Skip first entry in for loop in python?
Question:
In python, How do I do something like:
for car in cars:
# Skip first and last, do work for rest
Answers:
This code skips the first and the last element of the list:
for item in list_name[1:-1]:
#...do whatever
The other answers only work for a sequence.
For any iterable, to skip the first item:
itercars = iter(cars)
next(itercars)
for car in itercars:
# do work
If you want to skip the last, you could do:
itercars = iter(cars)
# add 'next(itercars)' here if you also want to skip the first
prev = next(itercars)
for car in itercars:
# do work on 'prev' not 'car'
# at end of loop:
prev = car
# now you can do whatever you want to do to the last one on 'prev'
Well, your syntax isn’t really Python to begin with.
Iterations in Python are over he contents of containers (well, technically it’s over iterators), with a syntax for item in container
. In this case, the container is the cars
list, but you want to skip the first and last elements, so that means cars[1:-1]
(python lists are zero-based, negative numbers count from the end, and :
is slicing syntax.
So you want
for c in cars[1:-1]:
do something with c
Here is a more general generator function that skips any number of items from the beginning and end of an iterable:
def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
for x in itertools.islice(it, at_start):
pass
queue = collections.deque(itertools.islice(it, at_end))
for x in it:
queue.append(x)
yield queue.popleft()
Example usage:
>>> list(skip(range(10), at_start=2, at_end=2))
[2, 3, 4, 5, 6, 7]
The best way to skip the first item(s) is:
from itertools import islice
for car in islice(cars, 1, None):
pass
# do something
islice
in this case is invoked with a start-point of 1
, and an end point of None
, signifying the end of the iterable
.
To be able to skip items from the end of an iterable
, you need to know its length (always possible for a list, but not necessarily for everything you can iterate on). for example, islice(cars, 1, len(cars)-1)
will skip the first and last items in cars
.
Based on @SvenMarnach ‘s Answer, but bit simpler and without using deque
>>> def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
it = itertools.islice(it, at_start, None)
it, it1 = itertools.tee(it)
it1 = itertools.islice(it1, at_end, None)
return (next(it) for _ in it1)
>>> list(skip(range(10), at_start=2, at_end=2))
[2, 3, 4, 5, 6, 7]
>>> list(skip(range(10), at_start=2, at_end=5))
[2, 3, 4]
Also Note, based on my timeit
result, this is marginally faster than the deque solution
>>> iterable=xrange(1000)
>>> stmt1="""
def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
it = itertools.islice(it, at_start, None)
it, it1 = itertools.tee(it)
it1 = itertools.islice(it1, at_end, None)
return (next(it) for _ in it1)
list(skip(iterable,2,2))
"""
>>> stmt2="""
def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
for x in itertools.islice(it, at_start):
pass
queue = collections.deque(itertools.islice(it, at_end))
for x in it:
queue.append(x)
yield queue.popleft()
list(skip(iterable,2,2))
"""
>>> timeit.timeit(stmt = stmt1, setup='from __main__ import iterable, skip, itertools', number = 10000)
2.0313770640908047
>>> timeit.timeit(stmt = stmt2, setup='from __main__ import iterable, skip, itertools, collections', number = 10000)
2.9903135454296716
An alternative method:
for idx, car in enumerate(cars):
# Skip first line.
if not idx:
continue
# Skip last line.
if idx + 1 == len(cars):
continue
# Real code here.
print car
I do it like this, even though it looks like a hack it works every time:
ls_of_things = ['apple', 'car', 'truck', 'bike', 'banana']
first = 0
last = len(ls_of_things)
for items in ls_of_things:
if first == 0
first = first + 1
pass
elif first == last - 1:
break
else:
do_stuff
first = first + 1
pass
The more_itertools
project extends itertools.islice
to handle negative indices.
Example
import more_itertools as mit
iterable = 'ABCDEFGH'
list(mit.islice_extended(iterable, 1, -1))
# Out: ['B', 'C', 'D', 'E', 'F', 'G']
Therefore, you can elegantly apply it slice elements between the first and last items of an iterable:
for car in mit.islice_extended(cars, 1, -1):
# do something
Here’s my preferred choice. It doesn’t require adding on much to the loop, and uses nothing but built in tools.
Go from:
for item in my_items:
do_something(item)
to:
for i, item in enumerate(my_items):
if i == 0:
continue
do_something(item)
Example:
mylist=['one','two','three','four','five']
for i in mylist[1:]:
print(i)
In python index start from 0, We can use slicing operator to make manipulations in iteration.
for i in range(1,-1):
Good solution for support of itertools.chain
is to use itertools.islice
in order to take a slice of an iterable:
your_input_list = ['list', 'of', 'things']
for i, variant in list(itertools.islice(enumerate(some_function_that_will_output_itertools_chain(your_input_list)), 1, None)):
"""
# No need for unnecessary conditions like this:
if i == 0:
continue
"""
variant = list(variant) # (optional) converting back to list
print(variant)
Similar to @maninthecomputer ‘s answer, when you need to skip the first iteration of a loop based on an int (self._model.columnCount()
in my case):
for col in range(self._model.columnCount()):
if col == 0:
continue
Put more simply:
test_int = 3
for col in range(test_int):
if col == 0:
continue
print(col)
Provides output:
1
2
3
In python, How do I do something like:
for car in cars:
# Skip first and last, do work for rest
This code skips the first and the last element of the list:
for item in list_name[1:-1]:
#...do whatever
The other answers only work for a sequence.
