python 第一天

ops/2025/2/7 9:03:13/

循环打印

        while循环

languages = ["Regional Assembly Language","Autocode","FORTRAN","IPL (LISP的先驱)","FLOW-MATIC (COBOL的先驱)","COMTRAN (COBOL的先驱)","LISP","ALGOL 58","FACT (COBOL的先驱)","COBOL","APL","Simula","SNOBOL","CPL (C的先驱)","BASIC","PL/I","BCPL (C的先驱)","Logo","Pascal","Forth","C语言","Smalltalk","Prolog","ML","Scheme","SQL","Ada","C++","Common Lisp","MATLAB","Eiffel","Objective-C","Erlang","Perl","Tcl","FL (Backus)","Haskell","Python","Visual Basic","HTML","Ruby","Lua","CLOS (part of ANSI Common Lisp)","Java","Delphi (Object Pascal)","JavaScript","PHP","REBOL","D","C#","Visual Basic .NET","F#","Scala","Factor","Windows PowerShell","Rust","Clojure","Go"]
years = [1951, 1952, 1954, 1954, 1955, 1957, 1958, 1958, 1959, 1959, 1962, 1962, 1962, 1963, 1964, 1964, 1967 ,1968 ,1970 ,1970 ,1972 ,1972 ,1972 ,1973 ,1975 ,1978 ,1980 ,1983 ,1984 ,1984 ,1985 ,1986 ,1986 ,1987 ,1988 ,1989 ,1990 ,1991 ,1991 ,1991 ,1993 ,1993 ,1994 ,1995 ,1995 ,1995 ,1995 ,1997 ,1999 ,2001 ,2001 ,2002 ,2003 ,2003 ,2006 ,2006 ,2007 ,2009]
i = 0
while i < len(years):language = languages[i]year = years[i]print(language, ':', year)i += 1

        for循环

languages = ["Regional Assembly Language","Autocode","FORTRAN","IPL (LISP的先驱)","FLOW-MATIC (COBOL的先驱)","COMTRAN (COBOL的先驱)","LISP","ALGOL 58","FACT (COBOL的先驱)","COBOL","APL","Simula","SNOBOL","CPL (C的先驱)","BASIC","PL/I","BCPL (C的先驱)","Logo","Pascal","Forth","C语言","Smalltalk","Prolog","ML","Scheme","SQL","Ada","C++","Common Lisp","MATLAB","Eiffel","Objective-C","Erlang","Perl","Tcl","FL (Backus)","Haskell","Python","Visual Basic","HTML","Ruby","Lua","CLOS (part of ANSI Common Lisp)","Java","Delphi (Object Pascal)","JavaScript","PHP","REBOL","D","C#","Visual Basic .NET","F#","Scala","Factor","Windows PowerShell","Rust","Clojure","Go"]
years = [1951, 1952, 1954, 1954, 1955, 1957, 1958, 1958, 1959, 1959, 1962, 1962, 1962, 1963, 1964, 1964, 1967 ,1968 ,1970 ,1970 ,1972 ,1972 ,1972 ,1973 ,1975 ,1978 ,1980 ,1983 ,1984 ,1984 ,1985 ,1986 ,1986 ,1987 ,1988 ,1989 ,1990 ,1991 ,1991 ,1991 ,1993 ,1993 ,1994 ,1995 ,1995 ,1995 ,1995 ,1997 ,1999 ,2001 ,2001 ,2002 ,2003 ,2003 ,2006 ,2006 ,2007 ,2009]
i = 0
for i in range(0, len(languages)):language = languages[i]year = years[i]print(language, ':',year)i += 1

        陌生语法:把for循环套在print函数里

languages = ["Regional Assembly Language","Autocode","FORTRAN","IPL (LISP的先驱)","FLOW-MATIC (COBOL的先驱)","COMTRAN (COBOL的先驱)","LISP","ALGOL 58","FACT (COBOL的先驱)","COBOL","APL","Simula","SNOBOL","CPL (C的先驱)","BASIC","PL/I","BCPL (C的先驱)","Logo","Pascal","Forth","C语言","Smalltalk","Prolog","ML","Scheme","SQL","Ada","C++","Common Lisp","MATLAB","Eiffel","Objective-C","Erlang","Perl","Tcl","FL (Backus)","Haskell","Python","Visual Basic","HTML","Ruby","Lua","CLOS (part of ANSI Common Lisp)","Java","Delphi (Object Pascal)","JavaScript","PHP","REBOL","D","C#","Visual Basic .NET","F#","Scala","Factor","Windows PowerShell","Rust","Clojure","Go"]
years = [1951, 1952, 1954, 1954, 1955, 1957, 1958, 1958, 1959, 1959, 1962, 1962, 1962, 1963, 1964, 1964, 1967 ,1968 ,1970 ,1970 ,1972 ,1972 ,1972 ,1973 ,1975 ,1978 ,1980 ,1983 ,1984 ,1984 ,1985 ,1986 ,1986 ,1987 ,1988 ,1989 ,1990 ,1991 ,1991 ,1991 ,1993 ,1993 ,1994 ,1995 ,1995 ,1995 ,1995 ,1997 ,1999 ,2001 ,2001 ,2002 ,2003 ,2003 ,2006 ,2006 ,2007 ,2009][print(languages[i], ':', years[i]) for i in range(0, len(languages))]

