(编辑:jimmy 日期: 2024/12/26 浏览:2)
利用切片
str1 = "hello world!" print(str1[::-1])
利用reduce函数实现
from functools import reduce str1 = "hello world!" print(reduce(lambda x, y : y+x, str1))
str1 = "123455" def fun(string): print("%s" % string == string[::-1] and "YES" or "NO") if __name__ == '__main__': fun(str1)
str1 = "i love you!" print(str1.title())# 单词首字母大写 print(str1.upper())# 所有字母大写 print(str1.lower())# 所有字母小写 print(str1.capitalize())# 字符串首字母大写
可以使用split()函数,括号内可添加拆分字符,默认空格,返回的是列表
str1 = "i love you!" print(str1.split()) # print(str1.split('\')) 则是以\为分隔符拆分
去除字符串两边的空格,返回的是字符串
str1 = " i love you! " print(str1.strip())
返回的是字符串类型
str1 = ["123", "123", "123"] print(''.join(str1))
str1 = "python" list1 = [1, 2, 3] # 乘法表述 print(str1 * 2) print(list1 * 2) # 输出 # pythonpython # [1, 2, 3, 1, 2, 3] #加法表述 str1 = "python" list1 = [1, 2, 3] str1_1 = "" list1_1 = [] for i in range(2): str1_1 += str1 list1_1.append(list1) print(str1_1) print(list1_1) # 输出同上
# 修改每个列表的值 list1 = [2, 2, 2, 2] print([x * 2 for x in list1]) # 展开列表 list2 = [[1, 2, 3], [4, 5, 6], [1]] print([i for k in list2 for i in k]) # 输出 [1, 2, 3, 4, 5, 6, 1]
x = 1 y = 2 x, y = y, x
调用collections中的Counter类
from collections import Counter list1 = ['1', '1', '2', '3', '1', '4'] count = Counter(list1) print(count) # 输出 Counter({'1': 3, '2': 1, '3': 1, '4': 1}) print(count['1']) # 输出 3 print(count.most_common(1))# 出现最多次数的 # [('1', 3)]
str1 = "123456" # 方法一 list_1 = list(map(int, str1)) #方法二 list_2 = [int(i) for i in str1]
str1 = "123456" list1 = [1, 2, 3, 4, 5] for i, j in enumerate(str1): print(i, j) ''' 输出 0 1 1 2 2 3 3 4 4 5 5 6 ''' str1 = "123456" list1 = [1, 2, 3, 4, 5] for i, j in enumerate(list1): print(i, j) # 输出同上
import time start = time.time() for i in range(1999999): continue end = time.time() print(end - start) # 输出 0.08042168617248535
sys.getsizeof()函数
import sys str1 = "123456" print(sys.getsizeof(str1)) # 输出 55
dirt1 = {'a':2, 'b': 3} dirt2 = {'c':3, 'd': 5} # 方法一 combined_dict = {**dirt1, **dirt2} print(combined_dict) # 输出 {'a': 2, 'b': 3, 'c': 3, 'd': 5} # 方法二 dirt1 = {'a':2, 'b': 3} dirt2 = {'c':3, 'd': 5} dirt1.update(dirt2) print(dirt1) # 输出同上
list1 = [1, 2, 3, 4, 5, 6] print('%s' % len(list1) == len(set(list1)) and "NO" or "YES")