Solution (thanks #Peter_Evan!) in case anyone coming across this question has a similar issue
(Original question is below)
## get all slopes (lm coefficients) first
# list of subfields of interest to loop through
sf <- c("left_presubiculum", "right_presubiculum",
"left_subiculum", "right_subiculum", "left_CA1", "right_CA1",
"left_CA3", "right_CA3", "left_CA4", "right_CA4", "left_GC-ML-DG",
"right_GC-ML-DG")
# dependent variables are sf, independent variable common to all models in the inner lm() call is ICV
# applies the lm(subfield ~ ICV, dataset = DF) to all subfields of interest (sf) specified previously
lm.results <- lapply(sf, function(dv) {
temp.lm <- lm(get(dv) ~ ICV, data = DF)
coef(temp.lm)
})
# returns a list, where each element is a vector of coefficients
# do.call(rbind, ) will paste them together
lm.coef <- data.frame(sf = sf,
do.call(rbind, lm.results))
# tidy up name of intercept variable
names(lm.coef)[2] <- "intercept"
lm.coef
## set up all components for the equation
# matrix to store output
out <- matrix(ncol = length(sf), nrow = NROW(DF))
# name the rows after each subject
row.names(out) <- DF$Subject
# name the columns after each subfield
colnames(out) <- sf
# nested for loop that goes by subject (j) and subfield (i)
for(j in DF$Subject){
for (i in sf) {
slope <- lm.coef[lm.coef$sf == i, "ICV"]
out[j,i] <- as.numeric( DF[DF$Subject == j, i] - (slope * (DF[DF$Subject == j, "ICV"] - mean(DF$ICV))) )
}
}
# check output
out
===============
Original Question:
I have a dataframe (DF) with 13 columns (12 different brain subfields, and one column containing total intracranial volume(ICV)) and 50 rows (each a different participant). I'm trying to automate an equation being looped over every column for each participant.
The data:
structure(list(Subject = c("sub01", "sub02", "sub03", "sub04",
"sub05", "sub06", "sub07", "sub08", "sub09", "sub10", "sub11",
"sub12", "sub13", "sub14", "sub15", "sub16", "sub17", "sub18",
"sub19", "sub20"), ICV = c(1.50813, 1.3964237, 1.6703585, 1.4641886,
1.6351018, 1.5524641, 1.4445532, 1.6384505, 1.6152434, 1.5278011,
1.4788126, 1.4373356, 1.4109637, 1.3634952, 1.3853583, 1.4855268,
1.6082085, 1.5644998, 1.5617522, 1.4304141), left_subiculum = c(411.225013,
456.168033, 492.968477, 466.030173, 533.95505, 476.465524, 448.278213,
476.45566, 422.617374, 498.995121, 450.773906, 461.989663, 549.805272,
452.619547, 457.545623, 451.988333, 475.885847, 490.127968, 470.686415,
494.06548), left_CA1 = c(666.893596, 700.982955, 646.21927, 580.864234,
721.170599, 737.413139, 737.683665, 597.392434, 594.343911, 712.781376,
733.157168, 699.820162, 701.640861, 690.942843, 606.259484, 731.198846,
567.70879, 648.887718, 726.219904, 712.367433), left_presubiculum = c(325.779458,
391.252815, 352.765098, 342.67797, 390.885737, 312.857458, 326.916867,
350.657957, 325.152464, 320.718835, 273.406949, 305.623938, 371.079722,
315.058313, 311.376271, 319.56678, 348.343569, 349.102678, 322.39908,
306.966008), `left_GC-ML-DG` = c(327.037756, 305.63224, 328.945065,
238.920358, 319.494513, 305.153183, 311.347404, 259.259723, 295.369164,
312.022281, 324.200989, 314.636501, 306.550385, 311.399107, 295.108592,
356.197094, 251.098248, 294.76349, 317.308576, 301.800253), left_CA3 = c(275.17038,
220.862237, 232.542718, 170.088695, 234.707172, 210.803287, 246.861975,
171.90896, 220.83478, 236.600832, 246.842024, 239.677362, 186.599097,
224.362411, 229.9142, 293.684776, 172.179779, 202.18936, 232.5666,
221.896625), left_CA4 = c(277.614028, 264.575987, 286.605092,
206.378619, 281.781858, 258.517989, 269.354864, 226.269982, 256.384436,
271.393257, 277.928824, 265.051581, 262.307377, 266.924683, 263.038686,
306.133918, 226.364556, 262.42823, 264.862956, 255.673948), right_subiculum = c(468.762375,
445.35738, 446.536018, 456.73484, 521.041823, 482.768261, 487.2911,
