统计序列中元素出现的频度并获取topK

将序列转换为计数字典{元素: 频度}，然后根据频度排序。

1、使用 dict.fromkeys() 构造计数字典

from random import randint

# 创建一个随机列表
L = [randint(0, 20) for _ in range(30)]
print(L)

# 创建一个所有key初始值为0的字典
d = dict.fromkeys(L, 0)
print(d)
# {20: 0, 3: 0, 9: 0, 7: 0, 6: 0, 14: 0, 8: 0, 19: 0, 15: 0, 18: 0, 12: 0, 4: 0, 17: 0, 5: 0, 1: 0, 0: 0, 2: 0}

# 统计频度
for i in L:
    d[i] += 1

print(d)
# {20: 2, 3: 3, 9: 2, 7: 3, 6: 1, 14: 2, 8: 2, 19: 2, 15: 2, 18: 1, 12: 1, 4: 2, 17: 2, 5: 2, 1: 1, 0: 1, 2: 1}

2、使用 dict.setdefault() 构造计数字典

from random import randint

L = [randint(0, 20) for _ in range(30)]

d = {}
for i in L:
    d[i] = d.setdefault(i, 0) + 1

print(d)
# {13: 2, 14: 2, 9: 2, 8: 3, 1: 2, 5: 2, 7: 1, 20: 2, 10: 3, 0: 2, 18: 1, 4: 1, 3: 1, 17: 2, 16: 1, 12: 2, 11: 1}

3、使用 heapq.nlargest(n, iterable, key=None) 进行频度统计

Equivalent to: sorted(iterable, key=key, reverse=True)[:n]

from random import randint
import heapq

# 根据频度进行排序，并取出排名前3个
L = [randint(0, 20) for _ in range(30)]
d = {}
for i in L:
    d[i] = d.setdefault(i, 0) + 1

s = sorted(d.items(), key=lambda x: x[1], reverse=True)[:3]
print(s)
# [(16, 4), (19, 3), (13, 3)]

# 使用堆，取出排名前3个
r = heapq.nlargest(3, d.items(), key=lambda x: x[1])
print(r)
# [(16, 4), (19, 3), (13, 3)]

4、使用 Counter 进行频度统计

• 一个 Counter 是一个 dict 的子类，用于计数可哈希对象。
• 元素像字典键(key)一样存储，它们的计数存储为值。
• 默认是降序。

这个算是频度统计最简单的姿势了，无需手动构造计数字典，可以直接操作一个可迭代对象。

from collections import Counter

c1 = Counter()                           # a new, empty counter
c2 = Counter('gallahad')                 # a new counter from an iterable
c3 = Counter({'red': 4, 'blue': 2})      # a new counter from a mapping
c4 = Counter(cats=4, dogs=8)             # a new counter from keyword args

from random import randint
from collections import Counter

L = [randint(0, 20) for _ in range(30)]
c = Counter(L)
print(c)
# Counter({16: 4, 19: 3, 13: 3, 3: 3, 1: 2, 18: 2, 14: 2, 10: 2, 9: 2, 4: 2, 7: 1, 20: 1, 15: 1, 5: 1, 8: 1})

# 使用most_common()方法获取topN，这里其实是基于heapq实现的
r = c.most_common(3)
print(r)
# [(16, 4), (19, 3), (13, 3)]

# 更新Counter，合并统计
c2 = Counter(L)
c.update(c2)
print(c)
# Counter({16: 8, 19: 6, 13: 6, 3: 6, 1: 4, 18: 4, 14: 4, 10: 4, 9: 4, 4: 4, 7: 2, 20: 2, 15: 2, 5: 2, 8: 2})

from collections import Counter
import re

# 词频统计，取出前5
with open('example.txt') as f:
    txt = f.read()
    w = re.split('\W+', txt)
    print(w)
    c2 = Counter(w)
    r = c2.most_common(5)
    print(r)
    # [('a', 21), ('the', 16), ('to', 15), ('and', 12), ('Service', 8)]

参考文档

• https://docs.python.org/zh-cn/3/library/heapq.html
• https://docs.python.org/zh-cn/3/library/collections.html#counter-objects

标签：20,randint,Counter,print,topK,频度,序列,import
来源： https://www.cnblogs.com/keithtt/p/15174267.html