 2009
## 15 0.750 2010
## 16 1.000 2011
## 17 0.400 2012
## 18 0.916 2013
## 19 1.000 2014
## 20 0.769 2015
## 21 0.357 2016
Then we can estimate the breakpoint(s) and afterwards transform from the "number" of observations back to the time scale. Note that I'm setting the minimal segment size h explicitly here because the default of 15% is probably somewhat small for this short series. 4 is still small but possibly enough for estimating of a constant mean.
bp <- breakpoints(success ~ 1, data = d, h = 4)
bp
## Optimal 2-segment partition:
##
## Call:
## breakpoints.formula(formula = success ~ 1, h = 4, data = d)
##
## Breakpoints at observation number:
## 6
##
## Corresponding to breakdates:
## 0.3333333
We ignore the break "date" at 1/3 of the observations but simply map back to the original time scale:
d$time[bp$breakpoints]
## [1] 2004
To re-estimate the model with nicely formatted factor levels, we could do:
lab <- c(
paste(d$time[c(1, bp$breakpoints)], collapse = "-"),
paste(d$time[c(bp$breakpoints + 1, nrow(d))], collapse = "-")
)
d$seg <- breakfactor(bp, labels = lab)
lm(success ~ 0 + seg, data = d)
## Call:
## lm(formula = success ~ 0 + seg, data = d)
##
## Coefficients:
## seg1996-2004 seg2005-2016
## 0.3125 0.6911
Or for visualization:
plot(success ~ time, data = d, type = "b")
lines(fitted(bp) ~ time, data = d, col = 4, lwd = 2)
abline(v = d$time[bp$breakpoints], lty = 2)
One final remark: For such short time series where just a simple shift in the mean is needed, one could also consider conditional inference (aka permutation tests) rather than the asymptotic inference employed in strucchange. The coin package provides the maxstat_test() function exactly for this purpose (= short series where a single shift in the mean is tested).
library("coin")
maxstat_test(success ~ time, data = d, dist = approximate(99999))
## Approximative Generalized Maximally Selected Statistics
##
## data: success by time
## maxT = 2.3953, p-value = 0.09382
## alternative hypothesis: two.sided
## sample estimates:
## "best" cutpoint: <= 2004
This finds the same breakpoint and provides a permutation test p-value. If however, one has more data and needs multiple breakpoints and/or further regression coefficients, then strucchange would be needed.
Sample Data
A B C D E
1 2016 94.49433733 2 81.28
5 2016 95.38104534 4 139.6944
7 2016 95.43885385 1 69.11
8 2016 94.91936704 1 7.23
9 2016 95.21859776 3 152.31
10 2016 95.15797535 1 86.32
11 2016 95.1830432 2 38.24
13 2016 94.01256633 2 33.3
Given the sample data and using R, I want to build a sequence that gives me the incremental impact from my predictor variable (C).
Expected Table (increment by 0.5):
C ANSWER
85
85.05
85.1
85.15
85.2
85.25
85.3
85.35
I am looking to understand for every delta change (increase) in C, what happens to D?
Here is what I tried with transform and apply
transform(df, volumen=unlist(tapply(C, D, function(x) c(0, diff(x)))))
fit <- lm(D ~ C, data = my_sample_data) #Fits a linear model
my_sequence <- seq(from = 85, to = 85.35, by = 0.05 ) # first column
result <- fit$coefficients[1] + my_sequence * fit$coefficients[2] #2nd column
df <- data.frame(C = my_sequence, ANSWER = result) #Makes a table
I have got 4 different data.frames that have observations that follow a binomial distribution and I need to calculate, for each one, the confidence intervals related to the means of a second column (Flow).
The number of successes are reported in the column Success and the total number of trials = 85.
How can I calculate confidence intervals?
How can I do it with R?
Here an example of my data.frames:
df1 <- read.table(text = 'Flow Success
725.661 4
25.54 4
318.481 4
230.556 4
2.823 3
12.6 3
9.891 3
11.553 1', header = TRUE)
> mean(df1$Flow)
[1] 167.1381
df2 <- read.table(text = 'Flow Success
725.661 3
25.54 3
318.481 3
230.556 2
2.823 2
12.6 1', header = TRUE)
> mean(df2$Flow)
[1] 219.2768
df3 <- read.table(text = 'Flow Success
725.661 2
25.54 2
318.481 1', header = TRUE)
> mean(df3$Flow)
[1] 356.5607
df4 <- read.table(text = 'Flow Success
725.661 2
25.54 2', header = TRUE)
> mean(df4$Flow)
[1] 375.6005
I need to calculate the confidence intervals of the above means.
I can give you more info about the data if needed.
Thanks for anyone who will help me.
