Python Cookbook (1-4) 內建演算法

1. Counter:

Counter 對 list 的元素進行計數，並可依照次數回傳前幾名的元素和該元素的次數。

from collections import Counter

animals= ["dog", "dog", "dog", "cat", "bird", "bird"]
animal_counts = Counter(animals)

print(animal_counts)
# Counter({'dog': 3, 'bird': 2, 'cat': 1})

print(animal_counts.most_common(2))
# [('dog', 3), ('bird', 2)] 

animal_counts.update(animals) # 追加一個 list
print(animal_counts)
# Counter({'dog': 6, 'bird': 4, 'cat': 2})

兩 Counter 的相互運算。

animals1= ["dog", "dog", "dog", "cat", "bird", "bird"]
a = Counter(animals1)

animals2= ["dog", "dog", "dog", "cat", "cat", "bird"]
b = Counter(animals2)

print(a + b)
print(a - b)

2. sorted, min and max

假設我們想對 list of dict 的特定鍵進行排序或是找出最大最小。

animals = [{"name":"doggy", "age":10},{"name":"kitty", "age":3},{"name":"ducky", "age":4}]
print(sorted(animals, key=lambda info: info["age"]))
print(min(animals, key=lambda info: info["age"]))
print(max(animals, key=lambda info: info["age"]))

另一種較快的方式是使用 itemgetter。

from operator import itemgetter
print(sorted(animals, key=itemgetter("age"))

3. 對 class 進行排序

class student:
    def __init__(self, name, sid):
        self.name=name
        self.sid=sid
    def __repr__(self):
        return "{},{}".format(self.name, self.sid)

lst = [student("alice", 3), student("bob", 1), student("cat", 2)]

from operator import attrgetter
sorted(lst, key=attrgetter("sid"))

也可將 attrgetter 用在 class 的 max 和 min 上。

4. groupby

將資料分群。

student = [{"name":"a", "class":1},{"name":"b", "class":2}, {"name":"c", "class":1}]

from itertools import groupby
from operator import itemgetter

for k,g in groupby(student, key=itemgetter("class")):
    print(k)
    for item in g:
        print(item)

k,g 分別是 key 和群組的內容。

執行結果如下，如果要整齊一點要先對資料進行排序，因為是線性對元素進行群組操作。

1
{'name': 'a', 'class': 1}
2
{'name': 'b', 'class': 2}
1
{'name': 'c', 'class': 1}

先對資料進行排序，讓線性操作整齊一點(應該說都排序完也算分群完了)。

student = [{"name":"a", "class":1},{"name":"b", "class":2}, {"name":"c", "class":1}]

from itertools import groupby

from operator import itemgetter
student.sort(key=itemgetter("class"))

for k,g in groupby(student, key=itemgetter("class")):
    print(k)
    for item in g:
        print(item)

執行結果如下。

1
{'name': 'a', 'class': 1}
{'name': 'c', 'class': 1}
2
{'name': 'b', 'class': 2}

5. filter

a = [i for i in range(10)]
print(list(filter(lambda x: x%3 == 0, a)))
# [0, 3, 6, 9]

Reference

[1] David Beazley, Brian K. Jones, Python Cookbook, 3e , O'Reilly Media (2014)