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Python in how to find out the sequence of the most frequent elements, many novices are not very clear about this, in order to help you solve this problem, the following editor will explain in detail for you, people with this need can come to learn, I hope you can gain something.

There is a sequence of elements. What is the element that appears most frequently in the sequence?

The Chinese Counter class of the collections module is designed for this kind of problem. It even has a very convenient most_common () method to tell us the answer. You can provide any hashable sequence of objects to the Counter object as input.

Example: suppose there is a list with a list of words, and we want to find out which words appear most frequently:

From collections import Counterwords= ['axiaophony', 'cantilever',''cantilever',','', 'a'] # use Counter to count the number of occurrences of each element words_counts=Counter (words) # the three elements with the highest number of occurrences top_three=words_counts.most_common (3) # return element and number of occurrences print (top_three) # Counter is a dictionary You can map elements to the number of times they appear, for example: # output the number of occurrences of element [f] print (words_counts ['f']) # if you want to increase the count manually Simply add words _ counts ['f'] + = 1print (words_counts ['f']) # if you want to increase the count manually, you can also use the update () method: # increase the count only once for the element [f]. Words_counts.update ('f') print (words_counts ['f']) # add one morewords= for all counts. 'f'] words_counts.update (morewords) print (words_counts [' f'])

Running result:

[('axiang, 5), (' baked, 4), ('crested, 3)] 2345

Another unknown feature of Counter objects is that they can be easily combined with a variety of mathematical operations.

From collections import Counterwords1= ['axiomagronomy', 'cantiledge', 'mendicant',''paraphrase,'', One=Counter (words1) two=Counter (words2) print (one) print (two) three=one+twoprint (three) four=one-twoprint (four)

Running result:

Counter ({'averse: 5,' baked: 4, 'caged: 3,' dumped: 2, 'eyed: 2,' favored: 2}) Counter ({'axed: 4,' baked: 3, 'clocked: 2,' dashed: 1, 'eyed: 1,' favored: 1}) Counter ({'axed: 9,' baked: 7, 'cased: 5,' dusted: 3, 'eBay: 3 ) Counter ({'averse: 1,' baked: 1, 'caged: 1,' dumped: 1, 'eyed: 1,' favored: 1})

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