I thought that match or which would be up to the task, but they appear to not be. For example, assume that I have the function divisibleByFive<-function(x){x%%5==0} and I want to find the first member of the vector (either its index or its value) 123:456 such that divisibleByFive returns true. Is there a single base function for this job, taking both divisibleByFive (possibly vectorized) and 123:456 as arguments? If not, what is the idiomatic way to solve these sorts of problems?
Yes, that function is Filter. Try this
Filter(divisibleByFive, 123:456)[[1L]]
However, I don't really recommend doing so as this function is too generic and thus a bit slow in practice. Usually, what you want can be easily achieved by something like this
(x <- 123:456)[[which(divisibleByFive(x))[[1L]]]]
, as pointed out by #SteveM. See the benchmark
Unit: microseconds
expr min lq mean median uq max neval cld
Filter(divisibleByFive, 123:456)[[1L]] 238.3 245.35 270.543 250.45 275.9 475.4 100 b
(x <- 123:456)[[which(divisibleByFive(x))[[1L]]]] 8.3 9.00 10.099 9.50 9.9 28.5 100 a
We can use which.max
x[which.max(divisibleByFive(x))]
#[1] 125
Find and Position.
divisibleByFive<-function(x){x%%5==0}
Find(divisibleByFive,123:456)#For the desired value.
Position(divisibleByFive,123:456)#For the index of the desired value
Output:
> Find(divisibleByFive,123:456)#For the desired value.
[1] 125
> Position(divisibleByFive,123:456)#For the index of the desired value
[1] 3
This question already has an answer here:
round number to the first 3 digits (start with digit != 0)
(1 answer)
Closed 3 years ago.
This is a really simple question but how do I round a number in R such that I only show 2 significant figures?
E.g. 326 rounds to 330 and 4999 rounds to 5000
Thanks
Use digits to indicate decimal places.
round(326, digits=-1)
[1] 330
Here is the difference between signif() and round(). Taken directly from documentation:
x2 <- pi * 100^(-1:3)
round(x2, 3)
signif(x2, 3)
[1] 0.031 3.142 314.159 31415.927 3141592.654
[1] 3.14e-02 3.14e+00 3.14e+02 3.14e+04 3.14e+06
Use the one that works for you.
Maybe this could help:
signif(4999,2)
5000
signif(326,2)
330
signif(326232,2)
330000
And as Jim O. pointed out, there is a difference between signif() and round(). Also in performances are different, due as pointed out by Gregor, this could be not too much useful to know but maybe interesting:
library(microbenchmark)
k <- sample(1:100000,1000000,replace=T)
microbenchmark(
round_ ={round(k, digits=-1)},
signif_ ={signif(k,2)}
)
Unit: milliseconds
expr min lq mean median uq max neval
round_ 68.56366 70.22595 74.02643 71.99918 75.32761 109.5727 100
signif_ 109.57957 111.86501 121.17458 114.13232 118.88837 495.0321 100
try this
round(3333,-1)
round(3333,-2)
and see what you get
I am a beginner and I would like to measure time of a spatial process and input it in a variable. Is there a way to do that with R. I have tried using library(tictoc) but I think the measurements are inaccurate when inputting them in a variable, because my time is 2 secs and using the toc() function I get the value 8320.
The microbenchmark package is pretty good if you just want to measure time for simple expressions. It is measuring time units much smaller than seconds and it will give you a data frame of times.
For example
> library(microbenchmark)
> (bench <- microbenchmark(mean(1:100), sum(1:100)/length(1:100)))
Unit: microseconds
expr min lq mean median uq max
mean(1:100) 3.771 3.9225 4.49793 4.013 4.1515 40.636
sum(1:100)/length(1:100) 1.023 1.1380 1.43525 1.217 1.3280 18.373
neval
100
100
will give you bench which is a data frame
> class(bench)
[1] "microbenchmark" "data.frame"
It contains the time measurements for all the runs. Use bench$expr to get the expression that was measured -- in this example the mean(1:100) or sum(1:100)/length(1:100) expression -- and bench$time gives you the time.
I got curious about the speed of string comparison in R, when's the right time to use != vs == and was wondering how much shortcutting they do.
