CODEWITHSUNDEEP
vectoriteratorjulia
baptiste
baptiste

Reputation: 77096

use a julia iterator as a regular vector

I was quite puzzled by the following,

sqrt(1:3) * [1 2 3]
# 3x3 Matrix, as expected
sqrt(1:3) * 1:3
# error `colon` has no method matching...

until I realised that 1:3 must be a different kind of beast, i.e not just a vector as I expected from Matlab. My current workaround is to use hcat to convert it to a vector, sqrt(1:3) * hcat(1:3...), is there a better approach?

Upvotes: 2

Views: 849

Answers (2)

IainDunning
IainDunning

Reputation: 11644

The main problem with the second version

sqrt(1:3) * 1:3

is actually operator precedence. The colon operator is very low precedence, so this translates to

(sqrt(1:3) * 1):3

which is nonsensical, hence the error

 ERROR: `colon` has no method matching colon(::Array{Float64,1}, ::Int64)`

Having said that, if you "fix it" with parentheses it doesn't work because the operator isn't defined. Hence you probably want sqrt(1:3) * [1:3]'.

Upvotes: 5

Pooya
Pooya

Reputation: 304

typeof(1:3) gives UnitRange{Int64} (constructor with 1 method), whereas typeof([1:3]) gives: Array{Int64,1}. Note that [1:3] is by default a column vector, so you need to transpose it: sqrt(1:3) * [1:3].'

Upvotes: 2

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