I want to create an R vector with two repeat elements. A length of the array is 200.
But each element can be either 'x' or 'y'.
an element can be x or y with equal chance.
Is there any grammatical function in R to do above task?
Please someone help.
A possible way to do it is to use rbinom. Step by step, generate first a vecotr of 0 and 1, then change it into x and y:
vec = ifelse(rbinom(200, 1, 0.5)==0,"x","y"))
We need a little bit more information to be helpful, but if you want a vector of 200 values, 100 x's and 100 y's, then just do this:
t <- rep(c('X','Y'), 100)
If you want this in a random order:
t <- sample(t)
Related
I am trying to create a vector from another vector where I multiply the numbers in the vector one more each time.
For example if I had (1,2,3) the new vector would be (1, 1 x 2, 1 x 2 x 3)=(1,2,6)
I tried to create a loop for this as seen below. It seems to work for whole numbers but not decimals. I am not sure why.
x <- c(0.99,0.98,0.97,0.96,0.95)
for(i in 1:5){x[i]=prod(x[1:i])}
The result given is 0.9900000 0.9702000 0.9316831 0.8590845 0.7303385
which is incorrect as prod(x) = 0.8582777. Which is not the same as the last element of the vector.
Does anyone know why this is the case? Or have a suggestion for improvement in my code to get the correct answer.
test<-c(1,2,3)
cumprod(test)
[1] 1 2 6
As #akrun suggests, one can achieve the same with:
Reduce("*", test, accumulate = TRUE)
I want to get all numbers that are greater than 0 and lesser than 1e6 and does not contain digit 4. How is that possible, please?
My try was:
library(prob)
A <- c(0:(1e6-1))
V <- subset(A, /*I don't know what to put here*/)
But I don't know how to state that I want all numbers that does not contain digit 4....
You could use grep to find out indices with numbers containing 4 and remove them with negative subsetting.
A = 0:1e6
V = A[-grep(4,A)]
I have a problem to solve in either Matlab or R (preferably in R).
Imagine I have a vector A with 10 elements.
I have also a vector B with 30 elements, of which 10 have value 'x'.
Now, I want to replace all the 'x' in B by the corresponding values taken from A, in the order that is established in A. Once a value in A is taken, the next one is ready to be used when the next 'x' in B is found.
Note that the sizes of A and B are different, it's the number of 'x' cells that coincides with the size of A.
I have tried different ways to do it. Any suggestion on how to program this?
As long as the number of x entries in B matches the length of A, this will do what you want:
B[B=='x'] <- A
(It should be clear that this is the R solution.)
MATLAB Solution
In MATLAB it's quite simple, use logical indexing:
B(B == 'x') = A;
This question already has an answer here:
Closed 10 years ago.
Possible Duplicate:
How to generate a vector containing a numeric sequence?
In R, how can I get the list of numbers from 1 to 100? Other languages have a function 'range' to do this. R's range does something else entirely.
> range(1,100)
[1] 1 100
Your mistake is looking for range, which gives you the range of a vector, for example:
range(c(10, -5, 100))
gives
-5 100
Instead, look at the : operator to give sequences (with a step size of one):
1:100
or you can use the seq function to have a bit more control. For example,
##Step size of 2
seq(1, 100, by=2)
or
##length.out: desired length of the sequence
seq(1, 100, length.out=5)
If you need the construct for a quick example to play with, use the : operator.
But if you are creating a vector/range of numbers dynamically, then use seq() instead.
Let's say you are creating the vector/range of numbers from a to b with a:b, and you expect it to be an increasing series. Then, if b is evaluated to be less than a, you will get a decreasing sequence but you will never be notified about it, and your program will continue to execute with the wrong kind of input.
In this case, if you use seq(), you can set the sign of the by argument to match the direction of your sequence, and an error will be raised if they do not match. For example,
seq(a, b, -1)
will raise an error for a=2, b=6, because the coder expected a decreasing sequence.
I have a data vector with 1024 values and need to count the number of negative entries. Is there an elegant way to do this without looping and checking if an element is <0 and incrementing a counter?
You want to read 'An Introduction to R'. Your answer here is simply
sum( x < 0 )
which works thanks to vectorisation. The x < 0 expression returns a vector of booleans over which sum() can operate (by converting the booleans to standard 0/1 values).
There is a good answer to this question from Steve Lianoglou How to identify the rows in my dataframe with a negative value in any column?
Let me just replicate his code with one small addition (4th point).
Imagine you had a data.frame like this:
df <- data.frame(a = 1:10, b = c(1:3,-4, 5:10), c = c(-1, 2:10))
This will return you a boolean vector of which rows have negative values:
has.neg <- apply(df, 1, function(row) any(row < 0))
Here are the indexes for negative numbers:
which(has.neg)
Here is a count of elements with negative numbers:
length(which(has.neg))
The above solutions prescribed need to be tweaked in-order to apply this for a df.
The below command helps get the count of negative or any other symbolic logical relationship.
Suppose you have a dataframe:
df <- data.frame(x=c(2,5,-10,NA,7), y=c(81,-1001,-1,NA,-991))
In-order to get count of negative records in x:
nrow(df[df$x<0,])