在python中检查type == list

Question

I can't figure out what's wrong with my code:

for key in tmpDict:
    print type(tmpDict[key])
    time.sleep(1)
    if(type(tmpDict[key])==list):
        print 'this is never visible'
        break

the output is <type 'list'> but the if statement never triggers. Can anyone spot my error here?

Answer 1

You should try using isinstance()

if isinstance(object, list):
       ## DO what you want

In your case

if isinstance(tmpDict[key], list):
      ## DO SOMETHING

To elaborate:

x = [1,2,3]
if type(x) == list():
    print "This wont work"
if type(x) == list:                  ## one of the way to see if it's list
    print "this will work"           
if type(x) == type(list()):
    print "lets see if this works"
if isinstance(x, list):              ## most preferred way to check if it's list
    print "This should work just fine"

The difference between isinstance() and type() though both seems to do the same job is that isinstance() checks for subclasses in addition, while type() doesn't.

Answer 2

Your issue is that you have re-defined list as a variable previously in your code. This means that when you do type(tmpDict[key])==list if will return False because they aren't equal.

That being said, you should instead use isinstance(tmpDict[key], list) when testing the type of something, this won't avoid the problem of overwriting list but is a more Pythonic way of checking the type.

Answer 3

This seems to work for me:

>>>a = ['x', 'y', 'z']
>>>type(a)
<class 'list'>
>>>isinstance(a, list)
True

Answer 4

Python 3.7.7

import typing
if isinstance([1, 2, 3, 4, 5] , typing.List):
    print("It is a list")

Answer 5

Although not as straightforward as isinstance(x, list) one could use as well:

this_is_a_list=[1,2,3]
if type(this_is_a_list) == type([]):
    print("This is a list!")

and I kind of like the simple cleverness of that