Skip iterations in enumerated list object (python)
Using the enumeration index
Similar to the accepted answer… except without using itertools
(IMHO islice
doesn't improve readability), plus enumerate()
already returns an iterator so you don't need the iter()
at all:
lines = [{str(x): x} for x in range(20)] # dummy data
it = enumerate(lines)
for i, line in it:
print(line)
if i == 10: # condition using enumeration index
[next(it, None) for _ in range(5)] # skip 5
That last line can optionally be expanded for readability:
for _ in range(5): # skip 5
next(it, None)
The None
argument in next()
avoids an exception if there aren't enough items to skip. (For the original question, it can be omitted as the OP wrote: "I can be sure that, if the condition is met, there are 5 or more objects left in the lines
object.")
Not using the enumeration index
If the skip condition isn't based on the enumeration index, simply treat the list as a FIFO queue and consume from it using pop()
:
lines = [{str(x): x} for x in range(20)] # dummy data
while lines:
line = lines.pop(0) # get first item
print(line)
if <condition>: # some other kind of condition
[lines.pop(0) for _ in range(5)] # skip 5
As before, that last line can optionally be expanded for readability:
for _ in range(5): # skip 5
lines.pop(0)
(For large lists, use collections.deque
for performance.)
You could use a functional programming style with recursion, first by putting the necessary parts of your for
loop into a function:
def my_function(iline, line, rest_of_lines, **other_args):
do_some_side_effects(iline, line, **other_args)
if rest_of_lines == []:
return <some base case>
increment = 5 if <condition> else 1
return my_function(iline+increment,
rest_of_lines[increment-1],
rest_of_lines[increment:],
**other_args)
Optionally, if it doesn't need to return anything, you can just adjust those lines of code to be function calls, and the return result will be None
.
Then some place you actually call it:
other_args = get_other_args(...)
my_function(0, lines[0], lines[1:], **other_args)
If you need the function to return something different for each index, then I would suggest modifying this slightly to account for the output data structure you want. In that case, you might want to pass the inner result of do_some_side_effects
back into the recursive function call so it can build up the result.
def my_function(iline, line, rest_of_lines, output, **other_args):
some_value = do_some_side_effects(iline, line, **other_args)
new_output = put_value_in_output(some_value, output)
# could be as simple as appending to a list/inserting to a dict
# or as complicated as you want.
if rest_of_lines == []:
return new_output
increment = 5 if <condition> else 1
return my_function(iline+increment,
rest_of_lines[increment-1],
rest_of_lines[increment:],
new_output,
**other_args)
Then to call
other_args = get_other_args(...)
empty_output = get_initial_data_structure(...)
full_output = my_function(0, lines[0], lines[1:], empty_output, **other_args)
Note that in Python, because of the way most of the basic data structures are implemented, this programming style is not going to gain you efficiency, and in the context of other object oriented code it may even be bad style to complicate things beyond the simple while
solution.
My advice: use the while
loop, though I would tend to structure my projects and APIs so that using a recursive functional approach will still be efficient and readable. I would also try not to allow side effects inside the loop.
The standard idiom for doing this is to make an iterator and then use one of the consumer patterns (see here in the itertools
docs.)
For example:
from itertools import islice
lines = list("abcdefghij")
lit = iter(enumerate(lines))
for iline, line in lit:
print(iline, line)
if line == "c":
# skip 3
next(islice(lit, 3,3), None)
produces
0 a
1 b
2 c
6 g
7 h
8 i
9 j
iline = 0
while iline < len(lines):
line = lines[iline]
if <condition>:
place_where_skip_happened = iline
iline += 5
iline += 1
If you are iterating over a file object you can skip lines using next or make lines an iterator:
lines = iter(range(20))
for l in lines:
if l == 10:
[next(lines) for _ in range(5)]
print(l)
0
1
2
3
4
5
6
7
8
9
10
16
17
18
19
It really depends on what you are iterating over and what you want to do.
Using your own code with iter and islice:
from itertools import islice
it = iter(enumerate(lines))
for iline, line in it:
if <condition>:
place_where_skip_happened = iline
next(islice(it,5 ,5), None)
print(line)