456.39996, 445.392976, 476.146498, 451.775611, 432.740085, 518.170065,
487.642399, 405.564237, 487.188989, 467.854363, 479.268714, 473.212833,
472.325916), right_CA1 = c(712.973011, 717.815214, 663.637105,
649.614586, 711.844375, 779.212704, 862.784416, 648.925038, 648.180611,
760.761704, 805.943016, 717.486756, 801.853608, 722.213109, 621.676321,
791.672796, 605.35667, 637.981476, 719.805053, 722.348921), right_presubiculum = c(327.285242,
364.937865, 288.322641, 348.30058, 341.309111, 279.429847, 333.096795,
342.184296, 364.245998, 350.707173, 280.389853, 276.423658, 339.439377,
321.534798, 302.164685, 328.365751, 341.660085, 305.366589, 320.04127,
303.83284), `right_GC-ML-DG` = c(362.391907, 316.853532, 342.93274,
282.550769, 339.792696, 357.867386, 342.512721, 277.797528, 309.585721,
343.770416, 333.524912, 302.505077, 309.063135, 291.29361, 302.510461,
378.682679, 255.061044, 302.545288, 313.93902, 297.167161), right_CA3 = c(307.007404,
243.839349, 269.063801, 211.336979, 249.283479, 276.092623, 268.183349,
202.947849, 214.642782, 247.844657, 291.206598, 235.864996, 222.285729,
201.427853, 237.654913, 321.338801, 199.035108, 243.204203, 236.305659,
213.386702), right_CA4 = c(312.164065, 272.905586, 297.99392,
240.765062, 289.98697, 306.459566, 284.533068, 245.965817, 264.750571,
296.149675, 290.66935, 264.821461, 264.920869, 246.267976, 266.07378,
314.205819, 229.738951, 274.152503, 256.414608, 249.162404)), row.names = c(NA,
-20L), class = c("tbl_df", "tbl", "data.frame"))
The equation:
adjustedBrain(participant1) = rawBrain(participant1) - slope*[ICV(participant1) - (mean of all ICV measures included in the calculation of the slope)]
The code (which is not working and I was hoping for some pointers):
adjusted_Brain <- function(DF, subject) {
subfields <- colnames(select(DF, "left_presubiculum", "right_presubiculum",
"left_subiculum", "right_subiculum", "left_CA1", "right_CA1",
"left_CA3", "right_CA3", "left_CA4", "right_CA4", "left_GC-ML-DG",
"right_GC-ML-DG"))
out <- matrix(ncol = length(subfields), nrow = NROW(DF))
for (i in seq_along(subfields)) {
DF[i] = DF[DF$Subject == "subject", "i"] -
slope * (DF[DF$Subject == "subject", "ICV"] -
mean(DF$ICV))
}
}
Getting this error:
Error: Can't subset columns that don't exist.
x Column `i` doesn't exist.
A few notes:
The slopes for each subject for each subfield will be different (and will come from a regression) -> is there a way to specify that in the function so the slope (coefficient from the appropriate regression equation) gets called in?
I have my nrow set to the number of participants right now in the output because I'd like to have this run through EVERY subject across EVERY subfield and spit out a matrix with all the adjusted brain volumes... But that seems very complicated and so for now I will just settle for running each participant separately.
Any help is greatly appreciated!
As others have noted in the comments, there are quite a few syntax issues that prevent your code from running, as well as a few unstated requirements. That aside, I think there is enough to recommend a few improvements that you can hopefully build on. Here are the top line changes:
You likely don't need this to be a function, but rather a nested for loop (if you want to do this with base R). As written, the code isn't flexible enough to merit a function. If you intend to apply this many times across different datasets, a function might make sense. However, it will require a much larger rewrite.
Assuming you are fitting a simple regression via lm, then you can pull out the coefficient of interest via the $ operator and indexing (see below). Some thought will need to go into how to handle different models in the loop. Here, we assume you only need one coefficient from one model.
