Loop that also accesses previous and next values
How can I iterate over a list of objects, accessing the previous, current, and next items? Like this C/C++ code, in Python?
foo = somevalue;
previous = next = 0;
for (i=1; i<objects.length(); i++) {
if (objects[i]==foo) {
previous = objects[i-1];
next = objects[i+1];
}
}
Solutions until now only deal with lists, and most are copying the list. In my experience a lot of times that isn't possible.
Also, they don't deal with the fact that you can have repeated elements in the list.
The title of your question says "Previous and next values inside a loop", but if you run most answers here inside a loop, you'll end up iterating over the entire list again on each element to find it.
So I've just created a function that. using the itertools
module, splits and slices the iterable, and generates tuples with the previous and next elements together. Not exactly what your code does, but it is worth taking a look, because it can probably solve your problem.
from itertools import tee, islice, chain, izip
def previous_and_next(some_iterable):
prevs, items, nexts = tee(some_iterable, 3)
prevs = chain([None], prevs)
nexts = chain(islice(nexts, 1, None), [None])
return izip(prevs, items, nexts)
Then use it in a loop, and you'll have previous and next items in it:
mylist = ['banana', 'orange', 'apple', 'kiwi', 'tomato']
for previous, item, nxt in previous_and_next(mylist):
print "Item is now", item, "next is", nxt, "previous is", previous
The results:
Item is now banana next is orange previous is None
Item is now orange next is apple previous is banana
Item is now apple next is kiwi previous is orange
Item is now kiwi next is tomato previous is apple
Item is now tomato next is None previous is kiwi
It'll work with any size list (because it doesn't copy the list), and with any iterable (files, sets, etc). This way you can just iterate over the sequence, and have the previous and next items available inside the loop. No need to search again for the item in the sequence.
A short explanation of the code:
-
tee
is used to efficiently create 3 independent iterators over the input sequence -
chain
links two sequences into one; it's used here to append a single-element sequence[None]
toprevs
-
islice
is used to make a sequence of all elements except the first, thenchain
is used to append aNone
to its end - There are now 3 independent sequences based on
some_iterable
that look like:-
prevs
:None, A, B, C, D, E
-
items
:A, B, C, D, E
-
nexts
:B, C, D, E, None
-
- finally
izip
is used to change 3 sequences into one sequence of triplets.
Note that izip
stops when any input sequence gets exhausted, so the last element of prevs
will be ignored, which is correct - there's no such element that the last element would be its prev
. We could try to strip off the last elements from prevs
but izip
's behaviour makes that redundant
Also note that tee
, izip
, islice
and chain
come from the itertools
module; they operate on their input sequences on-the-fly (lazily), which makes them efficient and doesn't introduce the need of having the whole sequence in memory at once at any time.
In python 3
, it will show an error while importing izip
,you can use zip
instead of izip
. No need to import zip
, it is predefined in python 3
- source
This should do the trick.
foo = somevalue
previous = next_ = None
l = len(objects)
for index, obj in enumerate(objects):
if obj == foo:
if index > 0:
previous = objects[index - 1]
if index < (l - 1):
next_ = objects[index + 1]
Here's the docs on the enumerate
function.
Using a list comprehension, return a 3-tuple with current, previous and next elements:
three_tuple = [(current,
my_list[idx - 1] if idx >= 1 else None,
my_list[idx + 1] if idx < len(my_list) - 1 else None) for idx, current in enumerate(my_list)]
I don't know how this hasn't come up yet since it uses only built-in functions and is easily extendable to other offsets:
values = [1, 2, 3, 4]
offsets = [None] + values[:-1], values, values[1:] + [None]
for value in list(zip(*offsets)):
print(value) # (previous, current, next)
(None, 1, 2)
(1, 2, 3)
(2, 3, 4)
(3, 4, None)