Related
Is there a simple way of testing if the generator has no items, like peek, hasNext, isEmpty, something along those lines?
Suggestion:
def peek(iterable):
try:
first = next(iterable)
except StopIteration:
return None
return first, itertools.chain([first], iterable)
Usage:
res = peek(mysequence)
if res is None:
# sequence is empty. Do stuff.
else:
first, mysequence = res
# Do something with first, maybe?
# Then iterate over the sequence:
for element in mysequence:
# etc.
The simple answer to your question: no, there is no simple way. There are a whole lot of work-arounds.
There really shouldn't be a simple way, because of what generators are: a way to output a sequence of values without holding the sequence in memory. So there's no backward traversal.
You could write a has_next function or maybe even slap it on to a generator as a method with a fancy decorator if you wanted to.
A simple way is to use the optional parameter for next() which is used if the generator is exhausted (or empty). For example:
_exhausted = object()
if next(some_generator, _exhausted) is _exhausted:
print('generator is empty')
Quick-dirty solution:
next(generator, None) is not None
Or replace None by whatever value you know it's not in your generator.
Edit: Yes, this will skip 1 item in the generator. Sometimes, however, I check whether a generator is empty only for validation purposes, then don't really use it. Or otherwise I do something like:
def foo(self):
if next(self.my_generator(), None) is None:
raise Exception("Not initiated")
for x in self.my_generator():
...
That is, this works if your generator comes from a function, as in my_generator().
The best approach, IMHO, would be to avoid a special test. Most times, use of a generator is the test:
thing_generated = False
# Nothing is lost here. if nothing is generated,
# the for block is not executed. Often, that's the only check
# you need to do. This can be done in the course of doing
# the work you wanted to do anyway on the generated output.
for thing in my_generator():
thing_generated = True
do_work(thing)
If that's not good enough, you can still perform an explicit test. At this point, thing will contain the last value generated. If nothing was generated, it will be undefined - unless you've already defined the variable. You could check the value of thing, but that's a bit unreliable. Instead, just set a flag within the block and check it afterward:
if not thing_generated:
print "Avast, ye scurvy dog!"
Just fell on this thread and realized that a very simple and easy to read answer was missing:
def is_empty(generator):
for item in generator:
return False
return True
If we are not suppose to consume any item then we need to re-inject the first item into the generator:
def is_empty_no_side_effects(generator):
try:
item = next(generator)
def my_generator():
yield item
yield from generator
return my_generator(), False
except StopIteration:
return (_ for _ in []), True
Example:
>>> g=(i for i in [])
>>> g,empty=is_empty_no_side_effects(g)
>>> empty
True
>>> g=(i for i in range(10))
>>> g,empty=is_empty_no_side_effects(g)
>>> empty
False
>>> list(g)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
I hate to offer a second solution, especially one that I would not use myself, but, if you absolutely had to do this and to not consume the generator, as in other answers:
def do_something_with_item(item):
print item
empty_marker = object()
try:
first_item = my_generator.next()
except StopIteration:
print 'The generator was empty'
first_item = empty_marker
if first_item is not empty_marker:
do_something_with_item(first_item)
for item in my_generator:
do_something_with_item(item)
Now I really don't like this solution, because I believe that this is not how generators are to be used.
Prompted by Mark Ransom, here's a class that you can use to wrap any iterator so that you can peek ahead, push values back onto the stream and check for empty. It's a simple idea with a simple implementation that I've found very handy in the past.
class Pushable:
def __init__(self, iter):
self.source = iter
self.stored = []
def __iter__(self):
return self
def __bool__(self):
if self.stored:
return True
try:
self.stored.append(next(self.source))
except StopIteration:
return False
return True
def push(self, value):
self.stored.append(value)
def peek(self):
if self.stored:
return self.stored[-1]
value = next(self.source)
self.stored.append(value)
return value
def __next__(self):
if self.stored:
return self.stored.pop()
return next(self.source)
All you need to do to see if a generator is empty is to try to get the next result. Of course if you're not ready to use that result then you have to store it to return it again later.
