编码练习5

最新推荐文章于 2022-10-19 17:17:37 发布

原创最新推荐文章于 2022-10-19 17:17:37 发布 · 189 阅读

0 ·

CC 4.0 BY-SA版权

本文包含三道Python编程练习题，涉及文件处理、多级目录遍历和日期计算。第一题演示如何从文件中去除所有数字，第二题展示如何在多级目录中去除文件内的所有字母，第三题教授计算指定日期与新中国成立日之间的天数。

习题1：把一个文件中的所有数字都去掉。
在这里插入图片描述

filtered_content=""
with open("f:\\a.txt","r",encoding="utf-8") as fp:
    content = fp.read()
    for i in content:
        if i >="0" and i <="9":
            continue
        else:
            filtered_content+=i

with open("f:\\a.txt","w",encoding="utf-8") as fp:
    fp.write(filtered_content)

with open("f:\\a.txt","r",encoding="utf-8") as fp:
    print(fp.read())

在这里插入图片描述
习题2：
多级目录的多个文件中，要把所有字母去掉。

import os.path
file_path = []
for root,dirs,files in os.walk("f:\\letter"):
    for file in files:
        file_path.append(os.path.join(root,file))

print(file_path)

for file in file_path:
    filter_content = ""
    with open(file) as fp:
        content = fp.read()
        for letter in content:
            if (letter >="a" and letter<="z") \
                or (letter >="A" and letter<="Z"):
                continue
            else:
                filter_content+=letter

    with open(file,"w") as fp:
        fp.write(filter_content)

在这里插入图片描述
习题3：
写一个python小程序，输入是年月日，计算距离1949-10-1的天数。

def is_leap_year(year):
    if (year%4==0 and year%100!=0) or year%400==0:
        return True
    else:
        return False


def count_days_interval():
    input_date = input("请输入日期，格式：year-month-day：")
    if input_date.count("-")!=2:
        print("输入日期格式有误，格式year-month-day，请重新输入日期：")
        return None
    year,month,day = input_date.split("-")
    if not (year.isdigit() and month.isdigit() and day.isdigit()):
        print("输入日期格式有误，格式year-month-day，请重新输入日期：")
    year = int(year)
    month = int(month)
    day = int(day)
    
    init_year = 1949
    init_month = 10
    init_day = 1

    if year - init_year ==0:
        if month == init_month:
            return day - init_day
        elif month - init_month ==1:
            return 31+day-1
        else:
            return 31+30+day-1
    elif year - init_year ==1:
        month_days = [31,28,31,30,31,30,31,31,30,31,30,31]
        leap_day = 0
        if month >2 and is_leap_year(year):
            leap_day = 1
        days = 0
        for i in month_days[:month-1]:
            days+=i
        return 92+leap_day+days+day
    else:
        month_days = [31,28,31,30,31,30,31,31,30,31,30,31]
        leap_day = 0
        if month >2 and is_leap_year(year):
            leap_day = 1
        days = 0
        for i in month_days[:month-1]:
            days+=i
        leap_days=0
        for i in range(1950,year):
            if is_leap_year(i):
                leap_days+=1
        years = year - init_year -1
        return 92+leap_day+days+day+years*365+leap_days

print(count_days_interval())

在这里插入图片描述