I have a vector file with 1000 values. All the values were generated using Random function between 0-1.
x <- runif(100,min=0,max=1)
x
[1] 0.84620011 0.82525410 0.31622827 0.08040362 0.12894525 0.23997187 0.57177296 0.91691368 0.65751720
[10] 0.39810175 0.60632205 0.26339035 0.93543618 0.09662383 0.35147739 0.51731042 0.29151612 0.54411769
[19] 0.73688309 0.26086586 0.37808273 0.19163366 0.62776847 0.70973345 0.31802726 0.69101574 0.50042561
[28] 0.20768256 0.23555818 0.21015820 0.18221151 0.85593725 0.12916935 0.52222127 0.62269135 0.51267707
[37] 0.60164023 0.30723904 0.81990231 0.61771762 0.02502631 0.47427724 0.21250040 0.88611710 0.88648546
[46] 0.92586513 0.57015942 0.33454379 0.03572245 0.68120369 0.48692522 0.76587764 0.55214917 0.31137200
[55] 0.47170307 0.48639510 0.68922858 0.73506033 0.23541740 0.81793240 0.17184666 0.06670039 0.55664270
[64] 0.10030533 0.94620061 0.58572228 0.53333567 0.80887841 0.55015406 0.82491114 0.81251132 0.06038019
[73] 0.10918904 0.84011824 0.33169617 0.03568364 0.07703029 0.15601158 0.31623253 0.25021777 0.77024833
[82] 0.88588620 0.49044305 0.10165930 0.55494697 0.17455070 0.94458467 0.43135868 0.99313733 0.04482747
[91] 0.53453604 0.52500493 0.35496966 0.06994880 0.11377845 0.71307042 0.35086237 0.04032254 0.23744845
[100] 0.81131033
Out of all these values in the vector, I need to find the most occurring value(Or close to that). I'm new to R and have no idea what this. Please help?
One approach I have - Divide all the values in a certain ranges and find the frequency distribution. But will it be helpful?
One possibility to analyze the distribution of the numbers could consist in plotting a histogram and adding an approximate probability density distribution.
This can be done with the ggplot2 library:
set.seed(123) # used here for reproducibility
x <- runif(100) # pseudo-random numbers between 0 and 1
library(ggplot2)
p <- ggplot(as.data.frame(x),aes(x=x, y=..density..)) +
geom_histogram(fill="lightblue",colour="grey60",bins=50) +
geom_density()
The value of bins specified in geom_histigram() is the number of bars in the histogram. You may want to try to change this value to obtain a different representation of the distribution.
OR
You could use base Rand plot a simple histogram:
hist(x)
There you can also change the bin width (see breaks), but the default might be sufficient to show the concept.
You can identify which bin in this histogram has the most entries with
> hist(x)$mids[which.max(hist(x)$counts)]
#[1] 0.45
Which in this case means that most values occur near a value of 0.45 (the middle of the bin describing the range between 0.4 and 0.5).
Hope this helps.
You can do this:
set.seed(12)
x <- runif(100,min=0,max=1)
n <- length(x)
x_cut<-cut(x, breaks = n/4)
which(table(x_cut)==max(table(x_cut)))
The result depends on the breaks value you set. This is an alternative to using a histogram if you don't need one.
To really get just the most occurrent value, or when using discrete data as input, you could simply create a table, sort the results and return the highest value:
values <- c("a", "a", "c", "c", "c")
names(sort(table(values), decreasing = TRUE)[1])
#> [1] "c"
Breaking it down:
# create a table of the values
table(values)
#> a c
#> 2 3
# sort the table descending on number of occurrences
sort(table(values), decreasing = TRUE)
#> c a
#> 3 2
# now only keep the first value
sort(table(values), decreasing = TRUE)[1]
#> c
#> 3
# so the final line:
names(sort(table(values), decreasing = TRUE)[1])
#> [1] "c"
If you're feeling like wanting to do fancy stuff, create a function that does this for you:
get_mode <- function(x) {
names(sort(table(values), decreasing = TRUE)[1])
}
get_mode(values)
#> [1] "c"
Related
I have the following code:
set.seed(1)
jitter(rep(0, 7))
#> [1] -0.009379653 -0.005115044 0.002914135 0.016328312 -0.011932723
#> [6] 0.015935587 0.017787011
Notice that the function jitter add negative values. How can I
parameterize the function so that it will return only positive values?
