Python删除列表中序列的内部项

Question

foo = [0,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,0,0,1,1]
bar = [x if x==0 else 'o' for x in foo]

bar:

[0, 0, 0, 0,'o', 'o', 'o', 'o', 'o', 'o', 0, 0, 0, 'o', 'o', 'o', 0, 0, 'o', 'o']

I would like to remove the inner 'o' at this point so that the result looks like so:

[0, 0, 0, 0, 'o','o', 0, 0, 0, 'o','o', 0, 0, 'o', 'o']

If possible, I'd like to do this within the list comprehension itself, and I would like to avoid anything with converting to a string (as my actual task for this involves dictionaries rather than 1's and 0's). Any ideas?

Answer 1

By keeping track of the previous and next element

We can perform a check on the previous and next element, and check if these are 'o' s as well, if so, we do not yield the element, otherwise we do, like:

nbar1 = len(bar) - 1
[ x for i, x in enumerate(bar) if not (0 < i < nbar1 and bar[i] == bar[i-1] == bar[i+1] == 'o') ]

The above can be made more elegant by using chain and zip :

from itertools import chain, islice

prev = chain((None,), bar)
nxt = islice(chain(bar, (None, )), 1, None)
result = [ x for p, x, n in zip(prev, bar, nxt) if not (p == x == n == 'o') ]

where p is the "previous item", x is the "current item", and n is the "next item".

This ten yields:

>>> [ x for p, x, n in zip(prev, bar, nxt) if not (p == x == n == 'o') ]
[0, 0, 'o', 'o', 0, 0, 'o', 'o', 0, 0, 'o']

The above will also work with elements that are not 0 s, for example:

>>> bar = [1, 3, 'o', 'o', 'o', 'o', 'o', 'o', 0, 0, 'o', 'o', 'o', 2, 0, 'o', 'o']
>>> prev = chain((None,), bar)
>>> nxt = islice(chain(bar, (None, )), 1, None)
>>> [ x for p, x, n in zip(prev, bar, nxt) if not (p == x == n == 'o') ]
[1, 3, 'o', 'o', 0, 0, 'o', 'o', 2, 0, 'o', 'o']

We can also easily change it to work with another element (than 'o' ), as long as it is not something that is equal to None . If that is the case, we can however chain other elements to the prev and nxt iterables.

The above works in linear time O(n) with n the length of the list to process.

By grouping and slicing

An alternative is to use itertools.groupby to detect "bursts" of characters, and in case the burst contains 'o' s, we islice(..) up to two elements:

from itertools import groupby, islice

[ x for k, g in groupby(bar) for x in (islice(g, 2) if k == 'o' else g) ]

again yielding:

>>> [ x for k, g in groupby(bar) for x in (islice(g, 2) if k == 'o' else g) ]
[0, 0, 0, 0, 'o', 'o', 0, 0, 0, 'o', 'o', 0, 0, 'o', 'o']

Answer 2

If you really want to do it with one comprehension list:

bar=[x if x==0 else 'o' for i,x in enumerate(foo) if (i==0 or i==len(foo)-1) or x==0 or 
foo[i-1]==0 or foo[i+1]==0]

should work for your example.

Answer 3

You can do:

>>> foo = [0,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,0,0,1,1]
>>> from itertools import groupby
>>> [ext for c, grp in groupby(foo) for ext in (grp if c==0 else ['o']*min(2,len(list(grp))))]
[0, 0, 0, 0, 'o', 'o', 0, 0, 0, 'o', 'o', 0, 0, 'o', 'o']

Answer 4

Using itertools.groupby if k is 0 we append all the items from that group, if k is 'o' we only append the first and last excluding the middle 'o' 's

from itertools import groupby

bar = [0, 0, 'o', 'o', 'o', 'o', 'o', 'o', 0, 0, 'o', 'o', 'o', 0, 0, 'o', 'o']
new = []
for k, g in groupby(bar):
    x = list(g)
    if k == 0:
        for i in x:
            new.append(i)
    elif k == 'o':
        new.append(x[0])
        new.append(x[-1])

print(new)
# [0, 0, 'o', 'o', 0, 0, 'o', 'o', 0, 0, 'o', 'o']