Map python ошибки

Что такое map?

Это функция, которая применит первый аргумент (функцию) к каждому элементу второго аргумента (коллекции), но в python3 это не происходит сразу, как в python2, где map возвращает список:

>>> map(int, "12345")
[1, 2, 3, 4, 5]

А постепенно, т.к. map возвращает итератор «map object»:

>>> map(int, "12345")
<map object at 0x000001D51030BB50>

И как все итераторы, мы можем итерировать по map с помощью next:

>>> m = map(int, "12345")
>>> next(m)
1
>>> next(m)
2
>>> next(m)
3
>>> next(m)
4
>>> next(m)
5
>>> next(m)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
StopIteration

Или же с помощь. for цикла:

>>> for i in map(int, "12345"):
...     print(i)
...
1
2
3
4
5

В чём проблема?

Вы пытаетесь получить первый элемент итератора, но т.к. у итератор нет элементов, он создаёт их «на лету», то и питон на это жалуется:

>>> map(int, "12345")[0]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'map' object is not subscriptable

Что же вы хотите сделать — превратить map в коллекцию, например list, тогда и по индексу можно будет обращаться:

>>> result = list(map(int, "12345"))
>>> result[0]
1
>>> for i in result[1:]:
...     print(i)
...
2
3
4
5

Альтернатива

В питоне также есть comprehensions — это способ инициализировать что-то, в нашем случае список:

>>> [int(i) for i in "12345"]
[1, 2, 3, 4, 5]

Здесь используется for цикл, чтобы создавать значения, а значение int(i) уже записывается в конечный список.

Хитрость

Вы использовали flag, чтобы проверить, вышли ли вы из цикла с помощью break, но в питоне есть специальный синтаксис для этого — for .. else ..:

>>> for i in 1, 2, 3:
...     if i == 2:
...             print("нашёл")
...             break
... else:
...     print("не нашёл")
...
нашёл

Если мы выходим из цикла через break, то else будет проигнорирован

>>> for i in 1, 3:
...     if i == 2:
...             print("нашёл")
...             break
... else:
...     print("не нашёл")
...
не нашёл

А если не выйдем, то выполнится.

Итог

Если мы применим наши новые знания, то получим:

a = [int(i) for i in input().split(" ")]

prev = a[0]

for i in a[1:]:
    if prev <= i:
        prev = i
    else:
        print("No")
        break
else:
    print("Yes")

или

a = list(map(int, input().split(" ")))

prev = a[0]

for i in a[1:]:
    if prev <= i:
        prev = i
    else:
        print("No")
        break
else:
    print("Yes")

в зависимости от того, что вы предпочитаете, map или comprehension.
(как по мне, comprehensions намного более интуитивные :] )

In Python 3, a map object is an iterator and is not subscriptable. If you try to access items inside a map object using the subscript operator [], you will raise the TypeError: ‘map’ object is not subscriptable.

This error typically occurs when using Python 2 syntax when using Python 3. In Python 2, calling the built-in map() function returns a list, which is subscriptable.

You can solve this error by converting the map object to a list using the built-in list function. For example,

new_list = list(map(int, old_list))

You can also iterate over the values in the iterator using a for loop

This tutorial will go through the error in detail and how to solve it with code examples.

TypeError: ‘map’ object is not subscriptable

A TypeError occurs when you perform an illegal operation for a specific data type.

What Does Subscriptable Mean?

The subscript operator, which is square brackets [], retrieves items from subscriptable objects like lists or tuples. The operator calls the __getitem__ method, for example, a[i] is equivalent to a.__getitem__(i).

All subscriptable objects have a __getitem__ method. Map objects are iterators and do not have a __getitem__ method. We can verify that map objects do not have the __getitem__ method by defining a creating a map object and passing it to the dir() method:

def square(i):
    res = i ** 2
    return res

lst = [2, 3, 4, 5]

squared = map(square, lst)

print(dir(squared))

['__class__', '__delattr__', '__dir__', '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__', '__init_subclass__', '__iter__', '__le__', '__lt__', '__ne__', '__new__', '__next__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__']

 A Python iterator object must implement two special methods, __iter__() and __next__(), collectively called the iterator protocol.

