Python ord 函数 - Python零基础入门教程

我不是女神ヾ 2023-01-23 13:52 96阅读 0赞

*  一.Python ord 函数介绍
 *  二.Python ord 函数使用
    
     *  案例 1
     *  案例 2
 *  三.猜你喜欢

> 零基础 Python 学习路线推荐 : [Python 学习目录][Python] >> [Python 基础入门][Python 1]

## 一.Python ord 函数介绍 ##

\*\* [ord 函数][ord]是以单个字符作为参数，返回对应的 ASCll 数值或者 Unicode 值，如果所给的 Unicode 字符超出了你的 [Python][Python 2] 定义范围，则会引发一个 TypeError 的异常\*\*。语法如下：

'''
    参数介绍：
        c — 是一个单字符，长度为1，例如：‘a’/‘b’/‘c’/‘d’/‘e‘等等
    
    返回值 — 对应的十进制整数（ASCll数值）
    '''
    
    ord(c)

**注意：ord 函数的参数是单个字符，并非多个，则会引发 TypeError 异常！！**

## 二.Python ord 函数使用 ##

### 案例 1 ###

>>>ord('a')
    97
    >>>ord('b')
    98
    >>>ord('c')
    99

### 案例 2 ###

# !usr/bin/env python
    # !usr/bin/env python
    # -*- coding:utf-8 _*-
    """
    @Author:猿说编程
    @Blog(个人博客地址): www.codersrc.com
    @File:Python ord 函数.py
    @Time:2021/04/21 07:37
    @Motto:不积跬步无以至千里，不积小流无以成江海，程序人生的精彩需要坚持不懈地积累！
    
    """
    
    
    
    str=input("请输入任意字符:")
    yin=shu=kong=qita=0
    for i in str:
    
        # 小写字母a~z的ascall码对应为：65-91
        # 大写字母A~Z的ascall码对应为：97-123
        if(ord(i)>=97 and ord(i)<=122) or (ord(i)>=65 and ord(i)<=90):
            yin=yin+1
        elif ord(i)>=48 and ord(i)<=57:
            shu=shu+1
        elif ord(i)==32:
            kong=kong+1
        else:
            qita=qita+1
    print("英文字母个数:{}个".format(yin))
    print("数字个数:{}个".format(shu))
    print("空格个数:{}个".format(kong))
    print("其他字符个数:{}个".format(qita))
    
    
    '''
    输出结果：
    
    请输入任意字符:dfsd fgdfghfhfg jh
    英文字母个数:16个
    数字个数:0个
    空格个数:3个
    其他字符个数:0个
    '''

## 三.**猜你喜欢** ##

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未经允许不得转载：[猿说编程][Python 2] » [Python ord 函数][ord]

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