Heim  >  Artikel  >  Backend-Entwicklung  >  详解Python中的元组与逻辑运算符

详解Python中的元组与逻辑运算符

WBOY
WBOYOriginal
2016-06-10 15:07:44989Durchsuche

Python元组
元组是另一个数据类型,类似于List(列表)。
元组用"()"标识。内部元素用逗号隔开。但是元素不能二次赋值,相当于只读列表。

#!/usr/bin/python
# -*- coding: UTF-8 -*-

tuple = ( 'abcd', 786 , 2.23, 'john', 70.2 )
tinytuple = (123, 'john')

print tuple # 输出完整元组
print tuple[0] # 输出元组的第一个元素
print tuple[1:3] # 输出第二个至第三个的元素 
print tuple[2:] # 输出从第三个开始至列表末尾的所有元素
print tinytuple * 2 # 输出元组两次
print tuple + tinytuple # 打印组合的元组

以上实例输出结果:

('abcd', 786, 2.23, 'john', 70.2)
abcd
(786, 2.23)
(2.23, 'john', 70.2)
(123, 'john', 123, 'john')
('abcd', 786, 2.23, 'john', 70.2, 123, 'john')

以下是元组无效的,因为元组是不允许更新的。而列表是允许更新的:

#!/usr/bin/python
# -*- coding: UTF-8 -*-

tuple = ( 'abcd', 786 , 2.23, 'john', 70.2 )
list = [ 'abcd', 786 , 2.23, 'john', 70.2 ]
tuple[2] = 1000 # 元组中是非法应用
list[2] = 1000 # 列表中是合法应用

Python逻辑运算符
Python语言支持逻辑运算符,以下假设变量a为10,变量b为20:

以下实例演示了Python所有逻辑运算符的操作:

#!/usr/bin/python

a = 10
b = 20
c = 0

if ( a and b ):
  print "Line 1 - a and b are true"
else:
  print "Line 1 - Either a is not true or b is not true"

if ( a or b ):
  print "Line 2 - Either a is true or b is true or both are true"
else:
  print "Line 2 - Neither a is true nor b is true"


a = 0
if ( a and b ):
  print "Line 3 - a and b are true"
else:
  print "Line 3 - Either a is not true or b is not true"

if ( a or b ):
  print "Line 4 - Either a is true or b is true or both are true"
else:
  print "Line 4 - Neither a is true nor b is true"

if not( a and b ):
  print "Line 5 - Either a is not true or b is not true or both are not true"
else:
  print "Line 5 - a and b are true"

以上实例输出结果:

Line 1 - a and b are true
Line 2 - Either a is true or b is true or both are true
Line 3 - Either a is not true or b is not true
Line 4 - Either a is true or b is true or both are true
Line 5 - Either a is not true or b is not true or both are not true

Stellungnahme:
Der Inhalt dieses Artikels wird freiwillig von Internetnutzern beigesteuert und das Urheberrecht liegt beim ursprünglichen Autor. Diese Website übernimmt keine entsprechende rechtliche Verantwortung. Wenn Sie Inhalte finden, bei denen der Verdacht eines Plagiats oder einer Rechtsverletzung besteht, wenden Sie sich bitte an admin@php.cn