cookbook_数据结构和算法

1.1将数据分解为单独的变量

list_a = [1,2,3,4,5,6,7,8,9]
a,b,c,d,e,f,g,h,i = list_a
print(a,b,c,d,e,f,g,h,i)
#使用相等数量的参数来接收
_,b,c,d,e,f,g,h,_ = list_a 
print(b,c,d,e,f,g,h)
#不要的数据使用一个没有用的变量接收

 

1.2从任意长度的可迭代对象中分解元素

使用 * XXX实现

list_a = range(20)
first,*middle,last = list_a
print(first,middle,last)
#使用*来接收任意数量，甚至没有，返回一个list#当一个元祖内有一个标志位时，一个较好的应用
records = [("foo",1,2),("bar","hello"),("foo",3,4)
]def do_foo(x,y):print("foo",x,y)def do_bar(s):print("bar",s)for tags,*args in records:if tags == "foo":do_foo(*args)elif tags == "bar":do_bar(*args)

 

1.3保存最后N个元素

collections.deque()

import collections#使用collections.deque(maxlen=5)来定义一个固定长度的list,有新数据写入时如果已经达到maxlen，会自动删除最早插入的数据
def search(lines,pattern,history = 5):previous_lines = collections.deque(maxlen=history)for line in lines:if pattern in line:yield line,previous_linesprevious_lines.append(line)if __name__ =="__main__":with open("test.txt","r",encoding="utf8") as f:for line,previous in search(f,"python",5):for pline in previous:print(pline,end="")print(line,end="")print("-"*20)#collections.deque使用简介
#一个更加强大的list

queue = collections.deque(["jiao","li",'hao',"yu"])
queue.appendleft("wu")
print(queue)
queue.append("haha")
print(queue)
queue.popleft()
print(queue)
print(queue[4])

1.4找到最大或最小的N个元素

heapq.nlargest(),heapq.nsmallest()

import heapqnums = [5,56,7,6,34,2,5,7,6,89,80,-90,0,9,-67,5,45,]print(min(nums))
print(max(nums))print(heapq.nlargest(3,nums))
print(heapq.nsmallest(3,nums))#可支持更加复杂的数据结构

portfolio = [{"name":"jiao","age":24},{"name":"jsdfo","age":2},{"name":"jisd","age":12},{"name":"jdo","age":36},{"name":"li","age":25},{"name":"jgd","age":50},
]print(heapq.nlargest(3,portfolio,key=lambda s:s['age']))
print(max(portfolio,key=lambda s:s['age']))

 

 

1.5实现优先级队列

heapq.heappush(),heapq.heappop()

import heapq#列表中实际存一个元组，（-priority,self._index,item）
class PriorityQueue:def __init__(self):self._queue = []self._index = 0def push(self,item,priority):heapq.heappush(self._queue,(-priority,self._index,item))self._index += 1def pop(self):return heapq.heappop(self._queue)[-1]def get(self):return self._queueq = PriorityQueue()
q.push("foo",2)
q.push("sdf",3)
q.push("sfasc",5)
q.push("fdsg",4)
print(q.pop())
print(q.get())

1.6在字典中将键映射到多个值上

collections.defaultdict(list),collections.defaultdict(set)

import collectionsd = collections.defaultdict(list)#自动初始化，不用判断是否存在
d["a"].append(1)
d["a"].append(1)
d["a"].append(1)
d["a"].append(1)
print(d['a'])

 

1.7让字典保持有序

collections.OrderedDict()

import collectionsd = collections.OrderedDict()#普通字典的两倍，大数据不应该使用
d['foo'] = 1
d["bar"] = 2
d["spam"] = 3
d["gork"] = 4
for i in d:print(i)

 

1.8与字典有关的计算问题

zip(),min(),sorted().max()

#字典进行大小运算时都是使用key值进行大小比较，而我们一般想要用value值比较，而且还想要得到该值的key

prices = {"ACME":23,"AAPL":345,"IBM":34,"FB":24
}#利用zip,zip返回一个迭代器，只能使用一次

min_price = min(zip(prices.values(),prices.keys()))
print(min_price)#排序
price_sorted = sorted(zip(prices.values(),prices.keys()))
print(price_sorted)

 

1.9在两个字典中寻找相同点

 

a = {"x":2,"y":5,"z":7
}b = {"x":2,"y":8,"w":4
}print(a.keys() & b.keys())#寻找相同的key
print(a.keys() - b.keys())#寻找a中有b中没有的key
print(a.items() & b.items())#寻找相同项

 

 

1.10从序列中移除重复项且保持元素间顺序不变

def dedupe(items,key = None):seen = set()for item in items:val = item if key is None else key(item)if val not in seen:yield itemseen.add(val)