标签:
一、生成器
1 def ran(): 2 print(‘Hello world‘) 3 yield ‘F1‘ 4 5 print(‘Hey there!‘) 6 yield ‘F2‘ 7 8 print(‘goodbye‘) 9 yield ‘F3‘ 10 11 ret = ran() # ran()称为生成器函数,ret才是生成器,仅仅具有一种生成能力,函数内部要有关键字yield 12 print(ret) 13 14 res = ret.__next__() #对生成器进行循环操作,遇到yield会停止操作,将yield的值返回给变量,并会记录保存位置 15 print(res) 16 17 res1 = ret.__next__() #下次再对生成器进行操作,会从停止出开始,直到下一个yield停止 18 print(res1) # 当__next__次数超过yield时,会报错 19 20 for i in ret: #进行__next__之后再进行for循环,也是从上次yield停止处开始 21 print(i)
二、字符串的格式化
① % 方法
1 s = ‘I am a %s guy‘ % (‘good‘) 2 print(s) 3 4 n = ‘I am a %s guy,%d years old‘ % (‘good‘,28) 5 print(n) 6 7 d = ‘I am a %(n1)s guy,%(n2)d years old‘ % {‘n1‘:"good",‘n2‘:28} 8 print(d) 9 10 f = ‘I am %f‘ % (28) # 浮点数占位符,默认保留小数点后6位,四舍五入 11 print(f) 12 13 f1 = ‘I am %.2f‘ % (28) #设置保留小数点后2位 14 print(f1) 15 16 # typecode 17 %s : 字符串 18 %d : 十进制数字 19 %f :浮点型 20 %% :% 21 %o : 将十进制转换成八进制返回 22 %x :将十进制转换成十六进制返回 23 %e :将数字转换成科学记数法
② format方法
1 tem = ‘I am {},age {},‘.format(‘Ethan‘,28) 2 print(tem) 3 4 tem = ‘I am {},age {},{}‘.format(*[‘Ethan‘,28,‘Ethan‘]) 5 print(tem) 6 7 tem = ‘I am {0},age {1},really {0}‘.format(‘Ethan‘,28) 8 print(tem) 9 10 tem = ‘I am {0},age {1},really {0}‘.format(*[‘Ethan‘,28]) 11 print(tem) 12 13 tem = ‘I am {name},age {age},really {name}‘.format(**{‘name‘:‘Ethan‘,"age":28}) 14 print(tem) 15 16 tem = ‘I am {name},age {age},really {name}‘.format(name = ‘Ethan‘,age = 28) 17 print(tem) 18 19 tem = ‘I am {0[0]},age {0[1]},really {0[0]}‘.format([‘Ethan‘,28],[‘Seven‘,27]) 20 print(tem) 21 22 tem = ‘I am {:s},age {:d},money {:f}‘.format(‘Ethan‘,28,8988.23) 23 print(tem) # I am Ethan,age 28,money 8988.230000 24 25 tem = ‘I am {:s},age {:d}‘.format(*[‘Ethan‘,28]) 26 print(tem) 27 28 tem = ‘I am {name:s},age {age:d}‘.format(age = 28,name = ‘Ethan‘) 29 print(tem) 30 31 tem = ‘I am {name:s},age {age:d}‘.format(**{‘name‘:‘Ethan‘,‘age‘:28}) 32 print(tem) 33 34 tem = ‘Numbers:{:b},{:o},{:d},{:x},{:X},{:%}‘.format(15,15,15,15,15,15.87623,2) 35 print(tem) # Numbers:1111,17,15,f,F,1587.623000%
标签:
原文地址:http://www.cnblogs.com/ethancui/p/5656757.html