If I have a vector with two levels, one which occurs frequently, and another which is rare, (trying to multiply my desired effect).
x <- sample(c('ALICE', 'HAL90000000000'), replace = TRUE, 1000, prob = c(0.05,0.95))
I would assume (if there is shortcutting) that the operation
x != 'ALICE'
would be considerably faster than:
x == 'HAL90000000000'
since to check equality in the latter case, I would assume I need to check every character, while the former would be invalidated by either the first or last character (depending on which side the algorithm checks)
but when I benchmark, it either does not seem to be the case (it was inconclusive in repeated trials, though with a very slight bias toward the == operation being faster ?!), or this isn't a fair trial:
> microbenchmark(x != 'ALICE', x == 'HAL90000000000')
Unit: microseconds
expr min lq mean median uq max neval
x != "ALICE" 4.520 4.5505 4.61831 4.5775 4.6525 4.970 100
x == "HAL90000000000" 3.766 3.8015 4.00386 3.8425 3.9200 13.766 100
Why is this?
EDIT:
I'm assuming it's because it's doing full string matching, but if so, is there a way to get R to optimize these ones? I don't get any gains from the obfuscation of the amount of time it takes to match long or short strings, no worries about passwords.
I am using R and attempting to recover frequencies (really, just a number close to the actual frequency) from a large number of sound waves (1000s of audio files) by applying Fast Fourier transforms to each of them and identifying the frequency with the highest magnitude for each file. I'd like to be able to recover these peak frequencies as quickly as possible. The FFT method is one method that I've learned about recently and I think it should work for this task, but I am open to answers that do not rely on FFTs. I have tried a few ways of applying the FFT and getting the frequency of highest magnitude, and I have seen significant performance gains since my first method, but I'd like to speed up the execution time much more if possible.
Here is sample data:
s.rate<-44100 # sampling frequency
t <- 2 # seconds, for my situation, I've got 1000s of 1 - 5 minute files to go through
ind <- seq(s.rate*t)/s.rate # time indices for each step
# let's add two sin waves together to make the sound wave
f1 <- 600 # Hz: freq of sound wave 1
y <- 100*sin(2*pi*f1*ind) # sine wave 1
f2 <- 1500 # Hz: freq of sound wave 2
z <- 500*sin(2*pi*f2*ind+1) # sine wave 2
s <- y+z # the sound wave: my data isn't this nice, but I think this is an OK example
The first method I tried was using the fpeaks and spec functions from the seewave package, and it seems to work. However, it is prohibitively slow.
library(seewave)
fpeaks(spec(s, f=s.rate), nmax=1, plot=F) * 1000 # *1000 in order to recover freq in Hz
[1] 1494
# pretty close, quite slow
After doing a bit more reading, I tried this next approach, wherein
spec(s, f=s.rate, plot=F)[which(spec(s, f=s.rate, plot=F)[,2]==max(spec(s, f=s.rate, plot=F)[,2])),1] * 1000 # again need to *1000 to get Hz
x
1494
# pretty close, definitely faster
After a bit more looking around, I found this approach to work reasonably well.
which(Mod(fft(s)) == max(abs(Mod(fft(s))))) * s.rate / length(s)
[1] 1500
# recovered the exact frequency, and quickly!
Here is some performance data:
library(microbenchmark)
microbenchmark(
WHICH.MOD = which(Mod(fft(s))==max(abs(Mod(fft(s))))) * s.rate / length(s),
SPEC.WHICH = spec(s,f=s.rate,plot=F)[which(spec(s,f=s.rate,plot=F)[,2] == max(spec(s,f=s.rate,plot=F)[,2])),1] * 1000, # this is spec from the seewave package
# to recover a number around 1500, you have to multiply by 1000
FPEAKS.SPEC = fpeaks(spec(s,f=s.rate),nmax=1,plot=F)[,1] * 1000, # fpeaks is from the seewave package... again, need to multiply by 1000
times=10)
Unit: milliseconds
expr min lq median uq max neval
WHICH.MOD 10.78 10.81 11.07 11.43 12.33 10
SPEC.WHICH 64.68 65.83 66.66 67.18 78.74 10
FPEAKS.SPEC 100297.52 100648.50 101056.05 101737.56 102927.06 10
Good solutions will be the ones that recover a frequency close (± 10 Hz) to the real frequency the fastest.
More Context
I've got many files (several GBs), each containing a tone that gets modulated several times a second, and sometimes the signal actually disappears altogether so that there is just silence. I want to identify the frequency of the unmodulated tone. I know they should all be somewhere less than 6000 Hz, but I don't know more precisely than that. If (big if) I understand correctly, I've got an OK approach here, it's just a matter of making it faster. Just fyi, I have no previous experience in digital signal processing, so I appreciate any tips and pointers related to the mathematics / methods in addition to advice on how better to approach this programmatically.
