python 创立并写入新文件，

python 统计特定文件夹下的 word 和 pdf 的数量

import glob,os

path 就是你说的特定文件夹

path = r”D:\linshi”

这里的 pdf 能够换成 docx

file=glob.glob(os.path.join(path, “*.pdf”))
count = 0
for i in file:

count = count + 1

print(count)

复制文件的残缺门路借助 python 对该文件夹的文件批量复制到另一个指定文件夹中。有两种模式，一种只复制文件。第二种复制文件的残缺门路

import os
import shutil
def get_all_file_by_type(path, type=()): # 取得以 type 类型结尾的所有文件，返回一个 list

filelist = []
for a, b, c in os.walk(path):
    for name in c:
        fname = os.path.join(a, name)
        if fname.endswith(type):
            filelist.append(fname)
return filelist

def get_all_file_by_string(path, string_list):

filelist = []
for a, b, c in os.walk(path):
    for name in c:
        fname = os.path.join(a, name)
        for string in string_list:  # 遍历 string_list，如果文件门路中蕴含 string，那么 append 进 filelist
            if string in fname:  # 如果只想要文件名符合条件，把 fname 换成 name 即可
                filelist.append(fname)
                break
return filelist

def copy_file_by_type(old_path, new_path, type=(‘doc’, ‘docx’), requird_dir=False):

try:
    file_list = get_all_file_by_type(old_path, type=type)  # 取得该门路下所有的 type 类型文件
    if not os.path.exists(new_path):  # 创立新的文件夹
        os.makedirs(new_path)
    if not requird_dir:  # 如果仅复制文件
        for file in file_list:
            name = file.split("\\")[-1]  # 取得文件名字
            new_paths = os.path.join(new_path, name)  # 与新门路拼接，取得残缺的新门路
            shutil.copy(file, new_paths)
            print(new_paths + "胜利")
    if requird_dir:
        for file in file_list:
            name = file.split("\\")[-1]  # 取得文件名字
            new_paths = file.replace(old_path, new_path)  # 将一个残缺门路中，开始的门路替换成新的门路
            dir = new_paths.split(name)[0]  [PerfectMoney 下载](https://www.gendan5.com/wallet/PerfectMoney.html)# 取得文件夹门路
            if not os.path.exists(dir):  # 创立新文件夹
                os.makedirs(dir)
            shutil.copy(file, new_paths)
            print(new_paths + "胜利")
except Exception as e:
    print(e)

def copy_file_by_string(old_path, new_path, string_list, requird_dir=False):

try:
    file_list = get_all_file_by_string(old_path, string_list=string_list)  # 与上述一样，只不过这里调用的是 get_all_file_by_string 办法
    if not os.path.exists(new_path):
        os.makedirs(new_path)
    if not requird_dir:
        for file in file_list:
            name = file.split("\\")[-1]
            new_paths = os.path.join(new_path, name)
            shutil.copy(file, new_paths)
            print(new_paths + "胜利")
    if requird_dir:
        for file in file_list:
            name = file.split("\\")[-1]
            new_paths = file.replace(old_path, new_path)
            print(new_paths)
            dir = new_paths.split(name)[0]
            if not os.path.exists(dir):
                os.makedirs(dir)
            shutil.copy(file, new_paths)
            print(new_paths + "胜利")
except Exception as e:
    print(e)

if name == ‘__main__’:

old_path = r"F:\aaaa"
new_path = r"F:\bbbb"
list = ["面试", "口试", "题库", "题目"]
copy_file_by_string(old_path=old_path, new_path=new_path, string_list=list, requird_dir=False)
# type = ('docx','doc',"pdf","md")
# copy_file_by_type(old_path=old_path, new_path=new_path, type=type, requird_dir=True)

python 压缩多个文件到 zip 格局 -zipfile 包实例

pip install zipfile
file=r’D:\test.zip’
out_path=r’D:\files’

遍历 files 文件夹下的文件，压缩发送

zip_1=zipfile.ZipFile(file,’w’)

for f in os.listdir(out_path):
    zip_1.write(os.path.join(out_path,f),f,zipfile.ZIP_DEFLATED)

zip_1.close()

python 批量删除文件名_Python 批量批改文件名

import os, re
while True:
keyword = input(“ 请输出你要删除的字符串:”)
if len(keyword)==0 or keyword.isspace():
print(“ 字符串不能为空!”)
else:
break
suffix = input(“ 须要筛选的文件名后缀 (Enter 代表所有):”)
fileNames = os.listdir() #获取当前目录下的所有文件
for file in fileNames:
check = os.path.join(os.path.abspath(‘.’),file)
if os.path.isfile(check):
if len(suffix)==0 or suffix.isspace():
if keyword in file:
print(file,” -> “,file.replace(keyword,”))
os.rename(file,file.replace(keyword,”))
else:

用正则表达式匹配后缀名

if re.match(‘.+?.’+suffix+’$’,file) != None and keyword in file:
print(file,” -> “,file.replace(keyword,”))
os.rename(file,file.replace(keyword,”))