I have a vector that contains fractional numbers:
a<-c(0.5,0.5,0.3,0.5,0.2)
I would like to determine the most frequent (i.e. majority) number in the vector and return that number.
table(a) doesn't work because it will return the whole table. I want it to return only 0.5.
In case of ties I would like to choose randomly.
I have a function that does this for integers:
function(x){
a<-tabulate(x,nbins=max(x)); b<-which(a==max(a))
if (length(b)>1) {a<-sample(b,1)} else{b}
}
However, this won't work for fractions.
Can someone help?
You can use
names(which.max(table(a)))
If you want the numeric one as in your case, then coerce it to numeric
as.numeric(names(which.max(table(a))))
To randomize the tie case, you can add randomize the table
as.numeric(names(which.max(sample(table(a))))) #note this works only if length(unique(a)) > 1
Related
I have a long vector of numbers that vary in the their sign (e.g.):
data <- c(1,-23,67,-21,10,32,64,-34,-6,10)
Working in R, how do I create a new vector that contains the same list of numbers, but give them a random sign (either positive or negative)? For each number, the probability of it being negative should be 0.5.
There are a bunch of options but
sample(c(-1,1), size=length(data), replace=TRUE) * abs(data)
should work. You could also multiply by sign(runif(length(data))-0.5) or sign(runif(length(data),-1,1)) [either of which should be a little more efficient than sample(), although in this case it hardly matters].
The problem
I would like to find a length of a list.
The expected output
I would like to find the length based on a condition.
Example
Suppose that I have a list of 4 elements as follows:
myve <–list(1,2,3,0)
Here I have 4 elements, one of them is zero. How can I find the length by extracting the zero values? Then, if the length is > 1I would like to substruct one. That is:
If the length is 4 then, I would like to have 4-1=3. So, the output should be 3.
Note
Please note that I am working with a problem where the zero values may be changed from one case to another. For example, For the first list may I have only one 0 value, while for the second list may I have 2 or 3 zero values.
The values are always positive or zero.
You just need to apply the condition to each element. This will produce a list of boolean, then you sum it to get the number of True elements (i.e. validation your condition).
In your case:
sum(myve != 0)
In a more complex case, where the confition is expressed by a function f:
sapply(myve, f)
Use sapply to extract the ones different to zeros and sum to count them
sum(sapply(myve, function(x) x!=0))
I would like to create a function that looks at a column of values. from those values look at each value individually, and asses which of the other data points value is closest to that data point.
I'm guessing it could be done by checking the length of the data frame, making a list of the respective length in steps of 1. Then use that list to reference which cell is being analysed against the rest of the column. though I don't know how to implement that.
eg.
data:
20
17
29
33
1) is closest to 2)
2) is closest to 1)
3) is closest to 4)
4) is closest to 3)
I found this example which tests for similarity but id like to know what letter is assigns to.
x=c(1:100)
your.number=5.43
which(abs(x-your.number)==min(abs(x-your.number)))
Also if you know how I could do this, could you expain the parts of the code and what they mean?
I wrote a quick function that does the same thing as the code you provided.
The code you provided takes the absolute value of the difference between your number and each value in the vector, and compares that the minimum value from that vector. This is the same as the which.min function that I use below. I go through my steps below. Hope this helps.
Make up some data
a = 1:100
yourNumber = 6
Where Num is your number, and x is a vector
getClosest=function(x, Num){
return(which.min(abs(x-Num)))
}
Then if you run this command, it should return the index for the value of the vector that corresponds to the closest value to your specified number.
getClosest(x=a, Num=yourNumber)
Hello I am new to R and I can't find the way to do exactly what I want to. I have a vector of x numbers, and what i want to do is order it in increasing order, and then start making subtractions like this (let's say the vecto has 100 numbers for example):
[x(100)-x(99)]+[x(99)-x(98)]+[x(98)-x(97)]+[x(97)-x(96)]+...[x(2)-x(1)]
and then divide all that sum by the number of elements the vector has, in this case 100.
The only thing that I am able to do at the moment is order the vector with:
sort(nameOfTheVector)
Sorry for my bad English.
diff returns suitably lagged and iterated differences. In your case you want the default single lag. sum will return the sum any arguments passed to it, so....
sum(diff(sort(nameOfTheVector))) / length(nameOfTheVector)
My question is: Suppose you have computed an algorithm that gives the number of iterations and you would like to print the number of iterations out. But the output always many decimal places, like the following:
64.00000000
Is it possible to get an integer by doing type casting in R ? How would you do it ??
There are some gotchas in coercing to integer mode. Presumably you have a variety of numbers in some structure. If you are working with a matrix, then the print routine will display all the numbers at the same precision. However, you can change that level. If you have calculated this result with an arithmetic process it may be actually less than 64 bit display as that value.
> 64.00000000-.00000099999
[1] 64
> 64.00000000-.0000099999
[1] 63.99999
So assuming you want all the values in whatever structure this is part of, to be displayed as integers, the safest would be:
round(64.000000, 0)
... since this could happen, otherwise.
> as.integer(64.00000000-.00000000009)
[1] 63
The other gotcha is that the range of value for integers is considerably less than the range of floating point numbers.
The function is.integer can be used to test for integer mode.
is.integer(3)
[1] FALSE
is.integer(3L)
[1] TRUE
Neither round nor trunc will return a vector in integer mode:
is.integer(trunc(3.4))
[1] FALSE
Instead of trying to convert the output into an integer, find out why it is not an integer in the first place, and fix it there.
Did you initialize it as an integer, e.g. num.iterations <- 0L or num.iterations <- integer(1) or did you make the mistake of setting it to 0 (a numeric)?
When you incremented it, did you add 1 (a numeric) or 1L (an integer)?
If you are not sure, go through your code and check your variable's type using the class function.
Fixing the problem at the root could save you a lot of trouble down the line. It could also make your code more efficient as numerous operations are faster on integers than numerics (an example).
The function as.integer() truncate the number up to 0 order, so you must add a 0.5 to get a proper approx
dd<-64.00000000
as.integer(dd+0.5)
If you have a numeric matrix you wish to coerce to an integer matrix (e.g., you are creating a set of dummy variables from a factor), as.integer(matrix_object) will coerce the matrix to a vector, which is not what you want. Instead, you can use storage.mode(matrix_object) <- "integer" to maintain the matrix form.