How to compare all list itetms in Python shortest

Question

I want to be able to get True if all the list's items are the same:

For example checking this list would return True:

myList = [1,1,1,1,1,1,1]

While checking this list would result to False:

myList = [2,2,2,2,2,2,1]

What would be a shortest solution without a need to declare any new variables?

shx2 · Accepted Answer · 2014-05-05T19:01:06.307

4

Using set would drop duplicates. Then you can chcek the length, to get the number of different values.

len(set(myList)) <= 1

This works if the values are hashable.

However, if you expect to run this on long lists and expect a negative answer often, short circuiting might prove faster:

def is_unique(myList):
   seen = set()
   for x in myList:
       seen.add(x)
       if len(seen) > 1:
          return False
   return True

edited May 05 '14 at 19:01

answered May 05 '14 at 18:55

shx2

Smart and simple! Thanks! – alphanumeric May 05 '14 at 18:56
My two cents: I would use a `collections.counter`, just because it gives a little more information about the distribution of the elements in the list, while maintaining the same time and space complexity (while still under the "hashable items" assumption) – inspectorG4dget May 05 '14 at 18:57
1

@inspectorG4dget `Collections.Counter` is implemented in Python, so it is no where near sets when it comes to speed. – Ashwini Chaudhary May 05 '14 at 19:00
@200OK: touché. I was thinking more of big-O complexity – inspectorG4dget May 05 '14 at 19:03

1 Answers1