map returning map object and converting it to list removes all elements in it [duplicate]

-1

This question already has an answer here:

Python: calling 'list' on a map object twice

1 answer

Just learned that the object returned from map() doesn't hold up once it has been used in a in expression or it was converted into a list.

What is causing b to get emptied at the end?

>>> a = [1, 2, 3]

>>> b = map(lambda x: x, a)

>>> b

<map object at 0x104d8ccc0>

>>> list(b)

[1, 2, 3]

>>> list(b)

edited Jan 20 at 4:18

asked Jan 20 at 3:05

msk

1004

marked as duplicate by Tomothy32, ShadowRanger, Austin, jpp python-3.x
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Jan 20 at 3:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

My bad. Thanks for the englightenment. Btw, isn't it more natural map returning the same collection(type)? Is there any merit in addition to some gain in performance of iteration immediately following it?

– msk
Jan 20 at 4:16

add a comment |

-1

This question already has an answer here:

Python: calling 'list' on a map object twice

1 answer

Just learned that the object returned from map() doesn't hold up once it has been used in a in expression or it was converted into a list.

What is causing b to get emptied at the end?

>>> a = [1, 2, 3]

>>> b = map(lambda x: x, a)

>>> b

<map object at 0x104d8ccc0>

>>> list(b)

[1, 2, 3]

>>> list(b)

edited Jan 20 at 4:18

asked Jan 20 at 3:05

msk

1004

marked as duplicate by Tomothy32, ShadowRanger, Austin, jpp python-3.x
Users with the python-3.x badge can single-handedly close python-3.x questions as duplicates and reopen them as needed.

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Jan 20 at 3:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

My bad. Thanks for the englightenment. Btw, isn't it more natural map returning the same collection(type)? Is there any merit in addition to some gain in performance of iteration immediately following it?

– msk
Jan 20 at 4:16

add a comment |

-1

This question already has an answer here:

Python: calling 'list' on a map object twice

1 answer

Just learned that the object returned from map() doesn't hold up once it has been used in a in expression or it was converted into a list.

What is causing b to get emptied at the end?

>>> a = [1, 2, 3]

>>> b = map(lambda x: x, a)

>>> b

<map object at 0x104d8ccc0>

>>> list(b)

[1, 2, 3]

>>> list(b)

edited Jan 20 at 4:18

asked Jan 20 at 3:05

msk

1004

This question already has an answer here:

Python: calling 'list' on a map object twice

1 answer

Just learned that the object returned from map() doesn't hold up once it has been used in a in expression or it was converted into a list.

What is causing b to get emptied at the end?

>>> a = [1, 2, 3]

>>> b = map(lambda x: x, a)

>>> b

<map object at 0x104d8ccc0>

>>> list(b)

[1, 2, 3]

>>> list(b)

This question already has an answer here:

Python: calling 'list' on a map object twice

1 answer

python python-3.x map-function

edited Jan 20 at 4:18

asked Jan 20 at 3:05

msk

1004

edited Jan 20 at 4:18

asked Jan 20 at 3:05

msk

1004

edited Jan 20 at 4:18

asked Jan 20 at 3:05

msk

1004

asked Jan 20 at 3:05

msk

1004

asked Jan 20 at 3:05

msk

1004

marked as duplicate by Tomothy32, ShadowRanger, Austin, jpp python-3.x
Users with the python-3.x badge can single-handedly close python-3.x questions as duplicates and reopen them as needed.

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Jan 20 at 3:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Tomothy32, ShadowRanger, Austin, jpp python-3.x
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Jan 20 at 3:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

My bad. Thanks for the englightenment. Btw, isn't it more natural map returning the same collection(type)? Is there any merit in addition to some gain in performance of iteration immediately following it?

– msk
Jan 20 at 4:16

add a comment |

My bad. Thanks for the englightenment. Btw, isn't it more natural map returning the same collection(type)? Is there any merit in addition to some gain in performance of iteration immediately following it?

– msk
Jan 20 at 4:16

My bad. Thanks for the englightenment. Btw, isn't it more natural map returning the same collection(type)? Is there any merit in addition to some gain in performance of iteration immediately following it?

– msk
Jan 20 at 4:16

add a comment |

3 Answers
3

active

oldest

votes

map outputs an iterator that applies a function (lambda x: x) over some iterators (a). As a result, b is an iterator. When calling list(b) for the first time, the iterator b is called several times until it reaches to its end. Afterward, b is an iterator which does not have any item left to produce. That's why when you call list(b) for the second time, it outputs an empty list.

edited Jan 20 at 3:22

wjandrea

1,0771328

answered Jan 20 at 3:17

noidsirius

1197

1

OK. it is iterator not iterable as I expected. Though not sure about its design idea, it is just as it is. Thanks.

– msk
Jan 20 at 4:06

add a comment |

The documentation for map specifies that it returns an "iterator", not an "iterable". Python defines an iterator to loop exactly once with no repetition; once the end is reached then it will never return another item.

