Reputation: 167
I am wanting to check if a list has a specific sequence of elements. I have sorted the list which contains 7 elements, I now want to check of the first 4 are the same as each other and the last 3 are the same as each other.
For what I want to achieve to be True the list would be like this:
list = ['1','1','1','1','2','2','2']
I hope this makes what I want to achieve clearer.
Upvotes: 11
Views: 311
Reputation: 5606
If you can transform your list into string, re
will do:
re.match(r'^(.)\1{3}(.)\2{2}$', ''.join(['1','1','1','1','2','2','2']))
Upvotes: 1
Reputation: 5606
If you want to check if list contains 3 items of one element, and 4 items of another, you can omit sorting by using collections.Counter
:
content = Counter(['1', '2', '2', '1', '1', '2', '1']).most_common()
print(content) # => [('1', 4), ('2', 3)]
if len(content) == 2 and content[0][1] == 4 and content[1][1] == 3 or
len(content) == 1 and content[0][1] == 7:
pass # Your list have desired structure
Upvotes: 2
Reputation: 85432
You can slice a list. Take the first four elements:
>>> L = ['1','1','1','1','2','2','2']
>>> L[:4]
['1', '1', '1', '1']
and the last three:
>>> L[-3:]
['2', '2', '2']
A set does not allow duplicates. Therefore:
>>> set(L[:4])
{1}
That means if he length of this set is 1, all elements in the sliced list are the same.
Putting this all together:
>>> len(set(L[:4])) == 1 and len(set(L[-3:])) == 1
True
shows you that your condition is met.
Upvotes: 8
Reputation: 908
Based on the extra details on the question, this can solve the problem:
def check_group_equal(inputList):
ref = inputList[0]
for e in inputList[1:]:
if e != ref:
return False
return True
list = some_random_list(length=7)
# Check first group
check_group_equal(list[0:3])
# Check second group
check_group_equal(list[4:7])
Upvotes: 1
Reputation: 10150
This should work, just pass each of your sublists into the function:
def all_same(items):
return all(x == items[0] for x in items)
The above was from the following post: Python: determine if all items of a list are the same item
Upvotes: 3