How to select elements from array in Julia matching predicate?

Question

Julia appears to have a lot of Matlab like features. I'd like to select from an array using a predicate. In Matlab I can do this like:

>> a = 2:7 ;
>> a > 4

ans =

     0     0     0     1     1     1

>> a(a>4)

ans =

     5     6     7

I found a kind of clunky seeming way to do part of this in Julia:

julia> a = 2:7
2:7

julia> [int(x > 3) for x in a]
6-element Array{Any,1}:
 0
 0
 1
 1
 1
 1

(Using what wikipedia calls list comprehension). I haven't figured out how to apply a set like this to select with in Julia, but may be barking up the wrong tree. How would one do a predicate selection from an array in Julia?

DSM · Accepted Answer · 2015-01-11 06:33:22Z

47

You can use a very Matlab-like syntax if you use a dot . for elementwise comparison:

julia> a = 2:7
2:7

julia> a .> 4
6-element BitArray{1}:
 false
 false
 false
  true
  true
  true

julia> a[a .> 4]
3-element Array{Int32,1}:
 5
 6
 7

Alternatively, you can call filter if you want a more functional predicate approach:

julia> filter(x -> x > 4, a)
3-element Array{Int32,1}:
 5
 6
 7

edited Jan 11, 2015 at 6:33

answered Jan 11, 2015 at 6:27

DSM

355k67 gold badges606 silver badges504 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Peeter Joot Over a year ago

For the record it looks like the elementwise method is about twice as fast as calling filter.

Zoe - Save the data dump · Accepted Answer · 2024-05-13 20:19:31Z

21

Array comprehension in Julia is somewhat more primitive than list comprehension in Haskell or Python. There are two solutions — you can either use a higher-order filtering function, or use broadcasting operations.

Higher-order filtering

filter(x -> x > 4, a)

This calls the filter function with the predicate x -> x > 4 (see Anonymous functions in the Julia manual).

Broadcasting and indexing

a[Bool[a[i] > 4 for i = 1:length(a)]]

This performs a broadcasting comparision between the elements of a and 4, then uses the resulting array of booleans to index a. It can be written more compactly using a broadcasting operator:

a[a .> 4]

edited May 13, 2024 at 20:19

Zoe - Save the data dump

28.4k22 gold badges130 silver badges163 bronze badges

answered Jan 11, 2015 at 6:36

jch

5,67927 silver badges42 bronze badges

4 Comments

lara Over a year ago

I have an array and would like to filter it on more than one thing. For example a > 3 && a < 5. However when I try to do this I get the error that non-boolean (BitArray{1}) used in boolean context. What is the problem here?

Fred Schoen Over a year ago

This works fine with filter: filter( x -> (x > 3 && x < 5), a) docs.julialang.org/en/latest/stdlib/collections/…

Fred Schoen Over a year ago

And for arrays, you have to use bitwise operators a .> 3 & a .< 5 Notice the single & for bitwise and

esel Over a year ago

Note that a=[1; 2; 3]; a[a.>0 & a.<1] will not yield the expected result! This gives 1 2 3, as the & operator has higher precedence. A correct solution is a[(a.>0) & (a.<1)].

Alexander Kooij · Accepted Answer · 2020-02-14 07:43:48Z

8

I'm currently using Julia 1.3.1 and some syntax has changed compared to earlier answers. To filter an array on multiple conditions I had to do:

x = range(0,1,length=100)
x[(x .> 0.4) .& (x .< 0.51)]

note the '.&' needed to do the AND operator.

answered Feb 14, 2020 at 7:43

Alexander Kooij

1211 silver badge3 bronze badges

Comments

KloppyToppy · Accepted Answer · 2021-02-02 22:56:51Z

3

I would like to add an aspect that has not been covered by the previous answers. If you want to filter the array by the index values (as opposed to the array values) you can do so bya[1:end ...] where instead of the dots you apply a broadcast operator to the index values. E.g. in order to remove the third element you would write

a[1:end .!= 3].

answered Feb 2, 2021 at 22:56

KloppyToppy

4163 silver badges7 bronze badges

1 Comment

Daniel Pinyol Over a year ago

This instance also allows a[1:end .!= [2,3]] But be careful because a[1:end .!= []] throws an exception

phyatt · Accepted Answer · 2019-02-20 14:19:55Z

To filter the keys in a dictionary, this worked for me:

mydict = Dict("key1" => 1.0, "key2" => 2.0, "a big string with a part of a string" => 3.0)
filter(x -> occursin("part of a string", string(x)), keys(mydict))

Here is what the output looks like on the REPL in Julia 1.0

julia> mydict = Dict("key1" => 1.0, "key2" => 2.0, "a big string with a part of a string" => 3.0)
Dict{String,Float64} with 3 entries:
  "key2"                                 => 2.0
  "key1"                                 => 1.0
  "a big string with a part of a string" => 3.0

julia> filter(x -> occursin("part of a string", string(x)), keys(mydict))
Set(["a big string with a part of a string"])

This in general is a great way to filter an array of strings.

Hope that helps.

Ted Dunning · Accepted Answer · 2024-07-22 00:34:21Z

0

11.572 μs ± 25.739 and 8.992 μs ± 15.098 are not so very different even after 10,000 samples. As a quick estimate, this means that the first one would be faster about 47% of the time which is a pretty good synonym for "about equal".

answered Jul 22, 2024 at 0:34

Ted Dunning

1,91715 silver badges12 bronze badges

Collectives™ on Stack Overflow

How to select elements from array in Julia matching predicate?

6 Answers 6

1 Comment

Higher-order filtering

Broadcasting and indexing

4 Comments

Comments

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

1 Comment

Higher-order filtering

Broadcasting and indexing

4 Comments

Comments

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related