using itertools.groupby()
, items are repeated so you might not be able to store all values in a dictionary, as item1
& item2
are repeated:
In [21]: l = ["item1", "item2", "item3", "item3", "item3", "item1", "item2", "item4", "item4", "item4"]
In [22]: for k,g in groupby(l):
print "{0}:{1}".format(k,len(list(g)))
....:
item1:1
item2:1
item3:3
item1:1
item2:1
item4:3