问题:编写一个程序,它将找到所有这些数字,可被7整除,但不是5的倍数,2000年至3200年(包括在内)。得到的数字应按逗号分隔的顺序打印在一行上。
提示:考虑使用range(#begin, #end)方法

l = []
for i in range(2000 , 3201):if (i%7==0) and (i%5!=0):l.append(str(i))print(','.join(l))

join的使用:将容器之中的各个元素之间用指定符号连接。

问题:编写一个可以计算给定数的阶乘的程序。结果应该以逗号分隔的顺序打印在一行上。假设向程序提供以下输入:8
则输出为:40320
提示:在为问题提供输入数据的情况下,应该假设它是控制台输入。

def fact(x):if x == 0:return 1return x * fact(x - 1)
x = int(input('请输入一个数字:'))
print(fact(x))

问题:使用给定的整数n,编写一个程序生成一个包含(i, i*i)的字典,该字典包含1到n之间的整数(两者都包含)。然后程序应该打印字典。
假设向程序提供以下输入:8
则输出为:
{1:1,2:4,3:9,4:16,5:25,6:36,,7:49,8:64}
提示:在为问题提供输入数据的情况下,应该假设它是控制台输入。考虑使用dict类型()


http://www.ppmy.cn/ops/45927.html

相关文章

深度学习入门

文章目录 深度学习基础前言深度学习应用计算机视觉 神经网络基础得分函数 f(x,W)损失函数Softmax分类器前向传播反向传播神经网络整体架构过拟合的解决办法激活函数 深度学习基础 前言 机器学习流程&#xff1a; 数据获取特征工程建立模型评估与应用 特征工程的作用&#x…

RabbitMQ(Direct 订阅模型-路由模式)的使用

文章目录 RabbitMQ&#xff08;Direct 订阅模型-路由模式&#xff09;一&#xff0c;Direct 订阅模型-路由模式介绍&#xff08;Routing&#xff09;二&#xff0c;使用1.添加依赖2.修改配置文件3.创建配置类4.注入RabbitMQ模版引擎5.消息的发送6.消息的接收(监听)7.设置回调函…

Java大厂面试题第2季

一、本课程前提要求和说明 面试题1&#xff1a; 面试题2&#xff1a; 面试题3&#xff1a; 面试题4&#xff1a; 面试题5&#xff1a; 高频最多的常见笔试面试题目 ArrayList HashMap 底层是什么东东 JVM/GC 多线程与高并发 java集合类

Adaboost 算法【python,机器学习,算法】

算法步骤 初始化样本数据&#xff0c;样本数据集大小为N&#xff0c;每个样本的权重设置为1/N。 相关公式&#xff1a; D 1 ( w 11 , w 12 , w 13 , w 14 , . . . , w 1 i , . . . , w 1 N ) , w 1 i 1 N , i 1 , 2 , 3 , 4 , . . . , N D_1(w_{11},w_{12},w_{13},w_{14},.…

自适应容积卡尔曼滤波|(自适应CKF)的MATLAB源代码

介绍 容积卡尔曼滤波在理论上拥有比UKF更高的精度和稳定性&#xff0c;本自适应算法通过对观测残差的计算&#xff0c;在观测协方差R不准确或无法获得时&#xff0c;对R进行调节&#xff0c;以起到降低估计误差的作用。 模型 使用的是三维的非线性模型&#xff0c;经过适当修…

Vue进阶之Vue无代码可视化项目(一)

Vue无代码可视化项目 项目搭建初始步骤拓展:工程项目从0-1项目规范化package.jsoncpell.jsoncustom-words.txtts-eslint规则.eslintrc.cjsgit钩子检查有没有问题type-checkspellchecklint:stylehusky操作安装pre-commitpnpm的commit规范package.json:commitlint.config.cjs安装…

chrome谷歌浏览器开启Gemini Nano模型

前提 确保您的操作系统语言设置为英语&#xff08;美国&#xff09; 可能还需要将 Chrome 浏览器的语言更改为英语&#xff08;美国&#xff09;。 下载dev或Canary版本Chrome Chrome Canary Chrome Dev 注意&#xff1a;确认您的版本高于 127.0.6512.0。 其中一个Chrome版本…

C++ 混合运算的类型转换

一 混合运算和隐式转换 257 整型2 浮点5 行吗&#xff1f;成吗&#xff1f;中不中&#xff1f; C 中允许相关的数据类型进行混合运算。 相关类型。 尽管在程序中的数据类型不同&#xff0c;但逻辑上进行这种运算是合理的相关类型在混合运算时会自动进行类型转换&#xff0c;再…