python简短实例
2022/6/12 1:20:28
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#重复元素判定 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] print(all_unique(x)) print(all_unique(y))
False True
#字符元素组成判定,两个字符串组成元素是否一样 from collections import Counter def anagram(first,second): return Counter(first)==Counter(second) anagram("abcd3","3acbd")
True
#内存占用 import sys variable = 30 print(sys.getsizeof(variable))
28
#字节占用 def byte_size(string): return len(string.encode('utf-8')) byte_size('Hello World')
11
#打印N次字符串 n = 2 s ="Programing" print(s*n)
ProgramingPrograming
#给定具体大小,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]]
#压缩 def compact(lst): return list(filter(bool,lst)) compact([0,1,False,2,'',3,'a','s',34])
[1, 2, 3, 'a', 's', 34]
#解包 unsuccessful array = [['a','b'],['c','d'],['e','f']] transposed = zip(*array) print(transposed)
<zip object at 0x00000172C944E6C0>
#链式对比 a =3 print(2<a<8)
True
#逗号链接 hobbies = ["basketball","football","swimming"] print("My hobbies are:"+",".join(hobbies))
My hobbies are:basketball,football,swimming
#元音统计 (aeiou) 通过正则表达式 import re def count_vowels(str): return len(re.findall(r'[aeiou]',str,re.IGNORECASE)) count_vowels('foobar')
3
def count_vowels(str): return re.findall(r'[aeiou]',str,re.IGNORECASE) count_vowels('foobar')
['o', 'o', 'a']
#展开列表 通过递归方式将列表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 deep_flatten([1,[2],[[3],4],5])
[1, 2, 3, 4, 5]
#列表的差 def difference(a,b): set_a = set(a) set_b = set(b) comparison = set_a.difference(set_b) #a有b无 return list(comparison) difference([1,2,3],[1,2,4])
[3]
#链式函数调用 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
#合并两个字典 def merge_dictionaries(a,b): return {**a,**b} a = {'x':1,'y':2} b = {'y':3,'z':4} print(merge_dictionaries(a,b))
{'x': 1, 'y': 3, 'z': 4}
#使用枚举 通常用for循环,但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
#try else try: 2*3 except TypeError: print("An exception was raised") else:print("Thank God, no exceptions were raised.")
Thank God, no exceptions were raised.
#元素频率 def most_frequent(list): return max(set(list),key=list.count) # 传入count函数之后,再set,再max取最多 list = [1,2,1,2,3,2,1,4,2] most_frequent(list)
2
#回文序列 from re import sub def palindrome(string): subs = sub('[\W_]', '', string.lower()) return subs == subs[::-1] palindrome('taco caT')
True
#不使用if-else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
#交换值 def swap(a,b): return b,a a,b=-1,14 swap(a,b)
(14, -1)
#字典默认值 d={'a':1,'b':2} print(d.get('c',3))
3
#展开列表 def spread(arg): ret =[] for i in arg: if isinstance(i, list): ret.extend(i) else: ret.append(i) return ret spread([1,2,3,[4,5,6],[7],8,9])
[1, 2, 3, 4, 5, 6, 7, 8, 9]
def most_frequent(lst): return max(set(lst),key=lst.count) lst = [1,2,1,2,3,2,1,4,2] most_frequent(lst)
2
#恢复关键字变量原有含义 from pydoc import locate list = locate('list')
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