im trying to write a function that returns the length of the longest run of repetition in a given list

Here is my code: `

def longest_repetition(a):
  longest = 0
  j = 0
  run2 = 0
  while j <= len(a)-1:
    for i in a:
      run = a.count(a[j] == i)
      if run == 1:
          run2  = 1
    if run2 > longest:
        longest = run2
    j  = 1
    run2 = 0
  return longest

print(longest_repetition([4,1,2,4,7,9,4]))
print(longest_repetition([5,3,5,6,9,4,4,4,4]))

3
0

`

The first test function works fine, but the second test function is not counting at all and I'm not sure why. Any insight is much appreciated

Edit: Just noticed that the question I was given and the expected results are not consistent. So what I'm basically trying to do is find the most repeated element in a list and the output would be the number of times it is repeated. That said, the output for the second test function should be 4 because the element '4' is repeated four times (elements are not required to be in one run as implied in my original question)

CodePudding user response：

First of all, let's check if you were consistent with your question (function that returns the length of the longest run of repetition): e.g.:

a = [4,1,2,4,7,9,4]

b = [5,3,5,6,9,4,4,4,4]

(assuming, you are only checking single position, e.g. c = [1,2,3,1,2,3] could have one repetition of sequence 1,2,3 - i am assuming that is not your goal)

So:

for a, there is no repetitions of same value, therefore length equals 0

for b, you have one, quadruple repetition of 4, therefore length equals 4

First, your max_amount_of_repetitions=0 and current_repetitions_run=0' So, what you need to do to detect repetition is simply check if value of n-1'th and n'th element is same. If so, you increment current_repetitions_run', else, you reset current_repetitions_run=0. Last step is check if your current run is longest of all:

max_amount_of_repetitions= max(max_amount_of_repetitions, current_repetitions_run)

to surely get both n-1 and n within your list range, I'd simply start iteration from second element. That way, n-1 is first element.

for n in range(1,len(a)):
    if a[n-1] == a[n]:
       print("I am sure, you can figure out the rest")

CodePudding user response：

you can use hash to calculate the frequency of the element and then get the max of frequencies.

using functional approach

from collections import Counter
def longest_repitition(array):
    return max(Counter(array).values())

other way, without using Counter

def longest_repitition(array):
    freq = {}
    for val in array:
        if val not in freq:
            freq[val] = 0
        freq[val]  = 1
    values = freq.values()
    return max(values)