Python: determine if all items of a list are the same item [duplicate]

This question already has answers here: Check if all elements in a list are identical (29 answers) Closed 6 years ago.

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 co开发者_JS百科de 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.

---

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

---

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

---

Best way to do this is to use Python sets.You need to define all_same like this:

def all_same(items):
    return len(set(items)) < 2

Test:

>>> def all_same(items):
...     return len(set(items)) < 2
... 
>>> 
>>> property_list = ["one", "one", "one"]
>>> all_same(property_list)
True
>>> property_list = ["one", "one", "two"]
>>> all_same(property_list)
False
>>> property_list = []
>>> all_same(property_list)
True

---

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.)

---

This works both for sequences and iterables:

def all_same(items):
  it = iter(items)
  first = next(it, None)
  return all(x == first for x in it)

---

This is likely to be faster if you know values are in a list.

def all_same(values):
    return values.count(values[0]) == len(values)

---

I created this snippet of code for that same issue after thinking it over. I'm not exactly sure if it works for every scenario though.

def all_same(list):
    list[0]*len(list) == list