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30个Python常用极简代码

时间:07-03来源:作者:点击数:

1.检查给定列表是不是存在重复元素

def all_unique(lst):
    return len(lst) == len(set(lst))


x = [1, 1, 2, 2, 3, 2, 3, 4, 5, 6]
y = [1, 2, 3, 4, 5]
all_unique(x)  # False
all_unique(y)  # True

2.检查两个字符串的组成元素是不是一样的

from collections import Counter


def anagram(first, second):
    return Counter(first) == Counter(second)


anagram("abcd3", "3acdb")  # True

3.内存占用

import sys

variable = 30
print(sys.getsizeof(variable))  # 28

4.检查字符串占用的字节数

def byte_size(string):
    return (len(string.encode('utf-8')))


byte_size('')  # 4
byte_size('Hello World')  # 11

5.打印 N 次字符串

n = 2
s = "Programming"
print(s * n)

6.大写第一个字母

s = "programming is awesome"
print(s.title())  # Programming Is Awesome

7.给定具体的大小,定义一个函数以按照这个大小切割列表

from math import ceil


def chunk(lst, size):
    return list(
        map(lambda x: lst[x * size:x * size + size],
            list(range(0, ceil(len(lst) / size)))))


chunk([1, 2, 3, 4, 5], 2)  # [[1,2],[3,4],5]

8.这个方法可以将布尔型的值去掉,例如(False,None,0,“”),它使用 filter() 函数

def compact(lst):
    return list(filter(bool, lst))


compact([0, 1, False, 2, '', 3, 'a', 's', 34])  # [ 1, 2, 3, 'a', 's', 34 ]

9.如下代码段可以将打包好的成对列表解开成两组不同的元组

array = [['a', 'b'], ['c', 'd'], ['e', 'f']]
transposed = zip(*array)
print(transposed)  # [('a', 'c', 'e'), ('b', 'd', 'f')]

10.我们可以在一行代码中使用不同的运算符对比多个不同的元素

a = 3
print(2 < a < 8)  # True
print(1 == a < 2)  # False

11.将列表连接成单个字符串,且每一个元素间的分隔方式设置为了逗号

hobbies = ["basketball", "football", "swimming"]
print("My hobbies are: " + ", ".join(hobbies))  # My hobbies are: basketball, football, swimming

12.统计字符串中的元音 (‘a’, ‘e’, ‘i’, ‘o’, ‘u’) 的个数,它是通过正则表达式做的

import re


def count_vowels(str):
    return len(re.findall(r'[aeiou]', str, re.IGNORECASE))


count_vowels('foobar')  # 3
count_vowels('gym')  # 0

13.给定字符串的第一个字符统一为小写

def decapitalize(string):
    return string[:1].lower() + string[1:]


decapitalize('FooBar')  # 'fooBar'
decapitalize('FooBar')  # 'fooBar'

14.通过递归的方式将列表的嵌套展开为单个列表

def spread(arg):
    ret = []
    for i in arg:
        if isinstance(i, list):
            ret.extend(i)
        else:
            ret.append(i)
    return ret


def deep_flatten(lst):
    result = []

    result.extend(
        spread(list(map(lambda x: deep_flatten(x) if type(x) == list else x, lst))))
    return result


print(deep_flatten([1, [2], [[3], 4], 5]))  # [1,2,3,4,5]

15.返回第一个列表的元素,其不在第二个列表内。如果同时要反馈第二个列表独有的元素,还需要加一句 set_b.difference(set_a)

def difference(a, b):
    set_a = set(a)
    set_b = set(b)
    comparison = set_a.difference(set_b)
    return list(comparison)


difference([1, 2, 3], [1, 2, 4])  # [3]

16.如下方法首先会应用一个给定的函数,然后再返回应用函数后结果有差别的列表元素

from math import floor


def difference_by(a, b, fn):
    b = set(map(fn, b))
    return [item for item in a if fn(item) not in b]


print(difference_by([2.1, 1.2], [2.3, 3.4], floor))  # [1.2]
print(difference_by([{'x': 2}, {'x': 1}], [{'x': 1}], lambda v: v['x']))  # [ { x: 2 } ]

