Question :
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?
Answer #1:
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 #2:
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 #3:
This seems to work for me:
>>>a = ['x', 'y', 'z']
>>>type(a)
<class 'list'>
>>>isinstance(a, list)
True
Answer #4:
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
Answer #5:
Python 3.7.7
import typing
if isinstance([1, 2, 3, 4, 5] , typing.List):
print("It is a list")
