L. Smith - 1 year ago 66

R Question

I've tried a couple ways of doing this problem but am having trouble with how to write it. I think I did the first three steps correctly, but now I have to fill the vector

`z`

`y`

`print`

This is different from that other question because I'm not using a while loop.

`#Step 1: Generate 1,000,000 random, uniformly distributed numbers between 0`

#and 1,000,000,000, and name as a vector x. With a seed of 1.

set.seed(1)

x=runif(1000000, min=0, max=1000000000)

#Step 2: Generate a rounded version of x with the name y

y=round(x,digits=0)

#Step 3: Empty vector named z

z=vector("numeric",length=0)

#Step 4: Create for loop that populates z vector with the numbers from y that are divisible by

#4, not divisible by 3, with an odd number of digits.

for(i in y) {

if(i%%4==0 && i%%3!=0 && nchar(i,type="chars",allowNA=FALSE,keepNA=NA)%%2!=0){

print(z,i)

}

}

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Answer Source

NOTE: As per @BenBolker's comment, a loop is an inefficient way to solve your problem here. Generally, in R, try to avoid loops where possible to maximise the efficiency of your code. @SymbolixAU has provided an example of doing so here in the comments. Having said that, in aid of helping you learn the ins-and-outs of loops and vectors, here's a solution which only requires a change to one line of your code:

You've got the vector created before the loop, that's a good start. Now, inside your loop, you need to populate that vector. To do so, you've currently got `print(z,i)`

, which won't really do too much. What you need to to change the vector itself:

```
z <- c( z, i )
```

Should work for you (just replace that `print`

line in your loop).

What's happening here is that we're taking the existing `z`

vector, binding `i`

to the end of it, and making that new vector `z`

again. So every time a value is added, the vector gets a little longer, such that you'll end up with a complete vector.

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