Here's a wrapper class that can be added to an existing iterator to add an __nonzero__ test, so you can see if the generator is empty with a simple if. It can probably also be turned into a decorator.
class GenWrapper:
def __init__(self, iter):
self.source = iter
self.stored = False
def __iter__(self):
return self
def __nonzero__(self):
if self.stored:
return True
try:
self.value = next(self.source)
self.stored = True
except StopIteration:
return False
return True
def __next__(self): # use "next" (without underscores) for Python 2.x
if self.stored:
self.stored = False
return self.value
return next(self.source)
Here's how you'd use it:
with open(filename, 'r') as f:
f = GenWrapper(f)
if f:
print 'Not empty'
else:
print 'Empty'
Note that you can check for emptiness at any time, not just at the start of the iteration.
I realize that this post is 5 years old at this point, but I found it while looking for an idiomatic way of doing this, and did not see my solution posted. So for posterity:
import itertools
def get_generator():
"""
Returns (bool, generator) where bool is true iff the generator is not empty.
"""
gen = (i for i in [0, 1, 2, 3, 4])
a, b = itertools.tee(gen)
try:
a.next()
except StopIteration:
return (False, b)
return (True, b)
Of course, as I'm sure many commentators will point out, this is hacky and only works at all in certain limited situations (where the generators are side-effect free, for example). YMMV.
Sorry for the obvious approach, but the best way would be to do:
for item in my_generator:
print item
Now you have detected that the generator is empty while you are using it. Of course, item will never be displayed if the generator is empty.
This may not exactly fit in with your code, but this is what the idiom of the generator is for: iterating, so perhaps you might change your approach slightly, or not use generators at all.
>>> gen = (i for i in [])
>>> next(gen)
Traceback (most recent call last):
File "<pyshell#43>", line 1, in <module>
next(gen)
StopIteration
At the end of generator StopIteration is raised, since in your case end is reached immediately, exception is raised. But normally you shouldn't check for existence of next value.
another thing you can do is:
>>> gen = (i for i in [])
>>> if not list(gen):
print('empty generator')
I found only this solution as working for empty iterations as well.
def is_generator_empty(generator):
a, b = itertools.tee(generator)
try:
next(a)
except StopIteration:
return True, b
return False, b
is_empty, generator = is_generator_empty(generator)
Or if you do not want to use exception for this try to use
def is_generator_empty(generator):
a, b = itertools.tee(generator)
for item in a:
return False, b
return True, b
is_empty, generator = is_generator_empty(generator)
In the marked solution you are not able to use it for empty generators like
def get_empty_generator():
while False:
yield None
generator = get_empty_generator()
If you need to know before you use the generator, then no, there is no simple way. If you can wait until after you have used the generator, there is a simple way:
was_empty = True
for some_item in some_generator:
was_empty = False
do_something_with(some_item)
if was_empty:
handle_already_empty_generator_case()
Simply wrap the generator with itertools.chain, put something that will represent the end of the iterable as the second iterable, then simply check for that.
Ex:
import itertools
g = some_iterable
eog = object()
wrap_g = itertools.chain(g, [eog])
Now all that's left is to check for that value we appended to the end of the iterable, when you read it then that will signify the end
for value in wrap_g:
if value == eog: # DING DING! We just found the last element of the iterable
pass # Do something
In my case I needed to know if a host of generators was populated before I passed it on to a function, which merged the items, i.e., zip(...). The solution is similar, but different enough, from the accepted answer:
Definition:
def has_items(iterable):
try:
return True, itertools.chain([next(iterable)], iterable)
except StopIteration:
return False, []
Usage:
def filter_empty(iterables):
for iterable in iterables:
itr_has_items, iterable = has_items(iterable)
if itr_has_items:
yield iterable
def merge_iterables(iterables):
populated_iterables = filter_empty(iterables)
for items in zip(*populated_iterables):
# Use items for each "slice"
My particular problem has the property that the iterables are either empty or has exactly the same number of entries.
Use the peek function in cytoolz.
from cytoolz import peek
from typing import Tuple, Iterable
def is_empty_iterator(g: Iterable) -> Tuple[Iterable, bool]:
try:
_, g = peek(g)
return g, False
except StopIteration:
return g, True
The iterator returned by this function will be equivalent to the original one passed in as an argument.