If you only want a positive jitter, then perhaps you could just take the absolute value of your call:
x <- rep(0, 7)
abs(jitter(x))
[1] 0.009379653 0.005115044 0.002914135 0.016328312 0.011932723 0.015935587
[7] 0.017787011
I am trying to generate 100 samples of Z, where Z is the summation of 8 independent uniformly distributed random variables in the interval [0;1]
I have the following code so far, but I'm not sure if it's correct. I am not sure if my loop is correct
eight<-runif(8,0,1) #Uses the runif function to generate 8 uniform 0-1 random variables
Z_1<-sum(eight) #caclulates the sum and stores it in the variable Z_1
sample <-NA
for (i in 1:100 ) { #Function continues the loop for 100 different values
eight<-runif(8,0,1); #Creates sum loop for 8 independent values uniform 0-1 random variables.
Z_1<-sum(eight); # stores in the sum loop in the variable Z
sample[i] = Z_1;
}`
Thanks
I would vectorize the whole thing. There is no real reason to run 100 iterations when you can just generate 800 observations in one run. Then just use matrix and colSums and you done
set.seed(123)
n <- 100
Z <- colSums(matrix(runif(8 * n), 8, n))
Z
# [1] 4.774425 4.671171 4.787691 4.804041 3.513257 2.330163 3.300135 3.568657 5.503481 2.861764 4.533283 3.353658
# [13] 4.230073 4.690739 4.364708 3.094156 4.933616 3.942834 3.712522 2.587036 3.731474 4.388749 4.484030 4.315968
# [25] 4.800758 4.252631 2.716972 5.159044 4.146881 3.244546 4.418618 4.350035 5.344182 3.176801 3.903337 2.261935
# [37] 3.646572 4.286075 3.074900 4.210506 3.172513 4.535665 4.245856 4.184848 4.532286 2.899883 4.473629 4.751224
# [49] 3.498038 3.337437 4.238989 3.349812 3.888696 4.748254 3.029479 4.246619 3.330434 3.879168 3.786216 3.839956
# [61] 3.878997 4.546531 2.863010 3.417072 4.266108 3.141875 4.960758 3.989613 4.373042 4.295742 4.459014 5.561066
# [73] 4.401990 4.121301 3.830575 3.412446 3.812347 5.591238 3.801587 4.454336 4.213343 5.222007 4.300991 2.765003
# [85] 3.293251 5.362586 2.954080 3.036312 3.655677 3.373603 5.575184 4.167740 3.904038 3.884440 2.901452 3.079311
# [97] 4.927770 3.930943 4.169907 2.922618
Suppose I have data like the following:
val <- .65
set.seed(1)
distr <- replicate(1000, jitter(.5, amount = .2))
d <- density(distr)
Since stats::density uses a specific bw, it does not include all possible values in the interval (becuase they're infinite):
d$x[ d$x > .64 & d$x < .66 ]
[1] 0.6400439 0.6411318 0.6422197 0.6433076 0.6443955 0.6454834 0.6465713 0.6476592 0.6487471
[10] 0.6498350 0.6509229 0.6520108 0.6530987 0.6541866 0.6552745 0.6563624 0.6574503 0.6585382
[19] 0.6596261
I would like to find a way to provide val to the density function, so that it will return its d$y estimate (I will then use it to color areas of the density plot).
I can't guess how silly this question is, but I can't find a fast solution.
I thought of obtaining it by a linear interpolation of the d$y corresponding to the two values of d$x that are closer to val. Is there a faster way?
This illustrates the use of approxfun:
> Af <- approxfun(d$x, d$y)
> Af(val)
[1] 2.348879
> plot(d(
+
> plot(d)
> points(val,Af(val) )
> png();plot(d); points(val,Af(val) ); dev.off()
I just discovered R and I am trying to work with it.
Here is what I am trying to achieve:
I have a vector of numbers, x, between 50 and 100 and with a size of 250 observations.
x = sample(seq(50, 100), 250, repeat = T)
Now, I want to generate another vector of numbers, y, between 0 and 100, which is the same size as vector x such that each element in y is less than or equal to its equivalent in x.
That is to say that if x[1] is 76, for example, the highest value y[1] could attain when generated is 76. But it could definitely be any other value below 76. In other words and more generally, I want vector y to be generated in such a way that y[i] <= x[i].
I hope I have made my request clearer.
