what is the difference in these two isinstances() definitions?

-2

1. isinstance(list(), type(mylist))




2. isinstance(mylist, list)

Is there any difference apart from the speed? I can see 2 is faster.

python python-3.x isinstance

share|improve this question

edited Nov 19 '18 at 15:12

asked Nov 19 '18 at 15:06

user2715898

124110

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  how did you define array? or do you mean the array.array?
  – Chris_Rands
  Nov 19 '18 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is array a type or an instance?
  – khelwood
  Nov 19 '18 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first one is making a new list and checking if is an instance of the type of array (whatever that is). The second is checking if array is an instance of list. They are not equivalent.
  – khelwood
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah it;s just a variable name. I'll edit the question
  – user2715898
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user2715898, The difference is you can easily use the first one incorrectly, see my example. It's not just about speed.
  – jpp
  Nov 19 '18 at 15:20
  
  
  

  
  
  
  

 | 
show 3 more comments

-2

1. isinstance(list(), type(mylist))




2. isinstance(mylist, list)

Is there any difference apart from the speed? I can see 2 is faster.

python python-3.x isinstance

share|improve this question

edited Nov 19 '18 at 15:12

asked Nov 19 '18 at 15:06

user2715898

124110

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  how did you define array? or do you mean the array.array?
  – Chris_Rands
  Nov 19 '18 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is array a type or an instance?
  – khelwood
  Nov 19 '18 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first one is making a new list and checking if is an instance of the type of array (whatever that is). The second is checking if array is an instance of list. They are not equivalent.
  – khelwood
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah it;s just a variable name. I'll edit the question
  – user2715898
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user2715898, The difference is you can easily use the first one incorrectly, see my example. It's not just about speed.
  – jpp
  Nov 19 '18 at 15:20
  
  
  

  
  
  
  

 | 
show 3 more comments

-2

-2

-2

1. isinstance(list(), type(mylist))




2. isinstance(mylist, list)

Is there any difference apart from the speed? I can see 2 is faster.

python python-3.x isinstance

share|improve this question

edited Nov 19 '18 at 15:12

asked Nov 19 '18 at 15:06

user2715898

124110

1. isinstance(list(), type(mylist))




2. isinstance(mylist, list)

Is there any difference apart from the speed? I can see 2 is faster.

python python-3.x isinstance

python python-3.x isinstance

share|improve this question

edited Nov 19 '18 at 15:12

asked Nov 19 '18 at 15:06

user2715898

124110

share|improve this question

edited Nov 19 '18 at 15:12

asked Nov 19 '18 at 15:06

user2715898

124110

share|improve this question

share|improve this question

edited Nov 19 '18 at 15:12

edited Nov 19 '18 at 15:12

edited Nov 19 '18 at 15:12

asked Nov 19 '18 at 15:06

user2715898

124110

asked Nov 19 '18 at 15:06

user2715898

124110

asked Nov 19 '18 at 15:06

user2715898

124110

124110

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  how did you define array? or do you mean the array.array?
  – Chris_Rands
  Nov 19 '18 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is array a type or an instance?
  – khelwood
  Nov 19 '18 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first one is making a new list and checking if is an instance of the type of array (whatever that is). The second is checking if array is an instance of list. They are not equivalent.
  – khelwood
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah it;s just a variable name. I'll edit the question
  – user2715898
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user2715898, The difference is you can easily use the first one incorrectly, see my example. It's not just about speed.
  – jpp
  Nov 19 '18 at 15:20
  
  
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  how did you define array? or do you mean the array.array?
  – Chris_Rands
  Nov 19 '18 at 15:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is array a type or an instance?
  – khelwood
  Nov 19 '18 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The first one is making a new list and checking if is an instance of the type of array (whatever that is). The second is checking if array is an instance of list. They are not equivalent.
  – khelwood
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah it;s just a variable name. I'll edit the question
  – user2715898
  Nov 19 '18 at 15:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user2715898, The difference is you can easily use the first one incorrectly, see my example. It's not just about speed.
  – jpp
  Nov 19 '18 at 15:20
  
  
  

  
  
  
  

4

4

how did you define array? or do you mean the array.array?
– Chris_Rands
Nov 19 '18 at 15:09

how did you define array? or do you mean the array.array?
– Chris_Rands
Nov 19 '18 at 15:09

Is array a type or an instance?
– khelwood
Nov 19 '18 at 15:10

Is array a type or an instance?
– khelwood
Nov 19 '18 at 15:10

The first one is making a new list and checking if is an instance of the type of array (whatever that is). The second is checking if array is an instance of list. They are not equivalent.
– khelwood
Nov 19 '18 at 15:12

The first one is making a new list and checking if is an instance of the type of array (whatever that is). The second is checking if array is an instance of list. They are not equivalent.
– khelwood
Nov 19 '18 at 15:12

ah it;s just a variable name. I'll edit the question
– user2715898
Nov 19 '18 at 15:12

ah it;s just a variable name. I'll edit the question
– user2715898
Nov 19 '18 at 15:12

1

1

@user2715898, The difference is you can easily use the first one incorrectly, see my example. It's not just about speed.
– jpp
Nov 19 '18 at 15:20

@user2715898, The difference is you can easily use the first one incorrectly, see my example. It's not just about speed.
– jpp
Nov 19 '18 at 15:20

 | 
show 3 more comments

2 Answers
2

active

oldest

votes

3

If used correctly, the results will be the same. But you should only use the second version.

