Is a map iterator object in Python3.x can only be used once and why? (codes and examples attached)

With this code:

strs = ["111", "1000", "1000", "1000"]
# count the numbers of '0' and '1' respectively for each string in strs 
counts = map(lambda x: [x.count('0'), x.count('1')], strs)

cntSortBy0 = sorted(counts, key=lambda x: x[0])  # sort with counts of '0'

cntSortBy1 = sorted(counts, key=lambda x: x[1])  # sort with counts of '1'

Here I have a list strs with strings as elements.

Then I count the numbers of '0' and '1' respectively for each string in strs and save the results in counts (I did this using map in python3, so count is a map object).

After that I sort counts for the first time, and it works fine, but when I sort counts for the second time, it returns an empty list(cntSortBy1 is empty),

I find that this is because counts became empty after the first sort:

>>> strs = ["111", "1000", "1000", "1000"]
>>>  counts = map(lambda x: [x.count('0'), x.count('1')], strs)
>>> cntSortBy0 = sorted(counts, key=lambda x: x[0])  # sort with counts of '0'
>>> list(counts)
[]

No wonder cntSortBy1 is empty, but why dose this happen?

Besides, if I just merely print list(counts), cntSortBy1 will also became empty, as shown below,

>>> strs = ["111", "1000", "1000", "1000"]
>>> counts = map(lambda x: [x.count('0'), x.count('1')], strs)
>>> list(counts)
[[0, 3], [3, 1], [3, 1], [3, 1]]
>>> cntSortBy0 = sorted(counts, key=lambda x: x[0])
>>> cntSortBy0
[]

Dose this mean a map object could only be used once, after that it will become an empty list?
Is there any other objects in Py3 with the same feature? (I have tried range(), it is not.)

Many Thanks!!!

Solution

Once an iterator has been exhausted (meaning, fully iterated over), there is nothing left to be iterated over, so iterating over it a second time won't yield anything. Take this example.

a = map(int,['1','2','3'])

for i in a:
    print(i)

for i in a:
    print(i)

This will output:

1
2
3

Whenever you need to iterate over an iterator multiple times, turn it into a list (or another type of sequence):

a = list(map(int,['1','2','3']))

This outputs:

When using sorted(), you're also iterating over the map iterator object:

a = map(int,['1','2','3'])
print(sorted(a)) # output: [1, 2, 3]
print(sorted(a)) # output: []

a = list(map(int,['1','2','3']))
print(sorted(a)) # output: [1, 2, 3]
print(sorted(a)) # output: [1, 2, 3]