There are a few areas where the syntax is incorrect and a review of sub setting in base R would be helpful. Others have pointed out in the comments were some of these are.
Here is one approach were we loop through each subject (j) through each feature or subfield (i) and store them in a matrix (out). This is just an approach and will almost certainly need tweaking on your end!
#NOTE: the dataset your provided is saved as x in this example.
#fit a linear model - here we assume there is only one coef. of interest, but you may need to alter
# depending on how the slope changes in each calculation
reg <- lm(ICV ~ right_CA3, x)
# view the coeff.
reg$coefficients
# pull out the slope by getting the coeff. of interest (via index) from the reg object
slope <- reg$coefficients[[1]]
# list of features/subfeilds to loop through
sf <- c("left_presubiculum", "right_presubiculum",
"left_subiculum", "right_subiculum", "left_CA1", "right_CA1",
"left_CA3", "right_CA3", "left_CA4", "right_CA4", "left_GC-ML-DG",
"right_GC-ML-DG")
# matrix to store output
out <- matrix(ncol = length(sf), nrow = NROW(x))
#name the rows after each subject
row.names(out) <- x$Subject
#name the columns after each sub feild
colnames(out) <- sf
# nested for loop that goes by subject (j) and features/subfeilds (i)
for(j in x$Subject){
for (i in sf) {
out[j,i] <- as.numeric( x[x$Subject == j, i] - (slope * (x[x$Subject == j, "ICV"] - mean(x$ICV))) )
}
}
# check output
out
I am relatively beginner in R and trying to figure out how to use cpquery function for bnlearn package for all edges of DAG.
First of all, I created a bn object, a network of bn and a table with all strengths.
library(bnlearn)
data(learning.test)
baynet = hc(learning.test)
fit = bn.fit(baynet, learning.test)
sttbl = arc.strength(x = baynet, data = learning.test)
Then I tried to create a new variable in sttbl dataset, which was the result of cpquery function.
sttbl = sttbl %>% mutate(prob = NA) %>% arrange(strength)
sttbl[1,4] = cpquery(fit, `A` == 1, `D` == 1)
It looks pretty good (especially on bigger data), but when I am trying to automate this process somehow, I am struggling with errors, such as:
Error in sampling(fitted = fitted, event = event, evidence = evidence, :
logical vector for evidence is of length 1 instead of 10000.
In perfect situation, I need to create a function that fills the prob generated variable of sttbl dataset regardless it's size. I tried to do it with for loop to, but stumbled over the error above again and again. Unfortunately, I am deleting failed attempts, but they were smt like this:
for (i in 1:nrow(sttbl)) {
j = sttbl[i,1]
k = sttbl[i,2]
sttbl[i,4]=cpquery(fit, fit$j %in% sttbl[i,1]==1, fit$k %in% sttbl[i,2]==1)
}
or this:
for (i in 1:nrow(sttbl)) {
sttbl[i,4]=cpquery(fit, sttbl[i,1] == 1, sttbl[i,2] == 1)
}
Now I think I misunderstood something in R or bnlearn package.
Could you please tell me how to realize this task with filling the column by multiple cpqueries? That would help me a lot with my research!
cpquery is quite difficult to work with programmatically. If you look at the examples in the help page you can see the author uses eval(parse(...)) to build the queries. I have added two approaches below, one using the methods from the help page and one using cpdist to draw samples and reweighting to get the probabilities.
Your example
library(bnlearn); library(dplyr)
data(learning.test)
baynet = hc(learning.test)
fit = bn.fit(baynet, learning.test)
sttbl = arc.strength(x = baynet, data = learning.test)
sttbl = sttbl %>% mutate(prob = NA) %>% arrange(strength)
This uses cpquery and the much maligned eval(parse(...)) -- this is the
approach the the bnlearn author takes to do this programmatically in the ?cpquery examples. Anyway,
# You want the evidence and event to be the same; in your question it is `1`
# but for example using learning.test data we use 'a'
state = "\'a\'" # note if the states are character then these need to be quoted
event = paste(sttbl$from, "==", state)
evidence = paste(sttbl$to, "==", state)
# loop through using code similar to that found in `cpquery`
set.seed(1) # to make sampling reproducible
for(i in 1:nrow(sttbl)) {
qtxt = paste("cpquery(fit, ", event[i], ", ", evidence[i], ",n=1e6", ")")
sttbl$prob[i] = eval(parse(text=qtxt))
}
I find it preferable to work with cpdist which is used to generate random samples conditional on some evidence. You can then use these samples to build up queries. If you use likelihood weighting (method="lw") it is slightly easier to do this programatically (and without evil(parse(...))).
