【Python教程】30个语言特性技巧整理

川长思鸟来 2022-09-12 06:48 181阅读 0赞

**1.分拆**

a, b, c = 1, 2, 3
    print(a,b,c) # 1 2 3
    a, b, c = [1, 2, 3]
    print(a,b,c) # 1 2 3
    a, b, c = (2 * i + 1 for i in range(3)) # range不能超出value的范围，超出too many values to unpack (expected 3)
    print(a,b,c) # 1 3 5
    
    a, (b, c), d = [1, (2, 3), 4]
    print(a,b,c) # 1 2 3
    print('-'*10,"分割线","-"*10)

**2.交换变量分拆**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    a, b = 1, 2
    a, b = b, a
    print(a,b) # 2 1
    print('-'*10,"分割线","-"*10)

拓展分拆 (Python 3下适用)

a, *b, c = [1, 2, 3, 4, 5] # *匹配多个值
    print(a,b,c) #1 [2, 3, 4] 5
    print('-'*10,"分割线","-"*10)

**3.负索引**

a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    print(a[-1]) # 10
    print('-'*10,"分割线","-"*10)

**4.列表切片 (a\[start:end\])**

a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    print(a[2:8]) #[2, 3, 4, 5, 6, 7]
    print('-'*10,"分割线","-"*10)

**5.使用负索引的列表切片**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    print(a[-4:-2]) # [7,8]
    print('-'*10,"分割线","-"*10)

**6.带步进值的列表切片 (`a[start:end:step]`)**

print(a[::2]) # [0, 2, 4, 6, 8, 10]
    print(a[::3]) # [0, 3, 6, 9]
    print(a[2:8:2]) # [2, 4, 6]
    print('-'*10,"分割线","-"*10)

**7.负步进值得列表切片**

print(a[::-1]) # [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
    print(a[::-2]) # [10, 8, 6, 4, 2, 0]
    print('-'*10,"分割线","-"*10)

**8.列表切片赋值**

a = [1, 2, 3, 4, 5]
    a[2:3] = [0, 0]
    print(a) # [1, 2, 0, 0, 4, 5]
    a[1:1] = [8, 9]
    print(a) # [1, 8, 9, 2, 0, 0, 4, 5]
    a[1:-1] = []
    print(a) # [1, 5]
    print('-'*10,"分割线","-"*10)

**9.命名切片 (slice(start, end, step))**

a = [0, 1, 2, 3, 4, 5]
    LASTTHREE = slice(-3, None)
    print(LASTTHREE) # slice(-3, None, None)
    print(a[LASTTHREE]) # [3, 4, 5]
    print('-'*10,"分割线","-"*10)

**10.zip打包解包列表和倍数(好奇)**

a = [1, 2, 3]
    b = ['a', 'b', 'c']
    z = zip(a, b) # 
    # <class 'zip'>类型是什么，打印z返回个内存地址,类似迭代器
    print(type(z))
    print(type(zip(*z)))

问题：  
zip(a,b)返回了什么对象,不知，但不能迭代，  
打印返回内存地址,不能用该赋值的对象用list直接转化为列表

zip(a,b)是从a,b种各自取出一个元素组成元组，再将依次组成的元组合成一个新的迭代器

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    print(list(zip(a,b))) # zip直接作用在参数上可以用list转化为列表 [(1, 'a'), (2, 'b'), (3, 'c')]
    print(list(z)) # []
    print(list(zip(*zip(a,b)))) #[(1, 2, 3), ('a', 'b', 'c')]
    print('-'*10,"分割线","-"*10)

**11.使用zip合并相邻的列表项**

a = [1, 2, 3, 4, 5, 6]
    b = zip(*([iter(a)] * 2))
    # 打印不出[(1, 2), (3, 4), (5, 6)]
    print(b) # <zip object at 0x000002387349DB48> ;zip属性还不能用下标，用了报错'zip' object is not subscriptable
    # zip属性迭代也打印不出
    print(i for i in b) # <generator object <genexpr> at 0x0000016DAE8C7A40>
    print('-'*10,"分割线","-"*10)
    group_adjacent = lambda a, k: zip(*([iter(a)] * k))
    print(group_adjacent(a, 3)) # <zip object at 0x000002251FBBDF48>
    print(list(group_adjacent(a,3))) # [(1, 2, 3), (4, 5, 6)]
    print('-'*10,"分割线","-"*10)

**12.使用zip和iterators生成滑动窗口 (n -grams) 不理解**

from itertools import islice
    def n_grams(a, n):
        z = (islice(a, i, None) for i in range(n))
        return list(zip(*z))
    a = [1, 2, 3, 4, 5, 6]
    print(n_grams(a, 3))
    print('-'*10,"分割线","-"*10)

