Some weird thing in python code, map() function

Question

arr = map(int, input().split())

for i in arr:
    print(i)

print(list(arr)[1])

When I run this code and give the input as,

1 2 3 4 5

it gives the output as follows,

1
2
3
4
5
Traceback (most recent call last):
  File "/home/muyustan/PycharmProjects/openCV/split.py", line 6, in <module>
    print(list(arr)[1])
IndexError: list index out of range

Then,

arr = map(int, input().split())

# for i in arr:
#     print(i)

print(list(arr)[1])

after commenting out the for loop and running the program and providing the same input, it gives the output at the terminal as follows;

Why using that for loop makes any changes?

Asked By: muyustan

||

Source

Answer 1

If you had printed arr before the line of the exception, you’d have seen:

c:srvtmp> py3 weirdmap.py
1 2 3 4 5
<map object at 0x03611770>
2

the map-object is an iterator (that you can iterate over exactly once).

You could fix your code by either using a list comprehension:

arr = [int(v) for v in input().split()]

or by converting the iterator to a list before doing anything else with it:

arr = list(map(int, input().split()))

to see what is going on in the last line, consider this re-write:

mapobj = map(int, input().split())
arr = list(mapobj)
print(arr[2])         # <=== return the 3rd element
print(mapobj[2])      # <=== TypeError: 'map' object is not subscriptable

changing the last line to

print(list(mapobj)[2])   # <=== IndexError: list index out of range

since the arr = list(mapobj) used up the iterator.

Answered By: thebjorn

Answer 2

Map function doesn’t return a list but an iterator.
To obtain a list

arr = list(map(int, input().split()))

You can only loop once over an iterator.
When you write

arr = map(int, input().split())

No operations are made, you need to call the next() method over arr to obtain the next item. For loop does that.

Answered By: BMarin

Some weird thing in python code, map() function

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