I'm setting up a script to extract the thickness and voltages from a single column text file and perform a Weibull distribution on it. When I try to use fitdistr() I get an error stating "'x' must be a non-empty numeric vector". R is supposed to interpret numbers in text files as numeric but that doesn't seem to be happening. Any thoughts?
filename <- "SampleBreakdownSet.txt"
d <- read.table(filename, header = FALSE, sep = "")
#Extract thickness from the dataset; set to variable t
t = d[1,1]
#Extract the breakdown voltages and toss into dataset, BDV
BDV = tail(d,(nrow(d)-1))
#Calculates the breakdown field from the thickness and BDV
BDF = (BDV*10000) / t
#Calculates the Weibull parameters from the input breakdown voltages.
fitdistr(BDF, densfun ="weibull", lower = 0)
fitdistr(BDF, densfun ="weibull", lower = 0)
Error in fitdistr(BDF, densfun = "weibull", lower = 0) :
'x' must be a non-empty numeric vector
Sample data I'm using:
2
200
250
450
320
100
400
200
403
502
203
420
120
342
304
253
423
534
534
243
253
423
123
433
534
234
633
432
342
543
532
123
453
231
532
342
213
243
You are passing a data.frame to fitdistr, but you should be passing the vector itself.
Try this:
d <- read.table(text='200
250
450
320
100
400
200
403
502
203
420
120
342
304
253
423
534
534
243
253
423
123
433
534
234
633
432
342
543
532
123
453
231
532
342
213
243', header=FALSE)
t <- d[1,1]
#Extract the breakdown voltages and toss into dataset, BDV
BDV <- d[-1, 1]
BDF <- (BDV*10000) / t
library(MASS)
fitdistr(BDF, densfun ="weibull", lower = 0)
You could also refer to the relevant column when calling fitdistr, e.g.:
fitdistr(BDF$V1, densfun ="weibull", lower = 0)
# shape scale
# 2.745485e+00 1.997509e+04
# (3.716797e-01) (1.283667e+03)
Related
library(MatchIt)
df <- data.frame(lalonde)
m.out1 <- matchit(treat ~ age + race + educ, data = lalonde,
method = "exact")
m.data1<-match.data(m.out1)
I would like to know how I can get the same size for both the control and treatment samples after running an exact matching with MatchIt package. Ideally, I would like to randomly pick a control if a treated unit has been matched to more than one control.
My real dataset is not lalonde. It is actually an extremely large one. So I might have many controls associated with a treated unit and I want to draw one randomly for each treated unit.
For exact matching you could use this code.
library(Matching)
data("lalonde")
Y <- lalonde$re78
Tr <- lalonde$treat
X <- lalonde[setdiff(names(lalonde), c('re78', 'treat'))]
set.seed(42) ## comment out for FIXING the ties
rmtch <- Match(Y=Y, Tr=Tr, X=X, exact=TRUE, ties=FALSE)
summary(rmtch)
# Estimate... 1678.6
# SE......... 981
# T-stat..... 1.7111
# p.val...... 0.087055
#
# Original number of observations.............. 445
# Original number of treated obs............... 185
# Matched number of observations............... 55
# Matched number of observations (unweighted). 55
#
# Number of obs dropped by 'exact' or 'caliper' 130
str(rmtch) ## what is stored in Match object
rmtch$index.control ## indices of control units
# [1] 261 254 188 279 288 317 323 280 186 311 305 234 337 302 219 345 234 328
# [19] 271 218 253 249 339 271 339 344 351 253 328 339 255 217 254 197 254 284
# [37] 266 252 253 280 208 226 209 354 204 282 350 296 202 247 219 330 347 280
# [55] 344
If you re-run the code, you will see that the IDs change slightly, which they would probably do more clearly if the dataset was larger.
To fix the randomization of the control units you may use set.seed(). For handling ties deterministically use ties=FALSE (see ?Match help page).
The easiest way is to do 1:1 nearest neighbor matching with exact matching constraints:
m.out1 <- matchit(treat ~ age + race + educ, data = lalonde,
method = "nearest",
exact = ~ age + race + educ)
If you are doing coarsened exact matching, there is an option already built in to request this which is by setting k2k = TRUE:
m.out1 <- matchit(treat ~ age + race + educ, data = lalonde,
method = "cem", k2k = TRUE,
cutpoints = 0)
Setting cutpoints = 0 requests exact matching (no coarsening).
I have a data named 'bicoal' which consists of annual bituminous coal production in the United States from 1920 to 1968.
`Time Series:
Start = 1920
End = 1968
Frequency = 1
[1] 569 416 422 565 484 520 573 518 501 505 468 382 310 334 359 372 439 446 349 395
[21] 461 511 583 590 620 578 534 631 600 438 516 534 467 457 392 467 500 493 410 412
[41] 416 403 422 459 467 512 534 552 545`
I made a time series, saved under the name time_series, and wanted to extract the seasonal effect using the code plot(decompose(time_series)) and plot(stl(time_series)), but got an error message
Error in stl(time_series) :
series is not periodic or has less than two periods
Error in decompose(time_series) :
time series has no or less than 2 periods
If stl nor decompose doesn't work, is there a way to extract the seasonal effect?
Without seeing how your time series is constructed I think this might be your problem.
data <- rep(seq(1,5),5)
ts.1 <- ts(data)
stl(ts.1)
Now to fix this issue the ts function has a frequency argument that defines the period of the data.
ts.2 <- ts(data, frequency = 5)
stl(ts.2, s.window = "periodic")
I am trying to get better with functions in R and I was working on a function to pull out every odd value from 100 to 500 that was divisible by 3. I got close with the function below. It keeps returning all of the values correctly but it also includes the first number in the sequence (101) when it should not. Any help would be greatly appreciated. The code I wrote is as follows:
Test=function(n){
if(n>100){
s=seq(from=101,to=n,by=2)
p=c()
for(i in seq(from=101,to=n,by=2)){
if(any(s==i)){
p=c(p,i)
s=c(s[(s%%3)==0],i)
}}
return (p)}else{
stop
}}
Test(500)
Here is a function that gets all non even multiples of 3. It's fully vectorized, no loops at all.