17.你可以在一行代码内调用多个函数

def add(a, b):
    return a + b


def subtract(a, b):
    return a - b


a, b = 4, 5
print((subtract if a > b else add)(a, b))  # 9

18.检查两个列表是不是有重复项

def has_duplicates(lst):
    return len(lst) != len(set(lst))


x = [1, 2, 3, 4, 5, 5]
y = [1, 2, 3, 4, 5]
print(has_duplicates(x))  # True
print(has_duplicates(y))  # False

19.合并两个字典

def merge_two_dicts(a, b):
    c = a.copy()  # make a copy of a
    c.update(b)  # modify keys and values of a with the once from b
    return c


a = {'x': 1, 'y': 2}
b = {'y': 3, 'z': 4}
print(merge_two_dicts(a, b))  # {'y':3,'x':1,'z':4}


def merge_dictionaries(a, b):
    return {**a, **b}


a = {'x': 1, 'y': 2}
b = {'y': 3, 'z': 4}
print(merge_dictionaries(a, b))  # {'y': 3, 'x': 1, 'z': 4}

20.两个列表转化为单个字典

def to_dictionary(keys, values):
    return dict(zip(keys, values))


keys = ["a", "b", "c"]
values = [2, 3, 4]
print(to_dictionary(keys, values))  # {'a': 2, 'c': 4, 'b': 3}

21.我们常用 For 循环来遍历某个列表,同样我们也能枚举列表的索引与值

list = ["a", "b", "c", "d"]
for index, element in enumerate(list):
    print("Value", element, "Index ", index, )
# ('Value', 'a', 'Index ', 0)
# ('Value', 'b', 'Index ', 1)
# ('Value', 'c', 'Index ', 2)
# ('Value', 'd', 'Index ', 3)

22.计算执行特定代码所花费的时间

import time

start_time = time.time()
a = 1
b = 2
c = a + b
print(c)  # 3
end_time = time.time()
total_time = end_time - start_time
print("Time: ", total_time)  # ('Time: ', 1.1205673217773438e-05)

23.我们在使用 try/except 语句的时候也可以加一个 else 子句,如果没有触发错误的话,这个子句就会被运行

try:
    2 * 3
except TypeError:
    print("An exception was raised")
else:
    print("Thank God, no exceptions were raised.")  # Thank God, no exceptions were raised.

24.根据元素频率取列表中最常见的元素

def most_frequent(list):
    return max(set(list), key=list.count)


list = [1, 2, 1, 2, 3, 2, 1, 4, 2]
print(most_frequent(list))  # 2

25.检查给定的字符串是不是回文序列,它首先会把所有字母转化为小写,并移除非英文字母符号。最后,它会对比字符串与反向字符串是否相等,相等则表示为回文序列

from re import sub


def palindrome(string):
    s = sub('[\W_]', '', string.lower())
    return s == s[::-1]


print(palindrome('taco cat'))  # True

26.不使用条件语句就实现加减乘除、求幂操作,它通过字典这一数据结构实现

import operator

action = {
    "+": operator.add,
    "-": operator.sub,
    "/": operator.truediv,
    "*": operator.mul,
    "**": pow
}
print(action['-'](50, 25))  # 25

27.该算法会打乱列表元素的顺序,它主要会通过 Fisher-Yates 算法对新列表进行排序

from copy import deepcopy
from random import randint


def shuffle(lst):
    temp_lst = deepcopy(lst)
    m = len(temp_lst)
    while (m):
        m -= 1
    i = randint(0, m)
    temp_lst[m], temp_lst[i] = temp_lst[i], temp_lst[m]
    return temp_lst


foo = [1, 2, 3]
print(shuffle(foo))  # [2,3,1] , foo = [1,2,3]

28.将列表内的所有元素,包括子列表,都展开成一个列表

def spread(arg):
    ret = []
    for i in arg:
        if isinstance(i, list):
            ret.extend(i)
        else:
            ret.append(i)
    return ret


print(spread([1, 2, 3, [4, 5, 6], [7], 8, 9]))  # [1,2,3,4,5,6,7,8,9]

29.不需要额外的操作就能交换两个变量的值

def swap(a, b):
    return b, a


a, b = -1, 14
print(swap(a, b))  # (14, -1)

30.通过 Key 取对应的 Value 值,可以通过以下方式设置默认值。如果 get() 方法没有设置默认值,那么如果遇到不存在的 Key,则会返回 None

d = {'a': 1, 'b': 2}
print(d.get('c', 3))  # 3
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