The evidence is added in a named list i.e. list(A='a').
# The following just gives a quick way to assign the same
# evidence state to all the evidence nodes.
evidence = setNames(replicate(nrow(sttbl), "a", simplify = FALSE), sttbl$to)
# Now loop though the queries
# As we are using likelihood weighting we need to reweight to get the probabilities
# (cpquery does this under the hood)
# Also note with this method that you could simulate from more than
# one variable (event) at a time if the evidence was the same.
for(i in 1:nrow(sttbl)) {
temp = cpdist(fit, sttbl$from[i], evidence[i], method="lw")
w = attr(temp, "weights")
sttbl$prob2[i] = sum(w[temp=='a'])/ sum(w)
}
sttbl
# from to strength prob prob2
# 1 A D -1938.9499 0.6186238 0.6233387
# 2 A B -1153.8796 0.6050552 0.6133448
# 3 C D -823.7605 0.7027782 0.7067417
# 4 B E -720.8266 0.7332107 0.7328657
# 5 F E -549.2300 0.5850828 0.5895373
I'm looking for a nicely formated markdown output of test results that are produced within a for loop and structured with headings. For example
df <- data.frame(x = rnorm(1000),
y = rnorm(1000),
z = rnorm(1000))
for (v in c("y","z")) {
cat("##", v, " (model 0)\n")
summary(lm(x~1, df))
cat("##", v, " (model 1)\n")
summary(lm(as.formula(paste0("x~1+",v)), df))
}
whereas the output should be
y (model 0)
Call:
lm(formula = x ~ 1, data = df)
Residuals:
Min 1Q Median 3Q Max
-3.8663 -0.6969 -0.0465 0.6998 3.1648
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.05267 0.03293 -1.6 0.11
Residual standard error: 1.041 on 999 degrees of freedom
y (model 1)
Call:
lm(formula = as.formula(paste0("x~1+", v)), data = df)
Residuals:
Min 1Q Median 3Q Max
-3.8686 -0.6915 -0.0447 0.6921 3.1504
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.05374 0.03297 -1.630 0.103
y -0.02399 0.03189 -0.752 0.452
Residual standard error: 1.042 on 998 degrees of freedom
Multiple R-squared: 0.0005668, Adjusted R-squared: -0.0004346
F-statistic: 0.566 on 1 and 998 DF, p-value: 0.452
z (model 0)
and so on...
There are several results discussing parts of the question like here or here suggesting the asis-tag in combination with the cat-statement. This one includes headers.
Closest to me request seems to be this question from two years ago. However, even though highly appreciated, some of suggestions are deprecated like the asis_output or I can't get them to work in general conditions like the formattable suggestion (e.g. withlm-output). I just wonder -- as two years have past since then -- if there is a modern approach that facilitates what I'm looking for.
Solution Type 1
You could do a capture.output(cat(.)) approach with some lapply-looping. Send the output to a file and use rmarkdown::render(.).
This is the R code producing a *.pdf.
capture.output(cat("---
title: 'Test Results'
author: 'Tom & co.'
date: '11 10 2019'
output: pdf_document
---\n\n```{r setup, include=FALSE}\n
knitr::opts_chunk$set(echo = TRUE)\n
mtcars <- data.frame(mtcars)\n```\n"), file="_RMD/Tom.Rmd") # here of course your own data
lapply(seq(mtcars), function(i)
capture.output(cat("# Model", i, "\n\n```{r chunk", i, ", comment='', echo=FALSE}\n\
print(summary(lm(mpg ~ ", names(mtcars)[i] ,", mtcars)))\n```\n"),
file="_RMD/Tom.Rmd", append=TRUE))
rmarkdown::render("_RMD/Tom.Rmd")
Produces:
Solution Type 2
When we want to automate the output of multiple model summaries in the rmarkdown itself, we could chose between 1. selecting chunk option results='asis' which would produce code output but e.g. # Model 1 headlines, or 2. to choose not to select it, which would produce Model 1 but destroys the code formatting. The solution is to use the option and combine it with inline code that we can paste() together with another sapply()-loop within the sapply() for the models.
