Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

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

python - Asterisk in function call


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 Answer

0 votes
by (71.8m points)

* is the "splat" operator: It takes a list as input, and expands it into actual positional arguments in the function call.

So if uniqueCrossTabs was [ [ 1, 2 ], [ 3, 4 ] ], then itertools.chain(*uniqueCrossTabs) is the same as saying itertools.chain([ 1, 2 ], [ 3, 4 ])

This is obviously different from passing in just uniqueCrossTabs. In your case, you have a list of lists that you wish to flatten; what itertools.chain() does is return an iterator over the concatenation of all the positional arguments you pass to it, where each positional argument is iterable in its own right.

In other words, you want to pass each list in uniqueCrossTabs as an argument to chain(), which will chain them together, but you don't have the lists in separate variables, so you use the * operator to expand the list of lists into several list arguments.

As Jochen Ritzel has pointed out in the comments, chain.from_iterable() is better-suited for this operation, as it assumes a single iterable of iterables to begin with. Your code then becomes simply:

uniqueCrossTabs = list(itertools.chain.from_iterable(uniqueCrossTabs))

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 OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

2.1m questions

2.1m answers

60 comments

57.0k users

...