The package binom provides methods for calculating binomial confidence intervals. One can choose to use all available methods, or specify a single method.
x gives the number of successes, and n the number of Bernouli trials.
library(binom)
binom.confint(x = 5, n = 10)
method x n mean lower upper
1 agresti-coull 5 10 0.5 0.2365931 0.7634069
2 asymptotic 5 10 0.5 0.1901025 0.8098975
3 bayes 5 10 0.5 0.2235287 0.7764713
4 cloglog 5 10 0.5 0.1836056 0.7531741
5 exact 5 10 0.5 0.1870860 0.8129140
6 logit 5 10 0.5 0.2245073 0.7754927
7 probit 5 10 0.5 0.2186390 0.7813610
8 profile 5 10 0.5 0.2176597 0.7823403
9 lrt 5 10 0.5 0.2176212 0.7823788
10 prop.test 5 10 0.5 0.2365931 0.7634069
11 wilson 5 10 0.5 0.2365931 0.7634069
binom.confint(x = 5, n = 10, method = "exact")
method x n mean lower upper
1 exact 5 10 0.5 0.187086 0.812914
I have the following data set:
dat<-as.data.frame(rbind(10,8,2,7,10,10,1,10,14,9,2,6,10,8,10,8,10,10,7,11,10))
colnames(dat)<-"Score"
print(dat)
Score
10
8
2
7
10
10
1
10
14
9
2
6
10
8
10
8
10
10
7
11
10
these are the test scores which students obtained, a student could get a maximum of 15 or a minimum of 0 in this test (by the way, nobody got the max or the min), however the lowest score obtained in this test was 1 and the highest was 14.
Now, I want to normalize/scale this data to the scale of 0 to 20.
How to achieve this in excel? or in R?
My final goal is to normalize the scores in this test to the above scale and to compare them with another set of data for which the max and min is 5 and 0 respectively.
How to compare these two different scaled data sets correctly against each other?
What I tried:
I went through many stuff on the internet, and came up with this:
which I got it from the wikipedia.
Is this method reliable?
In your case I would use the feature scale formula you posted on your question. The (x - min(x)) / (max(x) - min(x)) will essentially convert your test marks to the range between 0-1.
Since your edges are indeed 0 and 15 and not 2 and 14, your min(x)=0 and your max(x)=15. Once you have your marks between 0-1 using the above, you just multiply by 20.
i.e.
tests <- read.table(header=T, file='clipboard')
tests2 <- (tests - 0) / (15 - 0) #or equally tests / 15
And multiply by 20 to get marks between 0-20:
> tests2 * 20
Score
1 13.333333
2 10.666667
3 2.666667
4 9.333333
5 13.333333
6 13.333333
7 1.333333
8 13.333333
9 18.666667
10 12.000000
11 2.666667
12 8.000000
13 13.333333
14 10.666667
15 13.333333
16 10.666667
17 13.333333
18 13.333333
19 9.333333
20 14.666667
21 13.333333
The results are intuitive and the function is reliable. For example the person who scored 14/15 should get the highest mark (and very close to 20) which is the case here (after the transformation they scored 18.6666).
In Excel, if you want the normalized data to have a min of 0 and and max of 20, then we need to solve:
y = A * x + b
for two points.
Put the max of the raw data in C1:
=MAX(A:A)
Put the min of the raw data in C2:
=MIN(A:A)
Put the desired max in D1 and the desired min in D2. Put the formula for the A-coefficient in C3:
=($D$1-$D$2)/($C$1-$C$2)
and the formula for the B-coefficient in C4:
=$D$1-$C$3*$C$1
Finally put the scaling formula in B1:
=A1*$C$3+$C$4
and copy down:
Naturally, if you want the scaling to be independent of the raw max or min, you would use 15 in C1 and 0 in C2.
You can scale between 0 to 20 with this command in R:
newvalue <- 20/(max(score)-min(score))*(score-min(score))
The math way is fairly straightforward if the floor for all scales is 0.
new_value = new_ceiling * old_value / old_ceiling
The next formula will account for different floors on each scale:
new_value = new_floor + (new_ceiling - old_ceiling) * ((old_value-old_floor)/(old_ceiling-old_floor)) which is actually the formula you posted from Wikipedia. ;)
Hope this helps!
That is very simple. Due to the fact that both of those grades are linear, that a simple multiple ratio will do the work. Or in other word each grade in your set needs to be *20/15.
Here's a little r function which can help you run this if you need to repeat the operation and give you some flexibility on what you rescale to. Also one must be careful of NA values because min() and max() do not drop them by default which will then return NA. Therefore I provided an option on to handle NA values (drops them by default).
# function rescales data from 0 to 1 and optionally multiplies by new max
rescale <- function(x, new_max = 1, na.rm = T) {
as.vector(new_max * scale(x,
center = min(x, na.rm = na.rm),
scale = (max(x, na.rm = na.rm) - min(x, na.rm = na.rm))))
}
# old scores
scores <- c(10,8,2,7,10,10,1,10,14,9,2,6,10,8,10,8,10,10,7,11,10)
# new scores
data.frame(old = scores,
new = rescale(scores, new_max = 20))
#> old new
#> 1 10 13.846154
#> 2 8 10.769231
#> 3 2 1.538462
#> 4 7 9.230769
#> 5 10 13.846154
#> 6 10 13.846154
#> 7 1 0.000000
#> 8 10 13.846154
#> 9 14 20.000000
#> 10 9 12.307692
#> 11 2 1.538462
#> 12 6 7.692308
#> 13 10 13.846154
#> 14 8 10.769231
#> 15 10 13.846154
#> 16 8 10.769231
#> 17 10 13.846154
#> 18 10 13.846154
#> 19 7 9.230769
#> 20 11 15.384615
#> 21 10 13.846154
Created on 2022-03-10 by the reprex package (v2.0.1)