标签:word imp col print 获取 cti 序列 com 数学运算
需求:
如何从一个序列中快速获取出现次数最多的元素。
方法:
利用collections.Counter类可以解决这个问题,特别是他的most_common()方法更是处理此问题的最快途径。比如,现在有一个单词的序列,你想快速获取哪个单词出现频率最高,就可以这么做:
In [22]: words = [‘look‘, ‘into‘, ‘my‘, ‘eyes‘, ‘look‘, ‘into‘,
...: ‘my‘, ‘eyes‘, ‘the‘, ‘eye‘, ‘the‘, ‘eyes‘, ‘not‘,
...: ‘around‘, ‘the‘, ‘eyes‘, "don‘t", ‘look‘, ‘around‘,
...: ‘the‘, ‘eyes‘, ‘look‘, ‘into‘, ‘my‘, ‘eyes‘, "you‘re",
...: ‘under‘
...: ]
In [23]: from collections import Counter
In [24]: word_counts = Counter(words)
In [25]: print(word_counts.most_common(3))
[(‘eyes‘, 6), (‘look‘, 4), (‘the‘, 4)]
事实上,Counter对象是一个元素和其数目对应关系所构成的字典, 例如:
In [26]: word_counts[‘not‘]
Out[26]: 1
In [27]: word_counts[‘into‘]
Out[27]: 3
扩展:
如果你想手动扩展单词数目,可以使用下面的方式:
In [28]: more_words = [‘why‘, ‘are‘, ‘you‘, ‘not‘, ‘looking‘, ‘in‘,
...: ‘my‘, ‘eyes‘]
In [29]: for word in more_words:
...: word_counts[word] += 1
...: # word_counts.update(more_words)
In [30]: word_counts[‘eyes‘]
Out[30]: 7
Counter类还有一些类似于数学运算的方法,使用起来也是相当方便:
In [31]: a = Counter(words)
In [32]: b = Counter(more_words)
In [33]: a
Out[33]:
Counter({‘around‘: 2,
"don‘t": 1,
‘eye‘: 1,
‘eyes‘: 6,
‘into‘: 3,
‘look‘: 4,
‘my‘: 3,
‘not‘: 1,
‘the‘: 4,
‘under‘: 1,
"you‘re": 1})
In [34]: b
Out[34]:
Counter({‘are‘: 1,
‘eyes‘: 1,
‘in‘: 1,
‘looking‘: 1,
‘my‘: 1,
‘not‘: 1,
‘why‘: 1,
‘you‘: 1})
In [35]: c = a + b
In [36]: c
Out[36]:
Counter({‘are‘: 1,
‘around‘: 2,
"don‘t": 1,
‘eye‘: 1,
‘eyes‘: 7,
‘in‘: 1,
‘into‘: 3,
‘look‘: 4,
‘looking‘: 1,
‘my‘: 4,
‘not‘: 2,
‘the‘: 4,
‘under‘: 1,
‘why‘: 1,
‘you‘: 1,
"you‘re": 1})
In [37]: d = b - a
In [38]: d
Out[38]: Counter({‘are‘: 1, ‘in‘: 1, ‘looking‘: 1, ‘why‘: 1, ‘you‘: 1})
标签:word imp col print 获取 cti 序列 com 数学运算
原文地址:https://www.cnblogs.com/valorchang/p/11289098.html