Can't seem开发者_如何学Python to find a clue to this online and can't figure it out myself so:
How would I go about slicing a list so that I return a list of slices of contiguous non-zero integers. ie:
data = [3, 7, 4, 0, 1, 3, 7]
and I want to produce:
slices = [[3, 7, 4], [1, 3, 7]]
I have tried various methods of iterating through the list, have been leaning towards a generator that lets me know when the contiguous groups start and stop, by testing if there is a 0 before or after, but then I am a bit stumped.
import itertools
[ list(x[1]) for x in itertools.groupby(data, lambda x: x == 0) if not x[0] ]
Look at itertools.groupby:
>>> data = [3, 7, 4, 0, 1, 3, 7, 4, 0, 5]
>>> a=[list(i[1]) for i in itertools.groupby(data, key=lambda i:i==0)]
>>> a
[[3, 7, 4], [0], [1, 3, 7, 4], [0], [5]]
>>> [i for i in a if i != [0]]
[[3, 7, 4], [1, 3, 7, 4], [5]]
def split_on_zero(data):
start = 0
for (i, n) in enumerate(data):
if n == 0:
yield data[start:i]
start = i + 1
yield data[start:]
>>> list(split_on_zero([3, 7, 4, 0, 1, 3, 7]))
[[3, 7, 4], [1, 3, 7]]
>>> list(split_on_zero([0, 1, 2, 0, 3, 4, 5, 0]))
[[], [1, 2], [3, 4, 5], []]
Here the very base solution you can try to use too:
data = [1, 0, 3, 7, 4, 1, 3, 7]
def get_zero_sliced(inputList):
res = []
prev = 0
for i,x in enumerate(data):
if x == 0 and i != 0:
res.append(data[prev:i])
prev = i + 1
if i == len(data)-1 and prev != i:
res.append(data[prev:])
return res
get_zero_sliced(data)
A generator will work. Here's one, but there's probably other ways to do it. I called it x
because I don't have a good name right now (good naming takes some thought, so give it some thought).
def x(iterator):
result = []
for i in iterator:
if i != 0:
result.append(i)
elif result:
yield result
result = []
if result: yield result
精彩评论