python中统计一个文本中的重复行数

>>> d = {"third": 3, "first": 1, "fourth": 4, "second": 2}

>>> for k, v in d.items():
...     print "%s: %s" % (k, v)
...
second: 2
fourth: 4
third: 3
first: 1

>>> d
{'second': 2, 'fourth': 4, 'third': 3, 'first': 1}To make a new ordered dictionary from the original, sorting by the values:

>>> from collections import OrderedDict
>>> d_sorted_by_value = OrderedDict(sorted(d.items(), key=lambda x: x[1]))The OrderedDict behaves like a normal dict:

>>> for k, v in d_sorted_by_value.items():
...     print "%s: %s" % (k, v)
...
first: 1
second: 2
third: 3
fourth: 4

>>> d_sorted_by_value
OrderedDict([('first': 1), ('second': 2), ('third': 3), ('fourth': 4)])

f = open("f.txt")
count_dict = {}

for line in f.readlines():
    line = line.strip()
    count = count_dict.setdefault(line, 0)
    count += 1
    count_dict[line] = count

sorted_count_dict = sorted(count_dict.iteritems(), key=operator.itemgetter(1), reverse=True)

for item in sorted_count_dict:
    print "%s,%d" % (item[0], item[1])

Help on built-in function sorted in module __builtin__:

sorted(...)
    sorted(iterable, cmp=None, key=None, reverse=False) --> new sorted list