CODEWITHSUNDEEP
python-3.xiterator
Ser Jothan Chanes
Ser Jothan Chanes

Reputation: 85

Difference between map and list iterators in python3

I ran into unexpected behaviour when working with map and list iterators in python3. In this MWE I first generate a map of maps. Then, I want the first element of each map in one list, and the remaining parts in the original map:

# s will be be a map of maps
s=[[1,2,3],[4,5,6]]
s=map(lambda l: map(lambda t:t,l),s)

# uncomment to obtain desired output
# s = list(s) # s is now a list of maps

s1 = map(next,s)

print(list(s1))
print(list(map(list,s)))

Running the MWE as is in python 3.4.2 yields the expected output for s1:

s1 = ([1,4]),

but the empty list [] for s. Uncommenting the marked line yields the correct output, s1 as above, but with the expected output for s as well:

s=[[2,3],[5,6]].

The docs say that map expects an iterable. To this day, I saw no difference between map and list iterators. Could someone explain this behaviour?

PS: Curiously enough, if I uncomment the first print statement, the initial state of s is printed. So it could also be that this behaviour has something to do with a kind of lazy(?) evaluation of maps?

Upvotes: 0

Views: 2121

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1122222

A map() is an iterator; you can only iterate over it once. You could get individual elements with next() for example, but once you run out of items you cannot get any more values.

I've given your objects a few easier-to-remember names:

>>> s = [[1, 2, 3], [4, 5, 6]]
>>> map_of_maps = map(lambda l: map(lambda t: t, l), s)
>>> first_elements = map(next, map_of_maps)

Iterating over first_elements here will in turn iterate over map_of_maps. You can only do so once, so once we run out of elements any further iteration will fail:

>>> next(first_elements)
1
>>> next(first_elements)
4
>>> next(first_elements)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

list() does exactly the same thing; it takes an iterable argument, and will iterate over that object to create a new list object from the results. But if you give it a map() that is already exhausted, there is nothing to copy into the new list anymore. As such, you get an empty result:

>>> list(first_elements)
[]

You need to recreate the map() from scratch:

>>> map_of_maps = map(lambda l: map(lambda t: t, l), s)
>>> first_elements = map(next, map_of_maps)
>>> list(first_elements)
[1, 4]
>>> list(first_elements)
[]

Note that a second list() call on the map() object resulted in an empty list object, once again.

Upvotes: 2

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