[python] Checking if type == list in python

I may be having a brain fart here, but I really 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?

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

This seems to work for me:

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

Python 3.7.7

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

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.