In the main sapply we apply #G.Grothendieck's venerable solution to nicely substitute the Call: line of the output using do.call("lm", list(.)). We need to wrap an invisible(.) around it to avoid the unnecessary sapply() output [[1]] [[2]]... of the empty lists produced.
I included a ". " into the cat(), because leading white space like ` this` will be rendered to this in lines 6 and 10 of the summary outputs.
This is the rmarkdown script producing a *pdf that can also be executed ordinary line by line:
---
title: "Test results"
author: "Tom & co."
date: "15 10 2019"
output: pdf_document
---
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
# Overview
This is an example of an ordinary code block with output that had to be included.
```{r mtcars, fig.width=3, fig.height=3}
head(mtcars)
```
# Test results in detail
The test results follow fully automated in detail.
```{r mtcars2, echo=FALSE, message=FALSE, results="asis"}
invisible(sapply(tail(seq(mtcars), -2), function(i) {
fo <- reformulate(names(mtcars)[i], response="mpg")
s <- summary(do.call("lm", list(fo, quote(mtcars))))
cat("\n## Model", i - 2, "\n")
sapply(1:19, function(j)
cat(paste0("`", ". ", capture.output(s)[j]), "` \n"))
cat(" \n")
}))
```
***Note:*** This is a concluding remark to show that we still can do other stuff afterwards.
Produces:
(Note: Site 3 omitted)
Context
I was hit by the same need as that of OP when trying to generate multiple plots in a loop, but one of them would apparently crash the graphical device (because of unpredictable bad input) even when called using try() and prevent all the remaining figures from being generated. I needed really independent code blocks, like in the proposed solution.
Solution
I've thought of preprocessing the source file before it was passed to knitr, preferably inside R, and found that the jinjar package was a good candidate. It uses a dynamic template syntax based on the Jinja2 templating engine from Python/Django. There are no syntax clashes with document formats accepted by R Markdown, but the tricky part was integrating it nicely with its machinery.
My hackish solution was to create a wrapper rmarkdown::output_format() that executes some code inside the rmarkdown::render() call environment to process the source file:
preprocess_jinjar <- function(base_format) {
if (is.character(base_format)) {
base_format <- rmarkdown:::create_output_format_function(base_format)
}
function(...) {
# Find the markdown::render() environment.
callers <- sapply(sys.calls(), function(x) deparse(as.list(x)[[1]]))
target <- grep('^(rmarkdown::)?render$', callers)
target <- target[length(target)] # render may be called recursively
render_envir <- sys.frames()[[target]]
# Modify input with jinjar.
input_paths <- evalq(envir = render_envir, expr = {
original_knit_input <- sub('(\\.[[:alnum:]]+)$', '.jinjar\\1', knit_input)
file.rename(knit_input, original_knit_input)
input_lines <- jinjar::render(paste(input_lines, collapse = '\n'))
writeLines(input_lines, knit_input)
normalize_path(c(knit_input, original_knit_input))
})
# Add an on_exit hook to revert the modification.
rmarkdown::output_format(
knitr = NULL,
pandoc = NULL,
on_exit = function() file.rename(input_paths[2], input_paths[1]),
base_format = base_format(...),
)
}
}
Then I can call, for example:
rmarkdown::render('input.Rmd', output_format = preprocess_jinjar('html_document'))
Or, more programatically, with the output format specified in the source file metadata as usual:
html_jinjar <- preprocess_jinjar('html_document')
rmarkdown::render('input.Rmd')
Here is a minimal example for input.Rmd:
---
output:
html_jinjar:
toc: false
---
{% for n in [1, 2, 3] %}
# Section {{ n }}
```{r block-{{ n }}}
print({{ n }}**2)
```
{% endfor %}
Caveats
It's a hack. This code depends on the internal logic of markdown::render() and likely there are edge cases where it won't work. Use at your own risk.
For this solution to work, the output format contructor must be called by render(). Therefore, evaluating it before passing it to render() will fail:
render('input.Rmd', output_format = 'html_jinja') # works
render('input.Rmd', output_format = html_jinja) # works
render('input.Rmd', output_format = html_jinja()) # fails
This second limitation could be circumvented by putting the preprocessing code inside the pre_knit() hook, but then it would only run after other output format hooks, like intermediates_generator() and other pre_knit() hooks of the format.
