Yield multiple values
Question:
Can’t we yield more than one value in the python generator functions?
Example,
In [677]: def gen():
.....: for i in range(5):
.....: yield i, i+1
.....:
In [680]: k1, k2 = gen()
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-680-b21f6543a7e9> in <module>()
----> 1 k1, k2 = a()
ValueError: too many values to unpack
This works as follows:
In [678]: b = a()
In [679]: list(b)
Out[679]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
Same results even when I do this:
In [692]: def a():
for i in range(5):
yield i
yield i+1
Thanks.
Answers:
Because gen() returns a generator (a single item – so it can’t be unpacked as two), it needs to be advanced first to get the values…
g = gen()
a, b = next(g)
It works with list because that implicitly consumes the generator.
Can we further make this a generator? Something like this:
g = gen();
def yield_g():
yield g.next();
k1,k2 = yield_g();
and therefore list(k1) would give [0,1,2,3,4] and list(k2) would give [1,2,3,4,5].
Keep your existing generator, and use izip (or zip):
from itertools import izip
k1, k2 = izip(*gen())
Your function gen returns a generator and not values as you might expect judging from the example you gave. If you iterate over the generator the pairs of values will be yielded:
In [2]: def gen():
...: for i in range(5):
...: yield i, i+1
...:
In [3]: for k1, k2 in gen():
...: print k1, k2
...:
0 1
1 2
2 3
3 4
4 5
Use yield from
def gen():
for i in range(5):
yield from (i, i+1)
[v for v in gen()]
# [0, 1, 1, 2, 2, 3, 3, 4, 4, 5]
The python docs say:
When yield from <expr> is used, it treats the supplied expression as a subiterator.
Both Solution works just fine, in case you want to return multiple values from the generator.
def gen():
for i in range(5):
yield i, i+1
Solution 1:
for values in gen():
print(values[0],values[1])
Solution 2:
for value1, value2 in gen():
print(value1, value2)
Can’t we yield more than one value in the python generator functions?
Example,
In [677]: def gen():
.....: for i in range(5):
.....: yield i, i+1
.....:
In [680]: k1, k2 = gen()
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-680-b21f6543a7e9> in <module>()
----> 1 k1, k2 = a()
ValueError: too many values to unpack
This works as follows:
In [678]: b = a()
In [679]: list(b)
Out[679]: [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
Same results even when I do this:
In [692]: def a():
for i in range(5):
yield i
yield i+1
Thanks.
Because gen() returns a generator (a single item – so it can’t be unpacked as two), it needs to be advanced first to get the values…
g = gen()
a, b = next(g)
It works with list because that implicitly consumes the generator.
Can we further make this a generator? Something like this:
g = gen();
def yield_g():
yield g.next();
k1,k2 = yield_g();
and therefore list(k1) would give [0,1,2,3,4] and list(k2) would give [1,2,3,4,5].
Keep your existing generator, and use izip (or zip):
from itertools import izip
k1, k2 = izip(*gen())
Your function gen returns a generator and not values as you might expect judging from the example you gave. If you iterate over the generator the pairs of values will be yielded:
In [2]: def gen():
...: for i in range(5):
...: yield i, i+1
...:
In [3]: for k1, k2 in gen():
...: print k1, k2
...:
0 1
1 2
2 3
3 4
4 5
Use yield from
def gen():
for i in range(5):
yield from (i, i+1)
[v for v in gen()]
# [0, 1, 1, 2, 2, 3, 3, 4, 4, 5]
The python docs say:
When
yield from <expr>is used, it treats the supplied expression as a subiterator.
Both Solution works just fine, in case you want to return multiple values from the generator.
def gen():
for i in range(5):
yield i, i+1
Solution 1:
for values in gen():
print(values[0],values[1])
Solution 2:
for value1, value2 in gen():
print(value1, value2)