The second execution of list(b) attempts to build a list from an iterator that is already at the end, so it returns no items and an empty list is constructed.

answered Jan 20 at 3:18

lehiester

472211

add a comment |

-2

Whenever you are calling list(b) you are reassigning a new list to that variable b if you want to store b as a list use

a = [1, 2, 3]

b = list(map(lambda x: x, a))

print(b)

>>> [1, 2, 3]

print(*b)

>>> 1 2 3

Hope this helped.

answered Jan 20 at 3:14

Apex Predator

Calling list(b) doesn't assign to anything.

– wjandrea
Jan 20 at 3:23

add a comment |

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

edited Jan 20 at 3:22

wjandrea

1,0771328

answered Jan 20 at 3:17

noidsirius

1197

1

OK. it is iterator not iterable as I expected. Though not sure about its design idea, it is just as it is. Thanks.

– msk
Jan 20 at 4:06

add a comment |

edited Jan 20 at 3:22

wjandrea

1,0771328

answered Jan 20 at 3:17

noidsirius

1197

1

OK. it is iterator not iterable as I expected. Though not sure about its design idea, it is just as it is. Thanks.

– msk
Jan 20 at 4:06

add a comment |

edited Jan 20 at 3:22

wjandrea

1,0771328

answered Jan 20 at 3:17

noidsirius

1197

edited Jan 20 at 3:22

wjandrea

1,0771328

answered Jan 20 at 3:17

noidsirius

1197

edited Jan 20 at 3:22

wjandrea

1,0771328

edited Jan 20 at 3:22

wjandrea

1,0771328

edited Jan 20 at 3:22

wjandrea

1,0771328

answered Jan 20 at 3:17

noidsirius

1197

answered Jan 20 at 3:17

noidsirius

1197

answered Jan 20 at 3:17

noidsirius

1197

1

OK. it is iterator not iterable as I expected. Though not sure about its design idea, it is just as it is. Thanks.

– msk
Jan 20 at 4:06

add a comment |

1

OK. it is iterator not iterable as I expected. Though not sure about its design idea, it is just as it is. Thanks.

– msk
Jan 20 at 4:06

OK. it is iterator not iterable as I expected. Though not sure about its design idea, it is just as it is. Thanks.

– msk
Jan 20 at 4:06

add a comment |

The second execution of list(b) attempts to build a list from an iterator that is already at the end, so it returns no items and an empty list is constructed.

answered Jan 20 at 3:18

lehiester

472211

add a comment |

The second execution of list(b) attempts to build a list from an iterator that is already at the end, so it returns no items and an empty list is constructed.

answered Jan 20 at 3:18

lehiester

472211

add a comment |

The second execution of list(b) attempts to build a list from an iterator that is already at the end, so it returns no items and an empty list is constructed.

answered Jan 20 at 3:18

lehiester

472211

The second execution of list(b) attempts to build a list from an iterator that is already at the end, so it returns no items and an empty list is constructed.

answered Jan 20 at 3:18

lehiester

472211

answered Jan 20 at 3:18

lehiester

472211

answered Jan 20 at 3:18

lehiester

472211

answered Jan 20 at 3:18

lehiester

472211

add a comment |

-2

Whenever you are calling list(b) you are reassigning a new list to that variable b if you want to store b as a list use

a = [1, 2, 3]

b = list(map(lambda x: x, a))

print(b)

>>> [1, 2, 3]

print(*b)

>>> 1 2 3

Hope this helped.

answered Jan 20 at 3:14

Apex Predator

Calling list(b) doesn't assign to anything.

– wjandrea
Jan 20 at 3:23

add a comment |

-2

Whenever you are calling list(b) you are reassigning a new list to that variable b if you want to store b as a list use

a = [1, 2, 3]

b = list(map(lambda x: x, a))

print(b)

>>> [1, 2, 3]

print(*b)

>>> 1 2 3

Hope this helped.

answered Jan 20 at 3:14

Apex Predator

Calling list(b) doesn't assign to anything.

– wjandrea
Jan 20 at 3:23

add a comment |

-2

Whenever you are calling list(b) you are reassigning a new list to that variable b if you want to store b as a list use

a = [1, 2, 3]

b = list(map(lambda x: x, a))

print(b)

>>> [1, 2, 3]

print(*b)

>>> 1 2 3

Hope this helped.

answered Jan 20 at 3:14

Apex Predator

Whenever you are calling list(b) you are reassigning a new list to that variable b if you want to store b as a list use

a = [1, 2, 3]

b = list(map(lambda x: x, a))

print(b)

>>> [1, 2, 3]

print(*b)

>>> 1 2 3

Hope this helped.

answered Jan 20 at 3:14

Apex Predator

answered Jan 20 at 3:14

Apex Predator

answered Jan 20 at 3:14

Apex Predator

answered Jan 20 at 3:14

Apex Predator

Calling list(b) doesn't assign to anything.

– wjandrea
Jan 20 at 3:23

add a comment |

Calling list(b) doesn't assign to anything.

– wjandrea
Jan 20 at 3:23

Calling list(b) doesn't assign to anything.

– wjandrea
Jan 20 at 3:23

add a comment |

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