Thank you very much!
y <- x -1 # ...........................
y <- sapply( x, function(x) runif(n=1, max=x))
y
[1] 7.2713788 30.0008063 42.5205775 0.9271717 10.7114456 39.5199145 7.4109775
[8] 28.3464373 28.5840101 34.0654033 15.0675028 50.2836294 45.9031794 13.5931005
[15] 43.2751738 17.0560824 3.1507491 25.7619129 12.3391448 22.6203684 51.3334810
[22] 37.0481703 33.4733277 37.1304850 26.7984406 66.3844126 40.2775918 47.6379024
[29] 16.2480595 66.8358384 33.3513161 60.2673874 65.6204462 45.6951960 1.5729434
[36] 20.4850357 0.1345737 84.5334203 19.7997451 53.8025623 48.5528486 8.8992123
[43] 90.9651742 28.3584167 41.7728159 46.4790641 17.8129578 83.1906415 37.5114353
[50] 89.5685501 85.2499600
Suppose if I have a random time series that I want to interpolate over another time series. How would I do this in R?
# generate time interval series from exponential distribution
s = sort(rexp(10))
# scale between 0 and 1
scale01 = function(x){(x-min(x))/(max(x)-min(x))}
s = scale01(s)
> s
[1] 0.00000000 0.02804113 0.05715588 0.10630185 0.15778932 0.20391987 0.26066608 0.27265697 0.39100373
[10] 1.00000000
# generate random normal series
x = rnorm(20)
> x
[1] -0.82530658 0.92289557 0.39827984 -0.62416117 -1.69055539 -0.28164232 -1.32717654 -1.36992509
[9] -1.54352202 -1.09826247 -0.68260576 1.07307043 2.35298180 -0.41472811 0.38919315 -0.27325343
[17] -1.52592682 0.05400849 -0.43498544 0.73841106
# interpolate 'x' over 's' ?
> approx(x,xout=s)
$x
[1] 0.00000000 0.02804113 0.05715588 0.10630185 0.15778932 0.20391987 0.26066608 0.27265697 0.39100373
[10] 1.00000000
$y
[1] NA NA NA NA NA NA NA NA NA
[10] -0.8253066
>
I want to interpolate the series 'x' over the series 's'. Lets assume time interval series for the 'x' series has 20 elements distributed uniformly over the interval [0,1]. Now I want to interpolate those 10 elements from 'x' that occur at time intervals described by 's'.
EDIT:
I think this does the job.
> approx(seq(0,1,length.out=20), x, xout=s)
$x
[1] 0.00000000 0.02804113 0.05715588 0.10630185 0.15778932 0.20391987 0.26066608 0.27265697 0.39100373
[10] 1.00000000
$y
[1] -0.8253066 0.1061033 0.8777987 0.3781018 -0.6221134 -1.5566990 -0.3483466 -0.4703429 -1.4444105
[10] 0.7384111
Thanks for your help guys. I think I now understand how to use interpolation functions in R now. I should really use a time series data structure here.
This isn't meant as a direct answer to the OP's Q but rather to illustrate how approx() works so the OP can formulate a better Q
Your Q makes next to no sense. approx() works by taking a reference set of x, and y coordinates and then interpolating to find y at n locations over the range of x, or at the specified xout locations supplied by the user.
So in your call, you don't provide y and x doesn't contain a y component so I don't see how this can work.
If you want to interpolate s, so you can find time intervals for any value over range of s then:
> approx(s, seq_along(s), n = 20)
$x
[1] 0.00000000 0.05263158 0.10526316 0.15789474 0.21052632 0.26315789
[7] 0.31578947 0.36842105 0.42105263 0.47368421 0.52631579 0.57894737
[13] 0.63157895 0.68421053 0.73684211 0.78947368 0.84210526 0.89473684
[19] 0.94736842 1.00000000
$y
[1] 1.00000 26.25815 42.66323 54.79831 64.96162 76.99433 79.67388
[8] 83.78458 86.14656 89.86223 91.98513 93.36233 93.77353 94.19731
[15] 94.63652 95.26239 97.67724 98.74056 99.40548 100.00000
Here $y contains the interpolated values for s at n = 20 equally spaced locations on the range of s (0,1).
Edit: If x represents the series at unstated time intervals uniform on 0,1 and you want the interpolated values of x at the time intervals s, then you need something like this:
> set.seed(1)
> x <- rnorm(20)
> s <- sort(rexp(10))
> scale01 <- function(x) {
+ (x - min(x)) / (max(x) - min(x))
+ }
> s <- scale01(s)
>
> ## interpolate x at points s
> approx(seq(0, 1, length = length(x)), x, xout = s)
$x
[1] 0.00000000 0.04439851 0.11870795 0.14379236 0.20767388 0.21218632
[7] 0.25498856 0.29079300 0.40426335 1.00000000
$y
[1] -0.62645381 0.05692127 -0.21465011 0.94393053 0.39810806 0.29323742
[7] -0.64197207 -0.13373472 0.62763207 0.59390132
Is that closer to what you want?