Count the number of duplicate rows in a text in Python

For example, there is a file below

2
3
1
2

We are looking forward to

2,2
3,1
1,1
The way to solve the problem:
The occurrence of the text as a key, the number of occurrences as value, and then excluded by value after output

It is best to output from large to small according to value, you can refer to

The code is as follows

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In recent Python 2.7, we are have new ordereddict type, which remembers the order in which the items were added. >>> d = {"Third": 3, "a": 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, ' a ': 1} To make a new ordered dictionary from the original and 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 d Ict: >>> 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 ([(' I ': 1), (' Second ': 2), (' Third ': 3), (' Fourth ': 4)]

The code is as follows:

code is as follows	copy code
#coding = Utf-8 Import operator F = open ("F.txt") Count_dict = {} for line in F.readlines ():   & nbsp line = Line.strip ()     count = Count_dict.setdefault (line, 0)     count = 1 &N bsp;   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])

Supplementary Note:
two methods for the Dict object of 1.python:
The items method returns all dictionary entries as a list, each of which comes from (key, value)
The Iteritems method works roughly the same as items, but returns an iterator object instead of a list

2.python built-in function sorted

The code is as follows	Copy Code
>>> Help (sorted) Help on built-in function sorted in module __builtin__: Sorted (...) Sorted (iterable, Cmp=none, Key=none, Reverse=false)--> new sorted list