While I use the & all the time when doing a mapping, such as

someFunction[#] &/@ somelist

I don't really understand what it's doing.

I wanted to generate a list of RandomReal numbers. I realize that the RandomReal function itself could create such a list, but I wanted to see how the Array function handled it. So at first I tried:

Array[RandomReal[], 3]

and got

{0.211826[1], 0.211826[2], 0.211826[3]}

Then I tried:

Array[RandomReal[] &, 3]

This gave me what I wanted:

{0.748657, 0.422851, 0.247495}

I also saw that if I tried:

Array[RandomReal[#] &, 3]

It seemed to map RandomReal onto the list {1,2,3} - even though I didn't use /@.

Can someone explain to me what's happening with this syntax? Even though I can use it, I don't understand why, or how it works.

  • $\begingroup$ Array[expr, 3] generates {expr[1], expr[2], expr[3]} after evaluating expr, just like Map[expr, Range[3]] as you noticed. If expr is a function, then expr[1] etc. will be evaluated. If expr is RandomReal[], it first evaluates to a random number like 0.211826, which is then used as if the number were a function. You probably wanted Table[RandomReal[], 3]. Table holds its argument, so it will be evaluated three separate times, yielding three different random numbers. (Basically, I'm thinking it's Array that's confusing you, not &, but I could be wrong.) $\endgroup$ – Michael E2 Apr 13 '18 at 17:23

Array[f,3] is basically equivalent to Map[f,{1,2,3}]. Since Array does not have any Holding attribute, it first evaluates the first argument. Evaluating RandomReal[] yields a single random number. So Array[RandomReal[], 3] does this:



{0.211826[1], 0.211826[2], 0.211826[3]}

Next: Array[RandomReal[] &, 3]. Now, we have

f = RandomReal[]&.

FullForm[f] shows that this is f = Function[RandomReal[]]. Function has the HoldAll attribute so that RandomReal[] gets not evaluated until f is called with an argument. So f[1], f[2], f[3] will all yield a different random number (between 0 and 1). Hence Array[RandomReal[]&,3] returns a list of three random numbers.

The third one: Array[RandomReal[#]&, 3]

Well, RandomReal is already a function (not a Function, though) and so f = RandomReal[#]& is equivalent to Array[RandomReal[#1]&, 3]. That means no matter how many arguments you supply to f, it will only use the first one and hand it over to RandomReal. When called with a single argument like in RandomReal[3], RandomReal will return a random number between 0 and 3. So, Array[RandomReal, 3] returns a list of random numbers in which the first one is between 0 and 1, the second one is between 0 and 2, and the third one is between 0 and 3. I use SeedRandom in the following in order to highlight that:

Array[RandomReal, 3]

{0.455719, 1.95565, 2.82964}

What you probably want is obtaining a list of 3 random numbers between, say, -1 and 1. To this end, you should use the two-argument version of RandomReal:

  • $\begingroup$ Thank you - I didn't get that the & basically added function. I see now that If I do Function[RandomReal[]] it gives me what I want. $\endgroup$ – Mitchell Kaplan Apr 13 '18 at 17:29
  • $\begingroup$ You're welcome! $\endgroup$ – Henrik Schumacher Apr 13 '18 at 17:38

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