Question

If I have a list in Python like

[1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1]

How do I calculate the greatest number of repeats for any element? In this case 2 is repeated a maximum of 4 times and 1 is repeated a maximum of 3 times.

Is there a way to do this but also record the index at which the longest run began?

Answer 1

Use groupby, it group elements by value:

from itertools import groupby
group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1])
print max(group, key=lambda k: len(list(k[1])))

And here is the code in action:

>>> group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1])
>>> print max(group, key=lambda k: len(list(k[1])))
(2, <itertools._grouper object at 0xb779f1cc>)
>>> group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 3, 3])
>>> print max(group, key=lambda k: len(list(k[1])))
(3, <itertools._grouper object at 0xb7df95ec>)

From python documentation:

The operation of groupby() is similar
to the uniq filter in Unix. It
generates a break or new group every
time the value of the key function
changes

# [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
# [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D

If you also want the index of the longest run you can do the following:

group = groupby([1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 3, 3])
result = []
index = 0
for k, g in group:
   length = len(list(g))
   result.append((k, length, index))
   index += length

print max(result, key=lambda a:a[1])