Just to try to help with my "2 cents", I am going to describe my experience:
I have a generator that I need slicing it using itertools.islice into small generators. Then to check if my sub generators are empty or not, I just convert/consume them to a small list and I check if the list is empty or not.
For example:
from itertools import islice
def generator(max_yield=10):
a = 0
while True:
a += 1
if a > max_yield:
raise StopIteration()
yield a
tg = generator()
label = 1
while True:
itg = list(islice(tg, 3))
if not itg: # <-- I check if the list is empty or not
break
for i in itg:
print(f'#{label} - {i}')
label += 1
Output:
#1 - 1
#1 - 2
#1 - 3
#2 - 4
#2 - 5
#2 - 6
#3 - 7
#3 - 8
#3 - 9
#4 - 10
Maybe this is not the best approach, mainly because it consumes the generator, however it works to me.
Inspecting the generator before iterating over it conforms to the LBYL coding style. Another approach (EAFP) would be to iterate over it and then check whether it was empty or not.
is_empty = True
for item in generator:
is_empty = False
do_something(item)
if is_empty:
print('Generator is empty')
This approach also handles well infinite generators.
Here's a simple decorator which wraps the generator, so it returns None if empty. This can be useful if your code needs to know whether the generator will produce anything before looping through it.
def generator_or_none(func):
"""Wrap a generator function, returning None if it's empty. """
def inner(*args, **kwargs):
# peek at the first item; return None if it doesn't exist
try:
next(func(*args, **kwargs))
except StopIteration:
return None
# return original generator otherwise first item will be missing
return func(*args, **kwargs)
return inner
Usage:
import random
#generator_or_none
def random_length_generator():
for i in range(random.randint(0, 10)):
yield i
gen = random_length_generator()
if gen is None:
print('Generator is empty')
One example where this is useful is in templating code - i.e. jinja2
{% if content_generator %}
<section>
<h4>Section title</h4>
{% for item in content_generator %}
{{ item }}
{% endfor %
</section>
{% endif %}
peekable from more-itertools allows checking whether it's exhausted by checking its truth value. Demo with one empty and one non-empty iterator:
from more_itertools import peekable
for source in '', 'foobar':
it = iter(source)
if it := peekable(it):
print('values:', *it)
else:
print('empty')
Output:
empty
values: f o o b a r
This is an old and answered question, but as no one has shown it before, here it goes:
for _ in generator:
break
else:
print('Empty')
You can read more here
There's a very simple solution: if next(generator,-1) == -1 then the generator is empty!
I solved it by using the sum function. See below for an example I used with glob.iglob (which returns a generator).
def isEmpty():
files = glob.iglob(search)
if sum(1 for _ in files):
return True
return False
*This will probably not work for HUGE generators but should perform nicely for smaller lists
bool(generator) will return the correct result
I'm using yield from, but I don't know about the influence of while for yield. If I put yield from in a while loop, it works well, but when I cancel the loop at the mean time an exception occurs.
final_result = {}
def sales_sum(pro_name):
total = 0
nums = []
while True:
x = yield
print(pro_name+" Sales volume: ", x)
if not x:
break
total += x
nums.append(x)
return total, nums
def middle(key):
while True:
final_result[key] = yield from sales_sum(key)
def middle2(key):
final_result[key] = yield from sales_sum(key)
def main(fun):
data_sets = { "A": [1200, 1500], "B": [28,55,98]}
for key, data_set in data_sets.items():
m = fun(key)
m.send(None)
for value in data_set:
m.send(value)
m.send(None)
if __name__ == '__main__':
main(middle) # work well
main(middle2) # StopIteration
I expect main(middle2) to work well as main(middle), but there is a StopIteration exception.
The cause of the unexpected StopIteration exception in main is that your m.send(None) call causes your middle2 generator to be exhausted (after the sub-generator sales_sum breaks out of its loop in response to the falsey value it received). When a generator is exhausted, it raises StopIteration. Normally that's invisible because you consume iterators in for loops, but in this case, it breaks your code.
There are a few ways you could fix this. One would be to use a two-argument call to next instead of using m.send(None):
next(m, None)
This does the same thing as m.send(None), but has the added benefit of suppressing the StopIteration. Note that the None in the call to next is not really the same as the one in send. It's the default return value in the case of an exhausted iterator, not the value that gets sent in (which is always None when using next).