**13.使用zip反转字典**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    m = { 'a': 1, 'b': 2, 'c': 3, 'd': 4}
    print(m.items()) # dict_items([('a', 1), ('b', 2), ('c', 3), ('d', 4)])
    # [(1, 'a'), (2, 'b'), (3, 'c'), (4, 'd')]
    print( list(zip(m.values(), m.keys()))) # 由于zip返回的是内存地址，所以用list转化为列表，但list无法转化zip对象，转化的话返回空列表
    mi = dict(zip(m.values(), m.keys()))
    print(mi) # {1: 'a', 2: 'b', 3: 'c', 4: 'd'}
    print('-'*10,"分割线","-"*10)

**14.摊平列表**

import itertools
    a = [[1, 2], [3, 4], [5, 6]]
    # 使用itertools.chain.from_iterable使二维变一维
    print(list(itertools.chain.from_iterable(a))) # [1, 2, 3, 4, 5, 6]
    # sum使二维变一维
    b = sum(a, [])
    print(b) # [1, 2, 3, 4, 5, 6]
    # []内加双for
    c= [x for l in a for x in l]
    print(c) # [1, 2, 3, 4, 5, 6]
    a = [1, 2, [3, 4], [[5, 6], [7, 8]]]
    # 自定义函数这么长；我什么时候才有这么本事啊
    flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]
    print(flatten(a)) # [1, 2, 3, 4, 5, 6, 7, 8]
    print('-'*10,"分割线","-"*10)

**15.生成器表达式**

g = (x ** 2 for x in range(10))
    print(next(g)) # 0
    print(next(g)) # 1
    d = sum(x ** 3 for x in range(10))
    print(d) # 2025
    e = sum(x ** 3 for x in range(10) if x % 3 == 1) # 3的倍数
    print(e) # 408
    print('-'*10,"分割线","-"*10)

**16.迭代字典**

m = { x: x ** 2 for x in range(5)}
    print(m) # {0: 0, 1: 1, 2: 4, 3: 9, 4: 16}
    m = { x: 'A' + str(x) for x in range(10)} # {0: 'A0', 1: 'A1', 2: 'A2', 3: 'A3', 4: 'A4', 5: 'A5', 6: 'A6', 7: 'A7', 8: 'A8', 9: 'A9'}
    print(m)
    print('-'*10,"分割线","-"*10)

**17.通过迭代字典反转字典**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    m = { 'a': 1, 'b': 2, 'c': 3, 'd': 4}
    print(m) # {'a': 1, 'b': 2, 'c': 3, 'd': 4}
    m1 = { v: k for k, v in m.items()} 
    print(m1) # {1: 'a', 2: 'b', 3: 'c', 4: 'd'}
    print('-'*10,"分割线","-"*10)

**18.命名序列 (collections.namedtuple)**

import collections
    Point = collections.namedtuple('Point', ['x', 'y'])
    p = Point(x=1.0, y=2.0)
    print(p) # Point(x=1.0, y=2.0)
    print(p.x) # 1.0
    print(p.y) # 2.0
    print('-'*10,"分割线","-"*10)

**19.命名列表的继承:**

class Point(collections.namedtuple('PointBase', ['x', 'y'])):
        __slots__ = ()
        def __add__(self, other):
            return Point(x=self.x + other.x, y=self.y + other.y)
    
    p = Point(x=1.0, y=2.0)
    q = Point(x=2.0, y=3.0)
    print(p+q) # Point(x=3.0, y=5.0)
    print('-'*10,"分割线","-"*10)

**20.集合及集合操作**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    A = { 1, 2, 3, 3}
    B = { 3, 4, 5, 6, 7}
    print(A|B) # {1, 2, 3, 4, 5, 6, 7}
    print(A & B) # {3}
    print(A - B) # {1, 2}去除A中B所拥有的元素
    print(B - A) # {4, 5, 6, 7}去除B中A所拥有的元素
    # print(A + B) # 没有A+B
    print(A ^ B) # {1, 2, 4, 5, 6, 7}

^的作用是前面的集合去掉后面的集合剩下的部分或者成并集

print(set([1,2,3])^set([1])) # {2, 3}
    print(set([1])^set([1,2,3])) # {2, 3}
    print((A ^ B) == ((A - B) | (B - A))) # True
    print('-'*10,"分割线","-"*10)

**21.多重集及其操作 (collections.Counter**

A = collections.Counter([1, 2, 2])
    B = collections.Counter([2, 2, 3])
    print(A) # Counter({2: 2, 1: 1})
    print(B) # Counter({2: 2, 3: 1})
    print(A-B) # Counter({1: 1})
    print(B-A) # Counter({3: 1})
    print(A+B) # Counter({2: 4, 1: 1, 3: 1})
    print(A&B) # Counter({2: 2})
    print(A | B) # Counter({2: 2, 1: 1, 3: 1})
    print('-'*10,"分割线","-"*10)

**22.迭代中最常见的元素 (collections.Counter)**

A = collections.Counter([1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 6, 7])
    print(A) # Counter({3: 4, 1: 2, 2: 2, 4: 1, 5: 1, 6: 1, 7: 1})
    print(A.most_common(1)) # [(3, 4)]
    print(A.most_common(3)) # [(3, 4), (1, 2), (2, 2)]
    print('-'*10,"分割线","-"*10)

双端队列 (collections.deque) rotate的用法顺时针或逆时针旋转

Q = collections.deque()
    Q.append(1)
    Q.appendleft(2)
    Q.extend([3, 4])
    Q.extendleft([5, 6])
    print(Q) # deque([6, 5, 2, 1, 3, 4])
    print(Q.pop()) # 4
    print(Q.popleft()) # 6
    print(Q) # deque([5, 2, 1, 3])
    Q.rotate(3)  # 整个队列顺时针旋转3个单位
    print(Q) #deque([2, 1, 3, 5])
    Q.rotate(-3)
    print(Q) # deque([5, 2, 1, 3])
    print('-'*10,"分割线","-"*10)

**23.有最大长度的双端队列 (collections.deque)**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    last_three = collections.deque(maxlen=3)
    for i in range(10):
        last_three.append(i)
        # 这打印出每组3个元素有什么用么,每组3个，都是循环队列中最大的值
        print (','.join(str(x) for x in last_three))
    print('-'*10,"分割线","-"*10)

**24.缺省字典 (collections.defaultdict)**

m = dict()
    # m['a'] # KeyError: 'a'
    m = collections.defaultdict(int)
    print(m['a']) # 0
    m = collections.defaultdict(str)
    print(m['a']) # 空白换行
    m['b'] += 'a'
    print(m['b']) # a
    m = collections.defaultdict(lambda: '[default value]')
    print(m['a']) # [default value]
    
    print('-'*10,"分割线","-"*10)

用缺省字典表示简单的树

import json
    tree = lambda: collections.defaultdict(tree)
    root = tree()
    root['menu']['id'] = 'file'
    root['menu']['value'] = 'File'
    root['menu']['menuitems']['new']['value'] = 'New'
    root['menu']['menuitems']['new']['onclick'] = 'new();'
    root['menu']['menuitems']['open']['value'] = 'Open'
    root['menu']['menuitems']['open']['onclick'] = 'open();'
    root['menu']['menuitems']['close']['value'] = 'Close'
    root['menu']['menuitems']['close']['onclick'] = 'close();'
    # print(json.dumps(root, sort_keys=True, indent=4, separators=(',', ': '))) 
    print('-'*10,"分割线","-"*10)

**25.映射对象到唯一的序列数 (collections.defaultdict)**

import itertools, collections
    value_to_numeric_map = collections.defaultdict(itertools.count().__next__) # 这里使用__next__；不使用next
    print(value_to_numeric_map['a']) # 0
    print(value_to_numeric_map['b']) # 1
    print(value_to_numeric_map['a']) # 0
    print('-'*10,"分割线","-"*10)

**26.最大最小元素 (heapq.nlargest和heapq.nsmallest)**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    import random,heapq
    a = [random.randint(0, 100) for __ in range(100)]
    print(heapq.nsmallest(5, a)) # [0, 0, 0, 1, 1] 最小的5个元素
    print(heapq.nlargest(5, a)) # [100, 98, 97, 96, 92]
    print('-'*10,"分割线","-"*10)

**27.笛卡尔乘积 (itertools.product)**

for p in itertools.product([1, 2, 3], [4, 5]):
        print(p)
    # for p in itertools.product([0, 1], repeat=4):
    # print (''.join(str(x) for x in p))
    print('-'*10,"分割线","-"*10)

**28.组合的组合和置换 (itertools.combinations 和 itertools.combinations\_with\_replacement)**

# for c in itertools.combinations([1, 2, 3, 4, 5], 3):
    # print (''.join(str(x) for x in c))
    print('-'*10,"分割线","-"*10)
    
    # 排序 (itertools.permutations)
    # for p in itertools.permutations([1, 2, 3, 4]):
    # print (''.join(str(x) for x in p))

**29.链接的迭代 (itertools.chain)**

''' 学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：531509025 寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！ '''
    a = [1, 2, 3, 4]
    # for p in itertools.chain(itertools.combinations(a, 2), itertools.combinations(a, 3)):
    # print(p)
    print('-'*10,"分割线","-"*10)

**30.按给定值分组行 (itertools.groupby)**

# from operator import itemgetter
    # import itertools
    # with open('contactlenses.csv', 'r') as infile: # 我并没有准备这个文件
    # data = [line.strip().split(',') for line in infile]
    # data = data[1:]
    # def print_data(rows):
    # print ('\n'.join('\t'.join('{: <16}'.format(s) for s in row) for row in rows))
    
    # print_data(data)
    print('-'*10,"分割线","-"*10)