Python: determine if all items of a list are the same item [duplicate]
In some of my code I put a series of objects in a list and I build an additional list out of their attributes, which is a string. I need to determine if all the items in this second list have the exact same value, without knowing beforehand which value it is, and return a bool so that I can do different things in my code depending on the result.
I can't know the names of the properties beforehand, that is why I'm trying to make something as generic as possible.
To make the example clear, an ideal function, called "all_same" would work like this:
>>> property_list = ["one", "one", "one"]
>>> all_same(property_list)
True
>>> property_list = ["one", "one", "two"]
>>> all_same(property_list)
False
I was thinking of making a list of unique elements and then check if its length is 1, but I'm not sure if it's the most elegant solution out there.
Solution 1:
def all_same(items):
return all(x == items[0] for x in items)
Example:
>>> def all_same(items):
... return all(x == items[0] for x in items)
...
>>> property_list = ["one", "one", "one"]
>>> all_same(property_list)
True
>>> property_list = ["one", "one", "two"]
>>> all_same(property_list)
False
>>> all_same([])
True
Solution 2:
You could cheat and use set
:
def all_same( items ):
return len( set( items ) ) == 1 #== len( items )
or you could use:
def all_same( items ):
return all( map(lambda x: x == items[0], items ) )
or if you're dealing with an iterable instead of a list:
def all_same( iterable ):
it_copy = tee( iterable, 1 )
return len( set( it_copy) ) == 1
Solution 3:
I originally interpreted you to be testing identity ("the same item"), but you're really testing equality ("same value"). (If you were testing identity, use is instead of ==.)
def all_same(items):
it = iter(items)
for first in it:
break
else:
return True # empty case, note all([]) == True
return all(x == first for x in it)
The above works on any iterable, not just lists, otherwise you could use:
def all_same(L):
return all(x == L[0] for x in L)
(But, IMHO, you might as well use the general version—it works perfectly fine on lists.)