For any iterable, to skip the first item:
itercars = iter(cars)
next(itercars)
for car in itercars:
# do work
If you want to skip the last, you could do:
itercars = iter(cars)
# add 'next(itercars)' here if you also want to skip the first
prev = next(itercars)
for car in itercars:
# do work on 'prev' not 'car'
# at end of loop:
prev = car
# now you can do whatever you want to do to the last one on 'prev'
Well, your syntax isn’t really Python to begin with.
Iterations in Python are over he contents of containers (well, technically it’s over iterators), with a syntax for item in container
. In this case, the container is the cars
list, but you want to skip the first and last elements, so that means cars[1:-1]
(python lists are zero-based, negative numbers count from the end, and :
is slicing syntax.
So you want
for c in cars[1:-1]:
do something with c
Here is a more general generator function that skips any number of items from the beginning and end of an iterable:
def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
for x in itertools.islice(it, at_start):
pass
queue = collections.deque(itertools.islice(it, at_end))
for x in it:
queue.append(x)
yield queue.popleft()
Example usage:
>>> list(skip(range(10), at_start=2, at_end=2))
[2, 3, 4, 5, 6, 7]
The best way to skip the first item(s) is:
from itertools import islice
for car in islice(cars, 1, None):
pass
# do something
islice
in this case is invoked with a start-point of 1
, and an end point of None
, signifying the end of the iterable
.
To be able to skip items from the end of an iterable
, you need to know its length (always possible for a list, but not necessarily for everything you can iterate on). for example, islice(cars, 1, len(cars)-1)
will skip the first and last items in cars
.
Based on @SvenMarnach ‘s Answer, but bit simpler and without using deque
>>> def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
it = itertools.islice(it, at_start, None)
it, it1 = itertools.tee(it)
it1 = itertools.islice(it1, at_end, None)
return (next(it) for _ in it1)
>>> list(skip(range(10), at_start=2, at_end=2))
[2, 3, 4, 5, 6, 7]
>>> list(skip(range(10), at_start=2, at_end=5))
[2, 3, 4]
Also Note, based on my timeit
result, this is marginally faster than the deque solution
>>> iterable=xrange(1000)
>>> stmt1="""
def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
it = itertools.islice(it, at_start, None)
it, it1 = itertools.tee(it)
it1 = itertools.islice(it1, at_end, None)
return (next(it) for _ in it1)
list(skip(iterable,2,2))
"""
>>> stmt2="""
def skip(iterable, at_start=0, at_end=0):
it = iter(iterable)
for x in itertools.islice(it, at_start):
pass
queue = collections.deque(itertools.islice(it, at_end))
for x in it:
queue.append(x)
yield queue.popleft()
list(skip(iterable,2,2))
"""
>>> timeit.timeit(stmt = stmt1, setup='from __main__ import iterable, skip, itertools', number = 10000)
2.0313770640908047
>>> timeit.timeit(stmt = stmt2, setup='from __main__ import iterable, skip, itertools, collections', number = 10000)
2.9903135454296716
An alternative method:
for idx, car in enumerate(cars):
# Skip first line.
if not idx:
continue
# Skip last line.
if idx + 1 == len(cars):
continue
# Real code here.
print car
I do it like this, even though it looks like a hack it works every time:
ls_of_things = ['apple', 'car', 'truck', 'bike', 'banana']
first = 0
last = len(ls_of_things)
for items in ls_of_things:
if first == 0
first = first + 1
pass
elif first == last - 1:
break
else:
do_stuff
first = first + 1
pass
The more_itertools
project extends itertools.islice
to handle negative indices.
Example
import more_itertools as mit
iterable = 'ABCDEFGH'
list(mit.islice_extended(iterable, 1, -1))
# Out: ['B', 'C', 'D', 'E', 'F', 'G']
Therefore, you can elegantly apply it slice elements between the first and last items of an iterable:
for car in mit.islice_extended(cars, 1, -1):
# do something
Here’s my preferred choice. It doesn’t require adding on much to the loop, and uses nothing but built in tools.
Go from:
for item in my_items:
do_something(item)
to:
for i, item in enumerate(my_items):
if i == 0:
continue
do_something(item)
Example:
mylist=['one','two','three','four','five']
for i in mylist[1:]:
print(i)
In python index start from 0, We can use slicing operator to make manipulations in iteration.
for i in range(1,-1):
Good solution for support of itertools.chain
is to use itertools.islice
in order to take a slice of an iterable:
your_input_list = ['list', 'of', 'things']
for i, variant in list(itertools.islice(enumerate(some_function_that_will_output_itertools_chain(your_input_list)), 1, None)):
"""
# No need for unnecessary conditions like this:
if i == 0:
continue
"""
variant = list(variant) # (optional) converting back to list
print(variant)
Similar to @maninthecomputer ‘s answer, when you need to skip the first iteration of a loop based on an int (self._model.columnCount()
in my case):
for col in range(self._model.columnCount()):
if col == 0:
continue
Put more simply:
test_int = 3
for col in range(test_int):
if col == 0:
continue
print(col)
Provides output:
1
2
3