Another approach would be to change middle2 so that it doesn't end when the sales_sum generator does. You could add an extra yield statement at the end, so that it yields control one last time after doing its assignment to final_result when its sub-generator returns.
A final idea would be to replace m.send(None) with m.close(). This would require some changes in final_result, as the close call will throw a GeneratorExit exception into the generator. If you expect it, you could use that as your signal to be done instead of looking for a falsey value:
def sales_sum(pro_name):
total = 0
nums = []
while True:
try:
x = yield
except GeneratorExit:
return total, nums
print(pro_name+" Sales volume: ", x)
total += x
nums.append(x)
With this change, middle2 would not need any modification.
sales_sum is a finite iterator. middle2 iterates over it exactly once; middle tries to iterate over it multiple times.
What exactly happens, when yield and return are used in the same function in Python, like this?
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += len(sub) # use start += 1 to find overlapping matches
Is it still a generator?
Yes, it' still a generator. The return is (almost) equivalent to raising StopIteration.
PEP 255 spells it out:
Specification: Return
A generator function can also contain return statements of the form:
"return"
Note that an expression_list is not allowed on return statements in
the body of a generator (although, of course, they may appear in the
bodies of non-generator functions nested within the generator).
When a return statement is encountered, control proceeds as in any
function return, executing the appropriate finally clauses (if any
exist). Then a StopIteration exception is raised, signalling that the
iterator is exhausted. A StopIteration exception is also raised if
control flows off the end of the generator without an explict return.
Note that return means "I'm done, and have nothing interesting to
return", for both generator functions and non-generator functions.
Note that return isn't always equivalent to raising StopIteration:
the difference lies in how enclosing try/except constructs are
treated. For example,
>>> def f1():
... try:
... return
... except:
... yield 1
>>> print list(f1())
[]
because, as in any function, return simply exits, but
>>> def f2():
... try:
... raise StopIteration
... except:
... yield 42
>>> print list(f2())
[42]
because StopIteration is captured by a bare "except", as is any
exception.
Yes, it is still a generator. An empty return or return None can be used to end a generator function. It is equivalent to raising a StopIteration(see #NPE's answer for details).
Note that a return with non-None arguments is a SyntaxError in Python versions prior to 3.3.
As pointed out by #BrenBarn in comments starting from Python 3.3 the return value is now passed to StopIteration.
From PEP 380:
In a generator, the statement
return value
is semantically equivalent to
raise StopIteration(value)
There is a way to accomplish having a yield and return method in a function that allows you to return a value or generator.
It probably is not as clean as you would want but it does do what you expect.
Here's an example:
def six(how_many=None):
if how_many is None or how_many < 1:
return None # returns value
if how_many == 1:
return 6 # returns value
def iter_func():
for count in range(how_many):
yield 6
return iter_func() # returns generator
Note: you don't get StopIteration exception with the example below.
def odd(max):
n = 0
while n < max:
yield n
n = n + 1
return 'done'
for x in odd(3):
print(x)
The for loop catches it. That's its signal to stop
But you can catch it in this way:
g = odd(3)
while True:
try:
x = next(g)
print(x)
except StopIteration as e:
print("g return value:", e.value)
break
Is there a simple way of testing if the generator has no items, like peek, hasNext, isEmpty, something along those lines?
Suggestion:
def peek(iterable):
try:
first = next(iterable)
except StopIteration:
return None
return first, itertools.chain([first], iterable)
Usage:
res = peek(mysequence)
if res is None:
# sequence is empty. Do stuff.
else:
first, mysequence = res
# Do something with first, maybe?
# Then iterate over the sequence:
for element in mysequence:
# etc.
The simple answer to your question: no, there is no simple way. There are a whole lot of work-arounds.
There really shouldn't be a simple way, because of what generators are: a way to output a sequence of values without holding the sequence in memory. So there's no backward traversal.
You could write a has_next function or maybe even slap it on to a generator as a method with a fancy decorator if you wanted to.
