列表理解中“groupby”的“if”条件

Question

I have a sequence in which I want to count runs (ie consecutive identical entries), and return a list of the length of the runs. The code below

from itertools import groupby

S = [1,1,1,2,3,3,4,5,5]

grouped_S = [sum(1 for i in group) for key,group in groupby(S)]

Results, as it should, in

[3, 1, 2, 1, 2]

But I want to ignore spells that are 1-long, and get output just [3,2,2]. This

grouped_S = [sum(1 for i in group) for key,group in groupby(L) if sum(1 for i in g) >1]

gives

[0,0,0]. 

It clearly knows I want just the three sequences > 1, but won't return their length.

I don't understand this behavior, could someone please explain? Right now my solution is:

S = [sum(1 for i in g) for k,g in groupby(S)]
S = [i for i in S if i != 1]

and it works, but there has to be a pythonic one-liner I can't figure out.

Answer 1

As the documentation of groupby points out:

The returned group is itself an iterator that shares the underlying iterable with groupby() .

You can only iterate an iterator once, which you're doing in the if ; there's nothing left in the iterator to sum again then. It would be far easier to simply filter the 1 s out of the result:

grouped_S = list(filter(lambda s: s > 1, (sum(1 for i in g) for k,g in groupby(S))))

Answer 2

You could very well save your variable in a list comprehension and then check it afterwards:

from itertools import groupby

S = [1,1,1,2,3,3,4,5,5]
grouped_S = [group
             for k,g in groupby(S)
             for group in [sum(1 for i in g)]
             if group > 1]

print(grouped_S)

This yields

[3, 2, 2]

Answer 3

as @deceze already pointed your variable g is an iterator:

from the docs :

iterator

An object representing a stream of data. Repeated calls to the iterator's next () method (or passing it to the built-in function next()) return successive items in the stream. When no more data are available a StopIteration exception is raised instead. At this point, the iterator object is exhausted and any further calls to its next () method just raise StopIteration again print([len(g) for k, [*g] in groupby(S) if len(g) > 1])

you can make your g variable a list by using iterable unpacking operator :

print([len(g) for k, [*g] in groupby(S) if len(g) > 1])

output:

[3, 2, 2]