Python序列元素解压全面介绍

1、将序列元素解压到单独变量

1.1、简单元组解压

p = (4,5)
x,y = p
print(x,y)

输出： 4 5

1.2、复杂数据结构解压

data = ['Apple',100,4.5,(2021,1,28)]
name,weight,price,date = data
print(name,weight,price,date)

输出：Apple 100 4.5 (2021, 1, 28)

name,weight,price,(year,month,day) = data
print(name,weight,price,year,month,day)

输出：Apple 100 4.5 2021 1 28

**注意：**当变量个数与序列元素数量不匹配时，将出错

1.3、字符串解压

a,b,c = 'abc'
print(a,b,c)

输出：a b c

1.4、忽略变量

有时需要忽略一些变量，则可以通过下划线来实现。

name,_,price,_ = data
print(name,price)

输出：Apple 4.5

2、从任意长度可遍历序列解压元素

record = ['Jenson','[email protected]','[phone]','[phone]']
name,email,*phones = record
print(name,email,phones)

输出：Jenson [email protected] ['[phone]', '[phone]']

2.1、获取最后一个元素和第一个元素

*head,tail = [1,2,3,4,5,6,7,8,9,0]
print(tail)
first,*last = [1,2,2,3,4,5,6]
print(first)

输出：
    0
    1

2.2、过滤元素

fruits = [('apple',4.5),('pear',3.8),('banana',2.5),('grape',12.3)]
for name,*args in fruits:
    if name == 'apple':
        print('price of apple = ',args)

输出：price of apple =  [4.5]

2.3、忽略多个元素

data = ['Apple',100,4.5,(2021,1,28)]
name,*args,(year,_,_) = data
print(name,year)

输出：Apple 2021

2.4、递归调用

def sum(values):
    head,*tail = values
    return head + sum(tail) if tail else head

value = sum([1,2,3,4,5,6,7,8,9,0])
print(value)

输出45

3、保留最后N个元素

保留最后N个元素，可以使用collection.deque来实现

	from collections import deque
	def keep_last_n_elem(values,pattern,history=5):
        previous = deque(maxlen=history)
        for value in values:
            if value in pattern:
                yield value,previous
            previous.append(value)
        
    with open('datas/somefile.txt') as f:
        for line,prev in keep_last_n_elem(f,'python'):
            print(len(prev))
            for pline in prev:
                print(pline,end='')
            print(line,end='')
            print('-'*20)

4、查找最大和最小N个元素

可以通过使用heapq的nlargest和nsmallest实现。

import heapq
nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
print(heapq.nlargest(3, nums))
print(heapq.nsmallest(3, nums))

输出：
    [42, 37, 23]
    [-4, 1, 2]

对于复杂的数据，同样可以实现

    portfolio = [
    {'name': 'IBM', 'shares': 100, 'price': 91.1},
    {'name': 'AAPL', 'shares': 50, 'price': 543.22},
    {'name': 'FB', 'shares': 200, 'price': 21.09},
    {'name': 'HPQ', 'shares': 35, 'price': 31.75},
    {'name': 'YHOO', 'shares': 45, 'price': 16.35},
    {'name': 'ACME', 'shares': 75, 'price': 115.65}
    ]
    cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price'])
    print(cheap)
    expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price'])
    print(expensive)

输出：[{'name': 'YHOO', 'shares': 45, 'price': 16.35}, {'name': 'FB', 'shares': 200, 'price': 21.09}, {'name': 'HPQ', 'shares': 35, 'price': 31.75}]
    [{'name': 'AAPL', 'shares': 50, 'price': 543.22}, {'name': 'ACME', 'shares': 75, 'price': 115.65}, {'name': 'IBM', 'shares': 100, 'price': 91.1}]