A simple way is to use the optional parameter for next() which is used if the generator is exhausted (or empty). For example:
_exhausted = object()
if next(some_generator, _exhausted) is _exhausted:
print('generator is empty')
Quick-dirty solution:
next(generator, None) is not None
Or replace None by whatever value you know it's not in your generator.
Edit: Yes, this will skip 1 item in the generator. Sometimes, however, I check whether a generator is empty only for validation purposes, then don't really use it. Or otherwise I do something like:
def foo(self):
if next(self.my_generator(), None) is None:
raise Exception("Not initiated")
for x in self.my_generator():
...
That is, this works if your generator comes from a function, as in my_generator().
The best approach, IMHO, would be to avoid a special test. Most times, use of a generator is the test:
thing_generated = False
# Nothing is lost here. if nothing is generated,
# the for block is not executed. Often, that's the only check
# you need to do. This can be done in the course of doing
# the work you wanted to do anyway on the generated output.
for thing in my_generator():
thing_generated = True
do_work(thing)
If that's not good enough, you can still perform an explicit test. At this point, thing will contain the last value generated. If nothing was generated, it will be undefined - unless you've already defined the variable. You could check the value of thing, but that's a bit unreliable. Instead, just set a flag within the block and check it afterward:
if not thing_generated:
print "Avast, ye scurvy dog!"
Just fell on this thread and realized that a very simple and easy to read answer was missing:
def is_empty(generator):
for item in generator:
return False
return True
If we are not suppose to consume any item then we need to re-inject the first item into the generator:
def is_empty_no_side_effects(generator):
try:
item = next(generator)
def my_generator():
yield item
yield from generator
return my_generator(), False
except StopIteration:
return (_ for _ in []), True
Example:
>>> g=(i for i in [])
>>> g,empty=is_empty_no_side_effects(g)
>>> empty
True
>>> g=(i for i in range(10))
>>> g,empty=is_empty_no_side_effects(g)
>>> empty
False
>>> list(g)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
I hate to offer a second solution, especially one that I would not use myself, but, if you absolutely had to do this and to not consume the generator, as in other answers:
def do_something_with_item(item):
print item
empty_marker = object()
try:
first_item = my_generator.next()
except StopIteration:
print 'The generator was empty'
first_item = empty_marker
if first_item is not empty_marker:
do_something_with_item(first_item)
for item in my_generator:
do_something_with_item(item)
Now I really don't like this solution, because I believe that this is not how generators are to be used.
Prompted by Mark Ransom, here's a class that you can use to wrap any iterator so that you can peek ahead, push values back onto the stream and check for empty. It's a simple idea with a simple implementation that I've found very handy in the past.
class Pushable:
def __init__(self, iter):
self.source = iter
self.stored = []
def __iter__(self):
return self
def __bool__(self):
if self.stored:
return True
try:
self.stored.append(next(self.source))
except StopIteration:
return False
return True
def push(self, value):
self.stored.append(value)
def peek(self):
if self.stored:
return self.stored[-1]
value = next(self.source)
self.stored.append(value)
return value
def __next__(self):
if self.stored:
return self.stored.pop()
return next(self.source)
All you need to do to see if a generator is empty is to try to get the next result. Of course if you're not ready to use that result then you have to store it to return it again later.
Here's a wrapper class that can be added to an existing iterator to add an __nonzero__ test, so you can see if the generator is empty with a simple if. It can probably also be turned into a decorator.
class GenWrapper:
def __init__(self, iter):
self.source = iter
self.stored = False
def __iter__(self):
return self
def __nonzero__(self):
if self.stored:
return True
try:
self.value = next(self.source)
self.stored = True
except StopIteration:
return False
return True
def __next__(self): # use "next" (without underscores) for Python 2.x
if self.stored:
self.stored = False
return self.value
return next(self.source)
Here's how you'd use it:
with open(filename, 'r') as f:
f = GenWrapper(f)
if f:
print 'Not empty'
else:
print 'Empty'
Note that you can check for emptiness at any time, not just at the start of the iteration.
I realize that this post is 5 years old at this point, but I found it while looking for an idiomatic way of doing this, and did not see my solution posted. So for posterity:
import itertools
def get_generator():
"""
Returns (bool, generator) where bool is true iff the generator is not empty.
