本文实例讲述了python统计一个文本中重复行数的方法。分享给大家供大家参考。具体实现方法如下：

比如有下面一个文件

2

3

1

2

我们期望得到

2,2

3,1

1,1

解决问题的思路:

出现的文本作为key, 出现的数目作为value,然后按照value排除后输出

最好按照value从大到小输出出来,可以参照：

复制代码 代码如下:in recent Python 2.7, we have new OrderedDict type, which remembers the order in which the items were added.

>>> 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)])

代码如下:

复制代码 代码如下:#coding=utf-8

import operator

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])

补充说明:

1. python的dict对象的两个方法:

items方法将所有的字典项以列表的方式返回, 这些列表项中每一项都来自于(键, 值)

iteritems方法与items的作用大致相同, 但是返回一个迭代器对象而不是列表

2. python的内建函数sorted

复制代码 代码如下:>>> help(sorted)

Help on built-in function sorted in module __builtin__:

sorted(...)

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

希望本文所述对大家的Python程序设计有所帮助。