Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
525 views
in Technique[技术] by (71.8m points)

dataframe - Vectorized "in" function in julia?

I often want to loop over a long array or column of a dataframe, and for each item, see if it is a member of another array. Rather than doing

giant_list = ["a", "c", "j"]
good_letters = ["a", "b"]
isin = falses(size(giant_list,1))
for i=1:size(giant_list,1)
    isin[i] = giant_list[i] in good_letters
end

Is there any vectorized (doubly-vectorized?) way to do this in julia? In analogy with the basic operators I want to do something like

isin = giant_list .in good_letters

I realize this may not be possible, but I just wanted to make sure I wasn't missing something. I know I could probably use DefaultDict from DataStructures to do the similar but don't know of anything in base.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The indexin function does something similar to what you want:

indexin(a, b)

Returns a vector containing the highest index in b for each value in a that is a member of b. The output vector contains 0 wherever a is not a member of b.

Since you want a boolean for each element in your giant_list (instead of the index in good_letters), you can simply do:

julia> indexin(giant_list, good_letters) .> 0
3-element BitArray{1}:
  true
 false
 false

The implementation of indexin is very straightforward, and points the way to how you might optimize this if you don't care about the indices in b:

function vectorin(a, b)
    bset = Set(b)
    [i in bset for i in a]
end

Only a limited set of names may be used as infix operators, so it's not possible to use it as an infix operator.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...