"""
gen = (i for i in [0, 1, 2, 3, 4])
a, b = itertools.tee(gen)
try:
a.next()
except StopIteration:
return (False, b)
return (True, b)
Of course, as I'm sure many commentators will point out, this is hacky and only works at all in certain limited situations (where the generators are side-effect free, for example). YMMV.
Sorry for the obvious approach, but the best way would be to do:
for item in my_generator:
print item
Now you have detected that the generator is empty while you are using it. Of course, item will never be displayed if the generator is empty.
This may not exactly fit in with your code, but this is what the idiom of the generator is for: iterating, so perhaps you might change your approach slightly, or not use generators at all.
>>> gen = (i for i in [])
>>> next(gen)
Traceback (most recent call last):
File "<pyshell#43>", line 1, in <module>
next(gen)
StopIteration
At the end of generator StopIteration is raised, since in your case end is reached immediately, exception is raised. But normally you shouldn't check for existence of next value.
another thing you can do is:
>>> gen = (i for i in [])
>>> if not list(gen):
print('empty generator')
I found only this solution as working for empty iterations as well.
def is_generator_empty(generator):
a, b = itertools.tee(generator)
try:
next(a)
except StopIteration:
return True, b
return False, b
is_empty, generator = is_generator_empty(generator)
Or if you do not want to use exception for this try to use
def is_generator_empty(generator):
a, b = itertools.tee(generator)
for item in a:
return False, b
return True, b
is_empty, generator = is_generator_empty(generator)
In the marked solution you are not able to use it for empty generators like
def get_empty_generator():
while False:
yield None
generator = get_empty_generator()
If you need to know before you use the generator, then no, there is no simple way. If you can wait until after you have used the generator, there is a simple way:
was_empty = True
for some_item in some_generator:
was_empty = False
do_something_with(some_item)
if was_empty:
handle_already_empty_generator_case()
Simply wrap the generator with itertools.chain, put something that will represent the end of the iterable as the second iterable, then simply check for that.
Ex:
import itertools
g = some_iterable
eog = object()
wrap_g = itertools.chain(g, [eog])
Now all that's left is to check for that value we appended to the end of the iterable, when you read it then that will signify the end
for value in wrap_g:
if value == eog: # DING DING! We just found the last element of the iterable
pass # Do something
In my case I needed to know if a host of generators was populated before I passed it on to a function, which merged the items, i.e., zip(...). The solution is similar, but different enough, from the accepted answer:
Definition:
def has_items(iterable):
try:
return True, itertools.chain([next(iterable)], iterable)
except StopIteration:
return False, []
Usage:
def filter_empty(iterables):
for iterable in iterables:
itr_has_items, iterable = has_items(iterable)
if itr_has_items:
yield iterable
def merge_iterables(iterables):
populated_iterables = filter_empty(iterables)
for items in zip(*populated_iterables):
# Use items for each "slice"
My particular problem has the property that the iterables are either empty or has exactly the same number of entries.
Use the peek function in cytoolz.
from cytoolz import peek
from typing import Tuple, Iterable
def is_empty_iterator(g: Iterable) -> Tuple[Iterable, bool]:
try:
_, g = peek(g)
return g, False
except StopIteration:
return g, True
The iterator returned by this function will be equivalent to the original one passed in as an argument.
Just to try to help with my "2 cents", I am going to describe my experience:
I have a generator that I need slicing it using itertools.islice into small generators. Then to check if my sub generators are empty or not, I just convert/consume them to a small list and I check if the list is empty or not.
For example:
from itertools import islice
def generator(max_yield=10):
a = 0
while True:
a += 1
if a > max_yield:
raise StopIteration()
yield a
tg = generator()
label = 1
while True:
itg = list(islice(tg, 3))
if not itg: # <-- I check if the list is empty or not
break
for i in itg:
print(f'#{label} - {i}')
label += 1
Output:
#1 - 1
#1 - 2
#1 - 3
#2 - 4
#2 - 5
#2 - 6
#3 - 7
#3 - 8
#3 - 9
#4 - 10
Maybe this is not the best approach, mainly because it consumes the generator, however it works to me.
