import time
time01 = time.time()#现在时间
a ='abcdefghijklmnopqrstuvwxyz'print(a[-26])#打印第倒数26个字符
a = a.replace(a[3],'高')#更换第四个字符print('a is',a)
a = a.replace(a[3],'d')#更换第四个字符print(a[:-1])#打印到第倒数第一个字符
b ='to be or not to be'print(b[::-1])#打印从头到尾 倒着来
c ='sxtsxtsxtsxt'print(c[2::3])#打印第三个字符到最后一个,但是每打一个跳三个字符
d = b.split()#打印be to or not to be 字符,但是以空格分割print('d is ',d)
e = b.split("be")#打印字符串b, 然后吧be切割剔除print(e)
h =['to','be','or','not','to','be']
i =' '.join(h)print('I IS',i)#打印h 字符串,然后每一个字符连接起来,“分号中间有空格print('D IS',' '.join(d))
time01 = time.time()
j =''for i inrange(10):#定义j,然后循环10次sxt相加,从0到9,
j +='sxt'
time02 = time.time()print('运算时间:'+str(time02 - time01))
time04 = time.time()
h =[]for i inrange(10):
h.append('sxt')#重复添加10次,不是相加
k =''.join(h)#把添加好的新字符串定义成K,最后可以打印出来
time03 = time.time()print('运算时间2:'+str(time03-time04))
l = a.count("b")print("l is ",l)
m = a[1:25:2]#把字符串a ,从第二个字符串到第25个,然后每两步取一次,[1:5]意思是从第二个字符到第五个步长为1,[:]提取整个字符;print('m is',m)