这篇文章主要介绍了python统计一个文本中重复行数的方法,涉及针对Python中dict对象的使用及相关本文的操作,具有一定的借鉴价值,需要的朋友可以参考下
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: 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: 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 代码如下: 希望本文所述对大家的Python程序设计有所帮助。