Check if n is within the range [100, 500].
Create an integer vector N from 100 to n.
Create a logical index of the elements of N that are divisible by 3 but not by 2.
Extract the elements of N that match the index i.
The main work is done in 3 code lines.
Test <- function(n){
stopifnot(n >= 100)
stopifnot(n <= 500)
N <- seq_len(n)[-(1:99)]
i <- ((N %% 3) == 0) & ((N %% 2) != 0)
N[i]
}
Test(500)
Here is a vectorised one-liner which optionally allows you to change the lower bound from a default of 100 to anything you like. If the bounds are wrong, it returns an empty vector rather than throwing an error.
It works by creating a vector of 1:500 (or more generally, 1:n), then testing whether each element is greater than 100 (or whichever lower bound m you set), AND whether each element is odd AND whether each element is divisible by 3. It uses the which function to return the indices of the elements that pass all the tests.
Test <- function(n, m = 100) which(1:n > m & 1:n %% 2 != 0 & 1:n %% 3 == 0)
So you can use it as specified in your question:
Test(500)
# [1] 105 111 117 123 129 135 141 147 153 159 165 171 177 183 189 195 201 207 213 219
# [21] 225 231 237 243 249 255 261 267 273 279 285 291 297 303 309 315 321 327 333 339
# [41] 345 351 357 363 369 375 381 387 393 399 405 411 417 423 429 435 441 447 453 459
# [61] 465 471 477 483 489 495
Or play around with upper and lower bounds:
Test(100, 50)
# [1] 51 57 63 69 75 81 87 93 99
Here is a function example for your objective
Test <- function(n) {
if(n<100 | n> 500) stop("out of range")
v <- seq(101,n,by = 2)
na.omit(ifelse(v%%2==1 & v%%3==0,v,NA))
}
stop() is called when your n is out of range [100,500]
ifelse() outputs desired odd values + NA
na.omit filters out NA and produce the final results
I have a set of non linear data. The data is the X & Y coordinates of different objects/points in a video( that is the x&y pixel co-ordinates of same objects in all the frames in a video.) upon plotting the values in one frame, I am getting a nonlinear graph as shown in the picture.
I want to form an equation for this graph so that, if I have a known X coorrdinate in this frame, then the corresponding Y coordinate can be obtained using this equation.(kind of predicting the new position, I am not sure this idea is correct or not)
OR
If this idea is illogical, can you suggest something that will work so that I can predict the location of new object using these data.
Any help or new ideas is highly appreciated.
A sample of my data is given below:
X Y
----------
214 182
830 185
1451 173
219 554
1453 548
214 941
830 934
1455 942
213 190
829 193
1450 181
218 561
1452 555
214 945
830 938
1455 946
213 190
828 193
1451 182
219 560
1452 554
214 945
830 938
1455 946
213 190
829 193
1450 181
219 556
1453 550
215 936
830 929
1455 937
I have selected 9 objects in each frame, so the first 9 data set belongs to one frame, and so on..
Your XY data looks like this:
There are clusters located on corners and mid-edges.
and when the lines that connect successive points are added
The points should come in groups of 8, in the sequence shown above. You can predict the location of a point using the index
// predict location `(x,y)` of point based on index `i`
point = MOD(i-1,8)+1; // get number 1-8 of the point (as shown above)
select case point
case [1,4,6] : x = 215;
case [2,7] : x = 829;
case [3,5,8] : x = 1463;
end select
select case point
case [1,2,3] : y = 186;
case [4,5] : y = 555;
case [6,7,8] : y = 940;
end select
You have to cut this curve in lot of linear lines, so following the value of X, you will be on linear line and its easy to calculate the equation of line knowing 2 points of this line
So I have a corpus setup reading bunch of text file with paragraphs in them.
library('tm')
my.text.location <- "C:/Users//.../*/"
apapers <- VCorpus(DirSource(my.text.location))
Now I need to find the mean of the characters in each text. Running a
mean(nchar(apapers), na.rm =T) results in a very weird output, more than the number of characters.
Any other way to get the mean?
You didn't supply a reproducible example, but rowMeans(sapply(apapers, nchar)) will return the mean number of characters over all documents. "Content" is the column you need.
A longer version is running a sapply over the corpus counting the number of per document. Transpose this data and turn it into a data.frame. The data.frame will contain two columns, content and meta. Content is the one you need. Taking the mean of the content column will give you the average number of characters in a document. The advantage of this is that you have the table in case you need to report the numbers.
# your code
my_count <- data.frame(t(sapply(apapers, nchar)))
mean(my_count$content)
Reproducible example using the crude dataset:
library(tm)
data("crude")
crude <- as.VCorpus(crude)
# in one statement
rowMeans(sapply(crude, nchar))
content meta
1220.30 453.15
# longer version keeping intermediate results.
my_count <- data.frame(t(sapply(crude, nchar)))
mean(my_count$content)
[1] 1220.3
my_count
content meta
127 527 440
144 2634 458
191 330 444
194 394 441
211 552 441
236 2774 455
237 2747 477
242 930 453
246 2115 440
248 2066 466
273 2241 458
349 593 492
352 621 468
353 591 445
368 629 440
489 876 445
502 1166 446
543 463 447
704 1797 456
708 360 451