Checking a List for a Sequence

Question

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.

Possible duplicate of http://stackoverflow.com/questions/3787908/python-determine-if-all-items-of-a-list-are-the-same-item — Learner, Dec 06 '15 at 14:08
Possible duplicate of [How do you split a list into evenly sized chunks in Python?](http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python) — Vaibhav Mule, Dec 25 '15 at 13:31

Mike Müller · Accepted Answer · 2015-12-06T14:13:00.747

8

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.

edited Dec 06 '15 at 14:13

answered Dec 06 '15 at 14:03

Mike Müller

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Yap exactly what @Hugh Bothwell commented in the question. – Learner Dec 06 '15 at 14:07

score 3 · Answer 2 · edited May 23 '17 at 12:33

3

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

edited May 23 '17 at 12:33

Community

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answered Dec 06 '15 at 13:49

Simon

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GingerPlusPlus · Answer 3 · 2015-12-06T20:11:34.550

2

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

edited Dec 06 '15 at 20:11

answered Dec 06 '15 at 14:56

GingerPlusPlus

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SebasSBM · Answer 4 · 2015-12-06T13:54:59.060

1

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])

edited Dec 06 '15 at 13:54

answered Dec 06 '15 at 13:35

SebasSBM

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The last value helps a lot, so thanks a lot! However I don't think the expression will achieve what I want. I have edited the post to try to explain what I want to check better. – Frazer224 Dec 06 '15 at 13:42
this will never be true, a 3 lenght list is always diferent from a 2 lenght list – Netwave Dec 06 '15 at 13:43
Dear downvoter, please notice that my first answer's version was directed to the first question's version. After seeing extra information in the edit, I **adapted my answer accordingly**. Please reconsider what you have done – SebasSBM Dec 06 '15 at 13:52

GingerPlusPlus · Answer 5 · 2015-12-06T20:33:24.293

1

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']))

edited Dec 06 '15 at 20:33

answered Dec 06 '15 at 14:40

GingerPlusPlus

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Checking a List for a Sequence

5 Answers5