如何在python的字典中计算前10个最常见的值 [英] How to count top 10 most common values in a dict in python

问题描述

我是python和程序设计的新手，所以请客气.我正在尝试使用音乐信息分析一个csv文件，并返回收听次数最多的前n个乐队.在下面的代码中，每首歌曲监听的都是列表中的dict条目，格式如下:

I'm new to python and programming in general and so please be kind. I'm trying to analyze a csv file with music information and return the top n most listened to bands. From the code below, each song listen is a dict entry within a list formatted like this:

[{'album': 'Exile on Main Street', 'song': 'Happy', 'datetime': '3 Dec 2014 14:08', 'artist': 'The Rolling Stones'}, {'album': 'II', 'song': 'Black Dog', 'datetime': '1 Dec 2014 08:08', 'artist': 'Led Zepplin'}]

from collections import Counter

def count_artist_plays(filename):
    with open(filename, 'r') as data:
        header = data.readline().strip().split(',')

        entries = []
        for line in data:
            entry = line.strip().split(',')
            listens = {}
            for info, type in enumerate(header):
                listens[type] = entry[info]

            entries.append(listens)

    for d in entries:
        arts = d['artist']
        c = Counter(arts)
        print c.most_common(10)

如何获得最常见的字符串(带)而不是下面显示的字符分解符?

[('s', 2), ('a', 1), (' ', 1), ('E', 1), ('l', 1), ('o', 1), ('n', 1), ('S', 1), ('v', 1), ('y', 1)]

推荐答案

一次初始化计数器，让键成为美工，并在整个循环中每次增加一个键(美工):

Initialize the Counter once, let the keys be artists, and augment a key (artist) each time through the loop:

c = Counter()
for d in entries:
    arts = d['artist']
    c[arts] += 1
print(c.most_common(10))

当arts是字符串时，c = Counter(arts)计算arts中的字符:

When arts is a string, then c = Counter(arts) counts the characters in arts:

In [522]: collections.Counter('Led Zepplin')
Out[522]: Counter({'e': 2, 'p': 2, ' ': 1, 'd': 1, 'i': 1, 'L': 1, 'l': 1, 'n': 1, 'Z': 1})

相反:

In [523]: c = collections.Counter()

In [524]: c['Led Zepplin'] += 1

In [525]: c['The Rolling Stones'] += 1

In [526]: c.most_common()
Out[526]: [('Led Zepplin', 1), ('The Rolling Stones', 1)]

---

或者，正如乔恩·克莱门茨(Jon Clements)指出的那样，建立所有艺术家的列表，然后计算列表:

---

Alternatively, as Jon Clements points out, build a list of all the artists, and then count the list:

c = Counter(d['artist'] for d in entries)
print(c.most_common(10))

请注意，以上代码使用生成器表达式来避免构建(可能)大的临时列表，同时具有更简洁易读的语法.

Note that the above uses a generator expression to avoid building a (possibly) large temporary list, and at the same time has a much more succinct, readable syntax.

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