I have 2 datasets not very different to each other. Each dataset has 27 rows of actual and forecast values. When tested against Solver in Excel for minimization of the absolute error (abs(actual - par * forecast) they both give nearly equal values for the parameter 'par'. However, when each of these data sets are passed on to the same optimization function that I have written, it only works for one of them. For the other data set, the objective always gets evaluated to zero (0) with'par' assisgned the upper bound value.
This is definitely incorrect. What I am not able to understand is why is R doing so?
Here are the 2 data sets :-
test
dateperiod,usage,fittedlevelusage
2019-04-13,16187.24,17257.02
2019-04-14,16410.18,17347.49
2019-04-15,18453.52,17246.88
2019-04-16,18113.1,17929.24
2019-04-17,17712.54,17476.67
2019-04-18,15098.13,17266.89
2019-04-19,13026.76,15298.11
2019-04-20,13689.49,13728.9
2019-04-21,11907.81,14122.88
2019-04-22,13078.29,13291.25
2019-04-23,15823.23,14465.34
2019-04-24,14602.43,15690.12
2019-04-25,12628.7,13806.44
2019-04-26,15064.37,12247.59
2019-04-27,17163.32,16335.43
2019-04-28,17277.18,16967.72
2019-04-29,20093.13,17418.99
2019-04-30,18820.68,18978.9
2019-05-01,18799.63,17610.66
2019-05-02,17783.24,17000.12
2019-05-03,17965.56,17818.84
2019-05-04,16891.25,18002.03
2019-05-05,18665.49,18298.02
2019-05-06,21043.86,19157.41
2019-05-07,22188.93,21092.36
2019-05-08,22358.08,21232.56
2019-05-09,22797.46,22229.69
Optimization result from R
$minimum
[1] 1.018188
$objective
[1] 28031.49
test1
dateperiod,Usage,fittedlevelusage
2019-04-13,16187.24,17248.29
2019-04-14,16410.18,17337.86
2019-04-15,18453.52,17196.25
2019-04-16,18113.10,17896.74
2019-04-17,17712.54,17464.45
2019-04-18,15098.13,17285.82
2019-04-19,13026.76,15277.10
2019-04-20,13689.49,13733.90
2019-04-21,11907.81,14152.27
2019-04-22,13078.29,13337.53
2019-04-23,15823.23,14512.41
2019-04-24,14602.43,15688.68
2019-04-25,12628.70,13808.58
2019-04-26,15064.37,12244.91
2019-04-27,17163.32,16304.28
2019-04-28,17277.18,16956.91
2019-04-29,20093.13,17441.80
2019-04-30,18820.68,18928.29
2019-05-01,18794.10,17573.40
2019-05-02,17779.00,16969.20
2019-05-03,17960.16,17764.47
2019-05-04,16884.77,17952.23
2019-05-05,18658.16,18313.66
2019-05-06,21036.49,19149.12
2019-05-07,22182.11,21103.37
2019-05-08,22335.57,21196.23
2019-05-09,22797.46,22180.51
Optimization result from R
$minimum
[1] 1.499934
$objective
[1] 0
The optimization function used is shown below :-
optfn <- function(x)
{act <- x$usage
fcst <- x$fittedlevelusage
fn <- function(par)
{sum(abs(act - (fcst * par)))
}
adjfac <- optimize(fn, c(0.5, 1.5))
return(adjfac)
}
adjfacresults <- optfn(test)
adjfacresults <- optfn(test1)
Optimization result from R
adjfacresults <- optfn(test)
$minimum
[1] 1.018188
$objective
[1] 28031.49
Optimization result from R
adjfacresults <- optfn(test1)
$minimum [1]
1.499934
$objective
[1] 0
Can anyone help to identify why is R not doing the same process over the 2 data sets and outputting the correct results in both the cases.
The corresponding results using Excel Solver for the 2 datasets are as follows :-
For 'test' data set
par value = 1.018236659
objective function valule (min) : 28031
For 'test1' data set
par value = 1.01881062927878
objective function valule (min) : 28010
Best regards
Deepak
That's because the second column of test1 is named Usage, not usage. Therefore, act = x$usage is NULL, and the function fn returns sum(abs(NULL - something)) = sum(NULL) = 0. You have to rename this column to usage.