Question

我是python和编程的新手，所以请善待。我正在尝试分析带有音乐信息的csv文件，并返回收听最多的前n个乐队。从下面的代码中，每首歌曲是一个dict条目，其格式如下：

[{'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)

我如何获得最常见的字符串（波段）而不是下面的字符细分？

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

Answer 1

初始化计数器一次，让键成为艺术家，并在每次循环时增加一个键（艺术家）：

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

当arts为字符串时，c = Counter(arts)会对arts中的字符进行计数：

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Jon Clements指出的那样，建立一个所有艺术家的列表，然后计算列表：

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

请注意，上面使用generator expression来避免构建（可能）大型临时列表，同时具有更简洁，可读的语法。

如何计算python中dict中前10个最常见的值

1 个答案: