Python，使用列表，找到最大序列长度

Question

for example test_list:

test_list = ['a', 'a', 'a', 'b', 'b', 'a', 'c', 'b', 'a', 'a']

what tool or algorithm i need to use, to get max sequences count, for this example:

'a' = 3
'b' = 2
'c = 1

Answer 1

Using a dict to track max lengths, and itertools.groupby to group the sequences by consecutive value:

from itertools import groupby

max_count = {}

for val, grp in groupby(test_list):
    count = sum(1 for _ in grp)
    if count > max_count.get(val, 0):
        max_count[val] = count

Demo:

>>> from itertools import groupby
>>> test_list = ['a', 'a', 'a', 'b', 'b', 'a', 'c', 'b', 'a', 'a']
>>> max_count = {}
>>> for val, grp in groupby(test_list):
...     count = sum(1 for _ in grp)
...     if count > max_count.get(val, 0):
...         max_count[val] = count
... 
>>> max_count
{'a': 3, 'c': 1, 'b': 2}

Answer 2

Here is a direct way to do it:

Counts, Count, Last_item = {}, 0, None
test_list = ['a', 'a', 'a', 'b', 'b', 'a', 'c', 'b', 'a', 'a']
for item in test_list:
   if Last_item == item:
       Count+=1
   else:
       Count=1
       Last_item=item
   if Count>Counts.get(item, 0):
       Counts[item]=Count

print Counts
# {'a': 3, 'c': 1, 'b': 2}

Answer 3

You should read about what a dictionary is ( dict in Python ) and how you could store how many occurrences there are for a sequence.

Then figure out how to code the logic -

Figure out how to loop over your list.  As you go, for every item -
     If it isn't the same as the previous item
         Store how many times you saw the previous item in a row into the dictionary
     Else
         Increment how many times you've seen the item in the current sequence

Print your results

Answer 4

You can use re module for find all sequences of the character in a string composed by all the characters in your list. Then just pick the largest string for a single character.

import re

test_list = ['a', 'a', 'b', 'b', 'a', 'c', 'b', 'a', 'a', 'a']

# First obtain the characters.
unique  = set(test_list)

max_count = {}

for elem in unique:
    # Find all sequences for the same character.
    result = re.findall('{0}+'.format(elem), "".join(test_list))
    # Find the longest.
    maximun = max(result)
    # Save result.
    max_count.update({elem: len(maximun)})

print(max_count)

This will print: {'c': 1, 'b': 2, 'a': 3}

Answer 5

For Python, Martijn Pieters' groupby is the best answer.

That said, here is a 'basic' way to do it that could be translated to any language:

test_list = ['a', 'a', 'a', 'b', 'b', 'a', 'c', 'b', 'a', 'a']

hm={}.fromkeys(set(test_list), 0)
idx=0
ll=len(test_list)  
while idx<ll:
    item=test_list[idx]
    start=idx
    while idx<ll and test_list[idx]==item:
        idx+=1
    end=idx
    hm[item]=max(hm[item],end-start)    


print hm 
# {'a': 3, 'c': 1, 'b': 2}