After coming to a better understanding of this task and some of the terminology involved, I came across some additional approaches, which I'll present here. These additional approaches allow for window functions and a lot more, really, and the fastest approach in my question does not. I also just sped things up a bit by assigning the result of some of the functions to an object and indexing the object instead of running the function again
#i.e.
(ms<-meanspec(s,f=s.rate,wl=1024,plot=F))[which.max(ms[,2]),1]*1000
# instead of
meanspec(s,f=s.rate,wl=1024,plot=F)[which.max(meanspec(s,f=s.rate,wl=1024,plot=F)[,2]),1]*1000
I have my favorite approach, but I welcome constructive warnings, feedback, and opinions.
microbenchmark(
WHICH.MOD = which((mfft<-Mod(fft(s)))[1:(length(s)/2)] == max(abs(mfft[1:(length(s)/2)]))) * s.rate / length(s),
MEANSPEC = (ms<-meanspec(s,f=s.rate,wl=1024,plot=F))[which.max(ms[,2]),1]*1000,
DFREQ.HIST = (h<-hist(dfreq(s,f=s.rate,wl=1024,plot=F)[,2],200,plot=F))$mids[which.max(h$density)]*1000,
DFREQ.DENS = (dens <- density(dfreq(s,f=s.rate,wl=1024,plot=F)[,2],na.rm=T))$x[which.max(dens$y)]*1000,
FPEAKS.MSPEC = fpeaks(meanspec(s,f=s.rate,wl=1024,plot=F),nmax=1,plot=F)[,1]*1000 ,
times=100)
Unit: milliseconds
expr min lq median uq max neval
WHICH.MOD 8.119499 8.394254 8.513992 8.631377 10.81916 100
MEANSPEC 7.748739 7.985650 8.069466 8.211654 10.03744 100
DFREQ.HIST 9.720990 10.186257 10.299152 10.492016 12.07640 100
DFREQ.DENS 10.086190 10.413116 10.555305 10.721014 12.48137 100
FPEAKS.MSPEC 33.848135 35.441716 36.302971 37.089605 76.45978 100
DFREQ.DENS returns a frequency value farthest from the real value. The other approaches return values close to the real value.
With one of my audio files (i.e. real data) the performance results are a bit different (see below). One potentially relevant difference between the data being used above and the real data used for the performance data below is that above the data is just a vector of numerics and my real data is stored in a Wave object, an S4 object from the tuneR package.
library(Rmpfr) # to avoid an integer overflow problem in `WHICH.MOD`
microbenchmark(
WHICH.MOD = which((mfft<-Mod(fft(d#left)))[1:(length(d#left)/2)] == max(abs(mfft[1:(length(d#left)/2)]))) * mpfr(s.rate,100) / length(d#left),
MEANSPEC = (ms<-meanspec(d,f=s.rate,wl=1024,plot=F))[which.max(ms[,2]),1]*1000,
DFREQ.HIST = (h<-hist(dfreq(d,f=s.rate,wl=1024,plot=F)[,2],200,plot=F))$mids[which.max(h$density)]*1000,
DFREQ.DENS = (dens <- density(dfreq(d,f=s.rate,wl=1024,plot=F)[,2],na.rm=T))$x[which.max(dens$y)]*1000,
FPEAKS.MSPEC = fpeaks(meanspec(d,f=s.rate,wl=1024,plot=F),nmax=1,plot=F)[,1]*1000 ,
times=25)
Unit: seconds
expr min lq median uq max neval
WHICH.MOD 3.249395 3.320995 3.361160 3.421977 3.768885 25
MEANSPEC 1.180119 1.234359 1.263213 1.286397 1.315912 25
DFREQ.HIST 1.468117 1.519957 1.534353 1.563132 1.726012 25
DFREQ.DENS 1.432193 1.489323 1.514968 1.553121 1.713296 25
FPEAKS.MSPEC 1.207205 1.260006 1.277846 1.308961 1.390722 25
WHICH.MOD actually has to run twice to account for the left and right audio channels (i.e. my data is stereo), so it takes longer than the output indicates. Note: I needed to use the Rmpfr library in order for the WHICH.MOD approach to work with my real data, as I was having problems with integer overflow.
Interestingly, FPEAKS.MSPEC performed really well with my data, and it seems to return a pretty accurate frequency (based on my visual inspection of a spectrogram). DFREQ.HIST and DFREQ.DENS are quick, but the output frequency isn't as close to what I judge is the real value, and both are relatively ugly solutions. My favorite solution so far MEANSPEC uses the meanspec and which.max. I'll mark this as the answer since I haven't had any other answers, but feel free to provide an other answer. I'll vote for it and maybe select it as the answer if it provides a better solution.