There's no need to create an empty object, i.e. list(), and no reason for an extra function call via type.

There's also a danger you'll do something like this:

from collections import OrderedDict



array = dict()



res1 = isinstance(OrderedDict(), type(array))  # True

res2 = isinstance(array, OrderedDict)          # False

share|improve this answer

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a good point, the results are not always symmetric because of subclasses, of course we don't what they actually want in this case
  – Chris_Rands
  Nov 19 '18 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Chris_Rands, Yup, it's just an illustration of what's more likely to go wrong with the first method.
  – jpp
  Nov 19 '18 at 15:21
  
  
  

  
  
  
  

add a comment | 

1

The first one is very inefficient, as you are first creating a new list object, and then identifying the type of my_list and then checking if they both are same.

The second option is how the isinstance function is supposed to be used.

share|improve this answer

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

If used correctly, the results will be the same. But you should only use the second version.

There's no need to create an empty object, i.e. list(), and no reason for an extra function call via type.

There's also a danger you'll do something like this:

from collections import OrderedDict



array = dict()



res1 = isinstance(OrderedDict(), type(array))  # True

res2 = isinstance(array, OrderedDict)          # False

share|improve this answer

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a good point, the results are not always symmetric because of subclasses, of course we don't what they actually want in this case
  – Chris_Rands
  Nov 19 '18 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Chris_Rands, Yup, it's just an illustration of what's more likely to go wrong with the first method.
  – jpp
  Nov 19 '18 at 15:21
  
  
  

  
  
  
  

add a comment | 

3

If used correctly, the results will be the same. But you should only use the second version.

There's no need to create an empty object, i.e. list(), and no reason for an extra function call via type.

There's also a danger you'll do something like this:

from collections import OrderedDict



array = dict()



res1 = isinstance(OrderedDict(), type(array))  # True

res2 = isinstance(array, OrderedDict)          # False

share|improve this answer

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a good point, the results are not always symmetric because of subclasses, of course we don't what they actually want in this case
  – Chris_Rands
  Nov 19 '18 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Chris_Rands, Yup, it's just an illustration of what's more likely to go wrong with the first method.
  – jpp
  Nov 19 '18 at 15:21
  
  
  

  
  
  
  

add a comment | 

3

3

3

If used correctly, the results will be the same. But you should only use the second version.

There's no need to create an empty object, i.e. list(), and no reason for an extra function call via type.

There's also a danger you'll do something like this:

from collections import OrderedDict



array = dict()



res1 = isinstance(OrderedDict(), type(array))  # True

res2 = isinstance(array, OrderedDict)          # False

share|improve this answer

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

If used correctly, the results will be the same. But you should only use the second version.

There's no need to create an empty object, i.e. list(), and no reason for an extra function call via type.

There's also a danger you'll do something like this:

from collections import OrderedDict



array = dict()



res1 = isinstance(OrderedDict(), type(array))  # True

res2 = isinstance(array, OrderedDict)          # False

share|improve this answer

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

share|improve this answer

share|improve this answer

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

answered Nov 19 '18 at 15:17

jpp

92.2k2053103

92.2k2053103

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a good point, the results are not always symmetric because of subclasses, of course we don't what they actually want in this case
  – Chris_Rands
  Nov 19 '18 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Chris_Rands, Yup, it's just an illustration of what's more likely to go wrong with the first method.
  – jpp
  Nov 19 '18 at 15:21
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is a good point, the results are not always symmetric because of subclasses, of course we don't what they actually want in this case
  – Chris_Rands
  Nov 19 '18 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Chris_Rands, Yup, it's just an illustration of what's more likely to go wrong with the first method.
  – jpp
  Nov 19 '18 at 15:21
  
  
  

  
  
  
  

This is a good point, the results are not always symmetric because of subclasses, of course we don't what they actually want in this case
– Chris_Rands
Nov 19 '18 at 15:20

This is a good point, the results are not always symmetric because of subclasses, of course we don't what they actually want in this case
– Chris_Rands
Nov 19 '18 at 15:20

@Chris_Rands, Yup, it's just an illustration of what's more likely to go wrong with the first method.
– jpp
Nov 19 '18 at 15:21

@Chris_Rands, Yup, it's just an illustration of what's more likely to go wrong with the first method.
– jpp
Nov 19 '18 at 15:21

add a comment | 

1

The first one is very inefficient, as you are first creating a new list object, and then identifying the type of my_list and then checking if they both are same.

The second option is how the isinstance function is supposed to be used.

share|improve this answer

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

add a comment | 

1

The first one is very inefficient, as you are first creating a new list object, and then identifying the type of my_list and then checking if they both are same.

The second option is how the isinstance function is supposed to be used.

share|improve this answer

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

add a comment | 

1

1

1

The first one is very inefficient, as you are first creating a new list object, and then identifying the type of my_list and then checking if they both are same.

The second option is how the isinstance function is supposed to be used.

share|improve this answer

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

The first one is very inefficient, as you are first creating a new list object, and then identifying the type of my_list and then checking if they both are same.

The second option is how the isinstance function is supposed to be used.

share|improve this answer

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

share|improve this answer

share|improve this answer

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

answered Nov 19 '18 at 15:15

Aswin Murugesh

6,51652654

6,51652654

add a comment | 

add a comment | 

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