在列表中找到最大的重复数字序列

Question

The problem to solve is to find the biggest sequence of repeating numbers in a list. For example if the list was [2,3,4,4,4,5,4,7,6] and I was looking for the largest sequence of 4's, I would return 3.

So far, I've only come up with a loop that counts the number of the specified number in a row but you have to account for the other numbers and also have a way to compare say, 3 "4's" in a row versus 2 "4's". any help would be appreciated

def find(list):
    x = [2,3,4,4,5,5,5,5,6,7,8]  
    print(find(x))

the code above is just the basic layout. I don't need the answer to the whole function but just the main logic and explanations

Answer 1

If you're trying to implement this in Python, you should use groupby . itertools.groupby , by default, groups objects into tuples of (object_name, list_of_objects) , so:

a = [1, 0, 1, 1, 2, 2, 2, 3]
groupby(a) ~ [(1, [1]), (0, [0]), (1, [1, 1]), (2, [2, 2, 2]), (3, [3])]
# I use "~" here because it's not actually a list, it's a generator, but meh

You can easily use a list comprehension to grab what you need.

from itertools import groupby
repeat_lengths = [sum(1 for _ in group) for _, group in groupby(x)]
# the `sum(1 for _ in group)` is a workaround for the fact that `group` is not a list and has no length

Then just find the max of that list.

Answer 2

You need to keep track of two variables, the current count (count) of the consecutive values and the maximum count (max_count) so far. When you observe a different value, you reset count and update max_count and continue the loop.

def get_longest_seq(l, val):
  count = 0
  max_count = 0
  for e in l:
    if e == val:
      count += 1
    elif count > 0:
      max_count = max(max_count, count)
      count = 0
  max_count = max(max_count, max) # case when the sequence is at the end
  return max_count

l = [2,2,3,2,2,2,2,3,3,4,4,4,5,4,4,7,6]
print(get_longest_seq(l, 4))