Inspecting the generator before iterating over it conforms to the LBYL coding style. Another approach (EAFP) would be to iterate over it and then check whether it was empty or not.
is_empty = True
for item in generator:
is_empty = False
do_something(item)
if is_empty:
print('Generator is empty')
This approach also handles well infinite generators.
Here's a simple decorator which wraps the generator, so it returns None if empty. This can be useful if your code needs to know whether the generator will produce anything before looping through it.
def generator_or_none(func):
"""Wrap a generator function, returning None if it's empty. """
def inner(*args, **kwargs):
# peek at the first item; return None if it doesn't exist
try:
next(func(*args, **kwargs))
except StopIteration:
return None
# return original generator otherwise first item will be missing
return func(*args, **kwargs)
return inner
Usage:
import random
#generator_or_none
def random_length_generator():
for i in range(random.randint(0, 10)):
yield i
gen = random_length_generator()
if gen is None:
print('Generator is empty')
One example where this is useful is in templating code - i.e. jinja2
{% if content_generator %}
<section>
<h4>Section title</h4>
{% for item in content_generator %}
{{ item }}
{% endfor %
</section>
{% endif %}
peekable from more-itertools allows checking whether it's exhausted by checking its truth value. Demo with one empty and one non-empty iterator:
from more_itertools import peekable
for source in '', 'foobar':
it = iter(source)
if it := peekable(it):
print('values:', *it)
else:
print('empty')
Output:
empty
values: f o o b a r
This is an old and answered question, but as no one has shown it before, here it goes:
for _ in generator:
break
else:
print('Empty')
You can read more here
There's a very simple solution: if next(generator,-1) == -1 then the generator is empty!
I solved it by using the sum function. See below for an example I used with glob.iglob (which returns a generator).
def isEmpty():
files = glob.iglob(search)
if sum(1 for _ in files):
return True
return False
*This will probably not work for HUGE generators but should perform nicely for smaller lists
bool(generator) will return the correct result
I'm a big fan of Python's for...else syntax - it's surprising how often it's applicable, and how effectively it can simplify code.
However, I've not figured out a nice way to use it in a generator, for example:
def iterate(i):
for value in i:
yield value
else:
print 'i is empty'
In the above example, I'd like the print statement to be executed only if i is empty. However, as else only respects break and return, it is always executed, regardless of the length of i.
If it's impossible to use for...else in this way, what's the best approach to this so that the print statement is only executed when nothing is yielded?
You're breaking the definition of a generator, which should throw a StopIteration exception when iteration is complete (which is automatically handled by a return statement in a generator function)
So:
def iterate(i):
for value in i:
yield value
return
Best to let the calling code handle the case of an empty iterator:
count = 0
for value in iterate(range([])):
print value
count += 1
else:
if count == 0:
print "list was empty"
Might be a cleaner way of doing the above, but that ought to work fine, and doesn't fall into any of the common 'treating an iterator like a list' traps below.
There are a couple ways of doing this. You could always use the Iterator directly:
def iterate(i):
try:
i_iter = iter(i)
next = i_iter.next()
except StopIteration:
print 'i is empty'
return
while True:
yield next
next = i_iter.next()
But if you know more about what to expect from the argument i, you can be more concise:
def iterate(i):
if i: # or if len(i) == 0
for next in i:
yield next
else:
print 'i is empty'
raise StopIteration()
Summing up some of the earlier answers, it could be solved like this:
def iterate(i):
empty = True
for value in i:
yield value
empty = False
if empty:
print "empty"
so there really is no "else" clause involved.
As you note, for..else only detects a break. So it's only applicable when you look for something and then stop.
It's not applicable to your purpose not because it's a generator, but because you want to process all elements, without stopping (because you want to yield them all, but that's not the point).
So generator or not, you really need a boolean, as in Ber's solution.
If it's impossible to use for...else in this way, what's the best approach to this so that the print statement is only executed when nothing is yielded?
Maximum i can think of:
>>> empty = True
>>> for i in [1,2]:
... empty = False
... if empty:
... print 'empty'
...
>>>
>>>
>>> empty = True
>>> for i in []:
... empty = False
... if empty:
... print 'empty'
...
empty
>>>
What about simple if-else?
def iterate(i):
if len(i) == 0: print 'i is empty'
else:
for value in i:
yield value