I'm working with Python itertools and using groupby to sort a bunch of pairs by the last element. I've gotten it to sort and I can iterate through the groups just fine, but I would really love to be able to get the length of each group without having to iterate through each one, incrementing a counter.
The project is cluster some data points. I'm working with pairs of (numpy.array, int) where the numpy array is a data point and the integer is a cluster label
Here's my relevant code:
data = sorted(data, key=lambda (point, cluster):cluster) for cluster,clusterList in itertools.groupby(data, key=lambda (point, cluster):cluster): if len(clusterList) < minLen:
On the last line: if len(clusterList) < minLen:
, I get an error that
object of type 'itertools._grouper' has no len()
I've looked up the operations available for _groupers
, but can't find anything that seems to provide the length of a group.
itertools. count() makes an iterator that returns values that counts up or down infinitely. itertools.count() — Functions creating iterators for efficient looping — Python 3.9.7 documentation.
islice() - The islice() function allows the user to loop through an iterable with a start and stop , and returns a generator. map() - The map() function creates an iterable map object that applies a specified transformation to every element in a chosen iterable.
chain() function It is a function that takes a series of iterables and returns one iterable. It groups all the iterables together and produces a single iterable as output.
Just because you call it clusterList
doesn't make it a list! It's basically a lazy iterator, returning each item as it's needed. You can convert it to a list like this, though:
clusterList = list(clusterList)
Or do that and get its length in one step:
length = len(list(clusterList))
If you don't want to take up the memory of making it a list, you can do this instead:
length = sum(1 for x in clusterList)
Be aware that the original iterator will be consumed entirely by either converting it to a list or using the sum()
formulation.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With