Not really, it is "simply" because ``np.asarray(set([1, 2, 3]))`` returns an object array
Holy crap! To be pedantic, it looks like it turns it into a numpy scalar,
but still! I wouldn't have expected np.asarray() on a set (or dictionary,
for that matter) to work because order is not guaranteed. Is this expected
behavior?
Digging into the implementation of in1d(), I can see now how passing a
set() wouldn't be useful at all (as an aside, pretty clever algorithm). I
know sets aren't array-like, but the code that used this seemed to work at
first, and this problem wasn't revealed until I created some unit tests to
exercise some possible corner cases. Silently producing possibly erroneous
results is dangerous. Don't know if better documentation or some better
sanity checking would be called for here, though.
Ben Root
On Mon, Aug 10, 2015 at 1:10 PM, Sebastian Berg
On Mo, 2015-08-10 at 12:09 -0400, Benjamin Root wrote:
Just came across this one today:
np.in1d([1], set([0, 1, 2]), assume_unique=True) array([ False], dtype=bool)
np.in1d([1], [0, 1, 2], assume_unique=True)
array([ True], dtype=bool)
I am assuming this has something to do with the fact that order is not guaranteed with set() objects? I was kind of hoping that setting "assume_unique=True" would be sufficient to overcome that problem. Should sets be rejected as an error?
Not really, it is "simply" because ``np.asarray(set([1, 2, 3]))`` returns an object array and 1 is not the same as ``set([1, 2, 3])``.
I think earlier numpy versions may have had "short cuts" for short lists or something so this may have worked in some cases....
- Sebastian
This was using v1.9.0
Cheers!
Ben Root
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