Example

This error typically occurs when using Python 2 map operations in Python 3. First, let’s verify what the built-in map function returns in Python 2. We will use a virtual environment with Python 2 installed and confirm that we are using Python 2 using the sys library:

import sys

print(sys.version)

2.7.16 |Anaconda, Inc.| (default, Sep 24 2019, 16:55:38) 
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)]

Next, we will define a list of strings, which we will convert to a list of integers using the map() function. We will pass the object returned by the map function to the type() function to get the type.

string_list = ["2", "3", "4", "5", "6"]

int_list = map(int, string_list)

print(type(int_list))

Let’s run the code to get the result:

<type 'list'>

In Python 2, the map function returns a list, which is subscriptable. We access an element of the list using the subscript operator, for example:

print(int_list[0])

2

Let’s try to do this using Python 3. We will use a different virtual environment with Python 3 installed, which we will verify using sys.

import sys

print(sys.version)

3.8.12 (default, Oct 12 2021, 06:23:56) 
[Clang 10.0.0 ]

Next, we will define our list of strings and use the map() function to convert it to a list of integers. We will pass the object returned by the map function to the type function.

string_list = ["2", "3", "4", "5", "6"]

int_list = map(int, string_list)

print(type(int_list))

Let’s run the code to see the result:

<class 'map'>

In Python 3, the map function returns a map object, which is an iterator. If we try to access an element of the map object using the subscript operator, we will raise the TypeError.

print(int_list[0])

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
Input In [4], in <cell line: 1>()
----> 1 print(int_list[0])

TypeError: 'map' object is not subscriptable

Solution #1

We can convert map objects to lists using the built-in list() function. Let’s look at the revised code:

string_list = ["2", "3", "4", "5", "6"]

int_list = list(map(int, string_list))

print(type(int_list))

print(int_list[0])

Let’s run the code to get the result:

<class 'list'>
2

Solution #2

We can access values in an iterator using a for loop, which invokes the __next__() method of the map iterator. Let’s look at the revised code:

string_list = ["2", "3", "4", "5", "6"]
int_list = map(int, string_list)

for i in int_list:
    print(i)

Let’s run the code to get the result:

2
3
4
5
6

Summary

Congratulations on reading to the end of this tutorial! The TypeError: ‘map’ object is not subscriptable occurs when you try to access an item from a map object in Python 3. The Python 3 map object is an iterator and is not subscriptable. We can convert map objects to lists which are subscriptable.

For further reading on TypeErrors, go to the article: How to Solve Python TypeError: ‘function’ object is not subscriptable

Go to the online courses page on Python to learn more about Python for data science and machine learning.

Have fun and happy researching!

1. Cause of the TypeError: 'map' object is not subscriptable Error in Python
2. Fix the TypeError: 'map' object is not subscriptable Error in Python

Every programming language encounters many errors. Some occur at the compile time, some at run time.

This article will discuss the TypeError: 'map' object is not subscriptable, a sub-class of TypeError. We encounter a TypeError when we try to perform an operation incompatible with the type of object.

Cause of the TypeError: 'map' object is not subscriptable Error in Python

Python 2 Map Operations in Python 3

In Python 2, the map() method returns a list. We can access the list’s elements using their index through the subscriptor operator [].

In Python 3, the map() method returns an object which is an iterator and cannot be subscripted. If we try to access the item using the subscript operator [], the TypeError: 'map' object is not subscriptable will be raised.

Example Code:

#Python 3.x
my_list = ["1", "2", "3", "4"]
my_list = map(int, my_list)
print(type(my_list))
my_list[0]

Output:

#Python 3.x
<class 'map'>
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-1-07511913e32f> in <module>()
      2 my_list = map(int, my_list)
      3 print(type(my_list))
----> 4 my_list[0]

TypeError: 'map' object is not subscriptable

Fix the TypeError: 'map' object is not subscriptable Error in Python

Convert the Map Object to List in Python 3

If we convert the map object to a list using the list() method, we can use the subscript operator/list methods.

Example Code:

#Python 3.x
my_list = ["1", "2", "3", "4"]
my_list = list(map(int, my_list))
print(my_list[0])

Output:

Use the for Loop With Iterator in Python 3

We can access the items in the iterator using a for loop. It calls the __next__() method at the back and prints all the values.

Example Code:

#Python 3.x
my_list = ["1", "2", "3", "4"]
my_list = list(map(int, my_list))
for i in my_list:
    print(i)

Output: