35个不会也要知道的Python小技巧
2024-05-23 23:13:48
整个集合大概是按照难易程度排序，简单常见的在前面，比较少见的在最后。
1 拆箱
>>> a, b, c = 1, 2, 3
>>> a, b, c
(1, 2, 3)
>>> a, b, c = [1, 2, 3]
>>> a, b, c
(1, 2, 3)
>>> a, b, c = (2 * i + 1 for i in range(3))
>>> a, b, c
(1, 3, 5)
>>> a, (b, c), d = [1, (2, 3), 4]
>>> a
1
>>> b
2
>>> c
3
>>> d
42 拆箱变量交换
>>> a, b = 1, 2
>>> a, b = b, a
>>> a, b
(2, 1)
3 扩展拆箱（只兼容python3）
>>> a, *b, c = [1, 2, 3, 4, 5]
>>> a
1
>>> b
[2, 3, 4]
>>> c
5
4 负数索引
'''
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]
>>> a[-1]
10
>>> a[-3]
8
5 切割列表
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[2:8]
[2, 3, 4, 5, 6, 7]
6 负数索引切割列表
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[-4:-2]
[7, 8]
7指定步长切割列表
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[::2]
[0, 2, 4, 6, 8, 10]
>>> a[::3]
[0, 3, 6, 9]
>>> a[2:8:2]
[2, 4, 6]
8 负数步长切割列表
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> a[::-1]
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
>>> a[::-2]
[10, 8, 6, 4, 2, 0]
9 列表切割赋值
>>> a = [1, 2, 3, 4, 5]
>>> a[2:3] = [0, 0]
>>> a
[1, 2, 0, 0, 4, 5]
>>> a[1:1] = [8, 9]
>>> a
[1, 8, 9, 2, 0, 0, 4, 5]
>>> a[1:-1] = []
>>> a
[1, 5]
10 命名列表切割方式
>>> a = [0, 1, 2, 3, 4, 5]
>>> LASTTHREE = slice(-3, None)
>>> LASTTHREE
slice(-3, None, None)
>>> a[LASTTHREE]
[3, 4, 5]
11 列表以及迭代器的压缩和解压缩
>>> a = [1, 2, 3]
>>> b = ['a', 'b', 'c']
>>> z = zip(a, b)
>>> z
[(1, 'a'), (2, 'b'), (3, 'c')]
>>> zip(*z)
[(1, 2, 3), ('a', 'b', 'c')]
12 列表相邻元素压缩器
>>> a = [1, 2, 3, 4, 5, 6]
>>> zip(*([iter(a)] * 2))
[(1, 2), (3, 4), (5, 6)]>>> group_adjacent = lambda a, k: zip(*([iter(a)] * k))
>>> group_adjacent(a, 3)
[(1, 2, 3), (4, 5, 6)]
>>> group_adjacent(a, 2)
[(1, 2), (3, 4), (5, 6)]
>>> group_adjacent(a, 1)
[(1,), (2,), (3,), (4,), (5,), (6,)]>>> zip(a[::2], a[1::2])
[(1, 2), (3, 4), (5, 6)]>>> zip(a[::3], a[1::3], a[2::3])
[(1, 2, 3), (4, 5, 6)]>>> group_adjacent = lambda a, k: zip(*(a[i::k] for i in range(k)))
>>> group_adjacent(a, 3)
[(1, 2, 3), (4, 5, 6)]
>>> group_adjacent(a, 2)
[(1, 2), (3, 4), (5, 6)]
>>> group_adjacent(a, 1)
[(1,), (2,), (3,), (4,), (5,), (6,)]
13 在列表中用压缩器和迭代器滑动取值窗口
'''
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！
'''
>>> def n_grams(a, n):
... z = [iter(a[i:]) for i in range(n)]
... return zip(*z)
...
>>> a = [1, 2, 3, 4, 5, 6]
>>> n_grams(a, 3)
[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6)]
>>> n_grams(a, 2)
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
>>> n_grams(a, 4)
[(1, 2, 3, 4), (2, 3, 4, 5), (3, 4, 5, 6)]
14 用压缩器反转字典
>>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
>>> m.items()
[('a', 1), ('c', 3), ('b', 2), ('d', 4)]
>>> zip(m.values(), m.keys())
[(1, 'a'), (3, 'c'), (2, 'b'), (4, 'd')]
>>> mi = dict(zip(m.values(), m.keys()))
>>> mi
{1: 'a', 2: 'b', 3: 'c', 4: 'd'}
15 列表展开
>>> a = [[1, 2], [3, 4], [5, 6]]
>>> list(itertools.chain.from_iterable(a))
[1, 2, 3, 4, 5, 6]>>> sum(a, [])
[1, 2, 3, 4, 5, 6]>>> [x for l in a for x in l]
[1, 2, 3, 4, 5, 6]>>> a = [[[1, 2], [3, 4]], [[5, 6], [7, 8]]]
>>> [x for l1 in a for l2 in l1 for x in l2]
[1, 2, 3, 4, 5, 6, 7, 8]>>> a = [1, 2, [3, 4], [[5, 6], [7, 8]]]
>>> flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]
>>> flatten(a)
[1, 2, 3, 4, 5, 6, 7, 8]
16 生成器表达式
13
>>> g = (x ** 2 for x in xrange(10))
>>> next(g)
0
>>> next(g)
1
>>> next(g)
4
>>> next(g)
9
>>> sum(x ** 3 for x in xrange(10))
2025
>>> sum(x ** 3 for x in xrange(10) if x % 3 == 1)
408
17 字典推导
>>> m = {x: x ** 2 for x in range(5)}
>>> m
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}>>> m = {x: 'A' + str(x) for x in range(10)}
>>> m
{0: 'A0', 1: 'A1', 2: 'A2', 3: 'A3', 4: 'A4', 5: 'A5', 6: 'A6', 7: 'A7', 8: 'A8', 9: 'A9'}
18 用字典推导反转字典
>>> m = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
>>> m
{'d': 4, 'a': 1, 'b': 2, 'c': 3}
>>> {v: k for k, v in m.items()}
{1: 'a', 2: 'b', 3: 'c', 4: 'd'}
19 命名元组
>>> Point = collections.namedtuple('Point', ['x', 'y'])
>>> p = Point(x=1.0, y=2.0)
>>> p
Point(x=1.0, y=2.0)
>>> p.x
1.0
>>> p.y
2.0
20 继承命名元组
>>> 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)
>>> p + q
Point(x=3.0, y=5.0)
21 操作集合
>>> A = {1, 2, 3, 3}
>>> A
set([1, 2, 3])
>>> B = {3, 4, 5, 6, 7}
>>> B
set([3, 4, 5, 6, 7])
>>> A | B
set([1, 2, 3, 4, 5, 6, 7])
>>> A & B
set([3])
>>> A - B
set([1, 2])
>>> B - A
set([4, 5, 6, 7])
>>> A ^ B
set([1, 2, 4, 5, 6, 7])
>>> (A ^ B) == ((A - B) | (B - A))
True
22 操作多重集合
>>> A = collections.Counter([1, 2, 2])
>>> B = collections.Counter([2, 2, 3])
>>> A
Counter({2: 2, 1: 1})
>>> B
Counter({2: 2, 3: 1})
>>> A | B
Counter({2: 2, 1: 1, 3: 1})
>>> A & B
Counter({2: 2})
>>> A + B
Counter({2: 4, 1: 1, 3: 1})
>>> A - B
Counter({1: 1})
>>> B - A
Counter({3: 1})
23 统计在可迭代器中最常出现的元素
>>> A = collections.Counter([1, 1, 2, 2, 3, 3, 3, 3, 4, 5, 6, 7])
>>> A
Counter({3: 4, 1: 2, 2: 2, 4: 1, 5: 1, 6: 1, 7: 1})
>>> A.most_common(1)
[(3, 4)]
>>> A.most_common(3)
[(3, 4), (1, 2), (2, 2)]
24 两端都可操作的队列
>>> Q = collections.deque()
>>> Q.append(1)
>>> Q.appendleft(2)
>>> Q.extend([3, 4])
>>> Q.extendleft([5, 6])
>>> Q
deque([6, 5, 2, 1, 3, 4])
>>> Q.pop()
4
>>> Q.popleft()
6
>>> Q
deque([5, 2, 1, 3])
>>> Q.rotate(3)
>>> Q
deque([2, 1, 3, 5])
>>> Q.rotate(-3)
>>> Q
deque([5, 2, 1, 3])
25 有最大长度的双端队列
'''
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 xrange(10):
... last_three.append(i)
... print (', '.join(str(x) for x in last_three))
...
0
0, 1
0, 1, 2
1, 2, 3
2, 3, 4
3, 4, 5
4, 5, 6
5, 6, 7
6, 7, 8
7, 8, 9
26 可排序词典
>>> m = dict((str(x), x) for x in range(10))
>>> print (', '.join(m.keys()))
1, 0, 3, 2, 5, 4, 7, 6, 9, 8
>>> m = collections.OrderedDict((str(x), x) for x in range(10))
>>> print (', '.join(m.keys()))
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
>>> m = collections.OrderedDict((str(x), x) for x in range(10, 0, -1))
>>> print (', '.join(m.keys()))
10, 9, 8, 7, 6, 5, 4, 3, 2, 1
27 默认词典
>>> m = dict()
>>> m['a']
Traceback (most recent call last):File "<stdin>", line 1, in <module>
KeyError: 'a'
>>>
>>> m = collections.defaultdict(int)
>>> m['a']
0
>>> m['b']
0
>>> m = collections.defaultdict(str)
>>> m['a']
''
>>> m['b'] += 'a'
>>> m['b']
'a'
>>> m = collections.defaultdict(lambda: '[default value]')
>>> m['a']
'[default value]'
>>> m['b']
'[default value]'
28 默认字典的简单树状表达
>>> 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=(',', ': ')))
{"menu": {"id": "file","menuitems": {"close": {"onclick": "close();","value": "Close"},"new": {"onclick": "new();","value": "New"},"open": {"onclick": "open();","value": "Open"}},"value": "File"}
}
29 对象到唯一计数的映射
>>> import itertools, collections
>>> value_to_numeric_map = collections.defaultdict(itertools.count().next)
>>> value_to_numeric_map['a']
0
>>> value_to_numeric_map['b']
1
>>> value_to_numeric_map['c']
2
>>> value_to_numeric_map['a']
0
>>> value_to_numeric_map['b']
1
30 最大和最小的几个列表元素
>>> a = [random.randint(0, 100) for __ in xrange(100)]
>>> heapq.nsmallest(5, a)
[3, 3, 5, 6, 8]
>>> heapq.nlargest(5, a)
[100, 100, 99, 98, 98]
31 两个列表的笛卡尔积
'''
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！
'''
>>> for p in itertools.product([1, 2, 3], [4, 5]):
(1, 4)
(1, 5)
(2, 4)
(2, 5)
(3, 4)
(3, 5)
>>> for p in itertools.product([0, 1], repeat=4):
... print (''.join(str(x) for x in p))
...
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
32 列表组合和列表元素替代组合
>>> for c in itertools.combinations([1, 2, 3, 4, 5], 3):
... print (''.join(str(x) for x in c))
...
123
124
125
134
135
145
234
235
245
345
>>> for c in itertools.combinations_with_replacement([1, 2, 3], 2):
... print (''.join(str(x) for x in c))
...
11
12
13
22
23
33
33 列表元素排列组合
>>> for p in itertools.permutations([1, 2, 3, 4]):
... print (''.join(str(x) for x in p))
...
1234
1243
1324
1342
1423
1432
2134
2143
2314
2341
2413
2431
3124
3142
3214
3241
3412
3421
4123
4132
4213
4231
4312
4321
34 可链接迭代器
>>> a = [1, 2, 3, 4]
>>> for p in itertools.chain(itertools.combinations(a, 2), itertools.combinations(a, 3)):
... print (p)
...
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
(1, 2, 3)
(1, 2, 4)
(1, 3, 4)
(2, 3, 4)
>>> for subset in itertools.chain.from_iterable(itertools.combinations(a, n) for n in range(len(a) + 1))
... print (subset)
...
()
(1,)
(2,)
(3,)
(4,)
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
(1, 2, 3)
(1, 2, 4)
(1, 3, 4)
(2, 3, 4)
(1, 2, 3, 4)
35 根据文件指定列类聚
>>> import itertools
>>> with open('contactlenses.csv', 'r') as infile:
... 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)
young myope no reduced none
young myope no normal soft
young myope yes reduced none
young myope yes normal hard
young hypermetrope no reduced none
young hypermetrope no normal soft
young hypermetrope yes reduced none
young hypermetrope yes normal hard
pre-presbyopic myope no reduced none
pre-presbyopic myope no normal soft
pre-presbyopic myope yes reduced none
pre-presbyopic myope yes normal hard
pre-presbyopic hypermetrope no reduced none
pre-presbyopic hypermetrope no normal soft
pre-presbyopic hypermetrope yes reduced none
pre-presbyopic hypermetrope yes normal none
presbyopic myope no reduced none
presbyopic myope no normal none
presbyopic myope yes reduced none
presbyopic myope yes normal hard
presbyopic hypermetrope no reduced none
presbyopic hypermetrope no normal soft
presbyopic hypermetrope yes reduced none
presbyopic hypermetrope yes normal none>>> data.sort(key=lambda r: r[-1])
>>> for value, group in itertools.groupby(data, lambda r: r[-1]):
... print ('-----------')
... print ('Group: ' + value)
... print_data(group)
...
-----------
Group: hard
young myope yes normal hard
young hypermetrope yes normal hard
pre-presbyopic myope yes normal hard
presbyopic myope yes normal hard
-----------
Group: none
young myope no reduced none
young myope yes reduced none
young hypermetrope no reduced none
young hypermetrope yes reduced none
pre-presbyopic myope no reduced none
pre-presbyopic myope yes reduced none
pre-presbyopic hypermetrope no reduced none
pre-presbyopic hypermetrope yes reduced none
pre-presbyopic hypermetrope yes normal none
presbyopic myope no reduced none
presbyopic myope no normal none
presbyopic myope yes reduced none
presbyopic hypermetrope no reduced none
presbyopic hypermetrope yes reduced none
presbyopic hypermetrope yes normal none
-----------
Group: soft
young myope no normal soft
young hypermetrope no normal soft
pre-presbyopic myope no normal soft
pre-presbyopic hypermetrope no normal soft
presbyopic hypermetrope no normal soft
35个不会也要知道的Python小技巧相关推荐
- 【译】5 个你需要知道的 JavaScript 小技巧
JavaScript 是目前最流行的编程语言之一.就像其他任何编程语言一样,它也有很多小技巧,从今天开始你就可以使用它们 大多数程序员都应该每天训练这些小技巧,直到熟能生巧. 在这篇文章中,我们将一起 ...
- 前端开发者应该知道的 CSS 小技巧
一些小技巧让你的CSS技术更专业 使用:not()去除导航上不需要的边框 为body添加行高 垂直居中任何元素 逗号分离的列表 使用负nth-child选择元素 使用SVG图标 文本显示优化 在纯CS ...
- 一些你可能不知道的 Python 小技巧！
[CSDN 编者按]Python 风头正盛,未来一段时间内想必也会是热门编程语言之一.因此,熟练掌握 Python 对开发者来说极其重要,说不定能给作为开发者的你带来意想不到的财富. 作者 | Seb ...
- python 代码换行_你一定不知道的Python小技巧，提升你Python代码的可读性
Python的初学者,开发者都应该知道的代码可读性提高技巧,本篇主要介绍了如下内容: PEP 8是什么以及它存在的原因 为什么你应该编写符合PEP 8标准的代码 如何编写符合PEP 8的代码 为什么我 ...
- Python 程序员需要知道的 30 个技巧（转载）
Python 程序员需要知道的 30 个技巧 原创2017-04-14 伯乐专栏/阿喵 Python开发者 Python开发者 Python开发者 微信号 PythonCoder 功能介绍 人生苦短, ...
- Vue 开发必须知道的 36 个技巧【近1W字】
前言 Vue 3.x 的Pre-Alpha 版本.后面应该还会有 Alpha.Beta 等版本,预计至少要等到 2020 年第一季度才有可能发布 3.0 正式版; 所以应该趁还没出来加紧打好 Vue2 ...
- vue Mutation 必须是同步函数 为什么_Vue 开发必须知道的 36 个技巧「近1W字」
Vue 开发必须知道的 36 个技巧 说明:本文并非原创,来源于segmentfault,搜索文章标题可查看原文 前言 Vue 3.x 的Pre-Alpha 版本.后面应该还会有 Alpha.Beta ...
- Vue 开发必须知道的 36 个技巧
作者 | 火狼1 链接 | https://juejin.im/post/5d9d386fe51d45784d3f8637 前言 Vue 3.x 的Pre-Alpha 版本.后面应该还会有 Alpha ...
- 35 个提高千倍效率的 Java 代码小技巧
前言 代码优化 ,一个很重要的课题.可能有些人觉得没用,一些细小的地方有什么好修改的,改与不改对于代码的运行效率有什么影响呢?这个问题我是这么考虑的,就像大海里面的鲸鱼一样,它吃一条小虾米有用吗?没用 ...
最新文章
- 【Android 启动过程】Activity 启动源码分析 ( ActivityThread -＞ Activity、主线程阶段 一 )
- C#StreamWriter的操作解析
- 聊聊高并发（二十四）解析java.util.concurrent各个组件（六） 深入理解AQS（四）
- qpsk 锁相环_本科毕业设计课题—QPSK相干解调的MATLAB仿真（4）
- Intellij IDEA中maven更新不下来pom中的jar包，reimport失效
- VMware GSX Server 3.2.1 Build 19281免费下载
- 第六章、数组_黑马Java第57期个人学习笔记_个人笔记
- Linux-4.x_x _64 内核配置选项简介
- Python使用pyodbc访问数据库操作方法详解
- 应用Dubbo框架打造仿猫眼项目 理解微服务核心思想
- 基于 yolo_v5 训练识别人物
- LSCP: Locally Selective Combination in Parallel Outlier Ensembles-学习笔记
- 派蒙的奇妙冒险------石之海（C/C++）
- 关于ADB需要知道的一些知识
- C语言编程＞第一周 ⑧ 输入两个正整数m和n，求其最大公约数和最小公倍数。
- 学校的计算机是作文,未来的电脑学校作文
- 企业网盘到底应该怎么选？
- HDU - 1173 采矿
- 笑谈贝叶斯网络（干货）上
- 智邦国际ERP系统帮助企业控制成本