First, access to files and directories
<1>, in python, can accept path "/", "\", where the image of the prime metaphor ( 'posix') and downwards ( 'nt'). However, due to the "\" is used as an escape character in python, the use of "\" is to be written as "\\" in the path.
Therefore, in python, the following two versions are acceptable.
('posix') "c:/test/my doc"
('nt') "c:\\test\\my doc"
<Two>, the writing python absolute path, to be noted that the letter behind colon. For example: "c: \\ test / my doc"
<Three>, in python, the current directory, the parent directory, the root directory, subdirectory denote the following example:
File os.listdir () method returns the specified folder contains the file or folder name list. The list in alphabetical order . It does not include the '.' And '..' even if it is in the same folder. Supports only be used under Unix, Windows.
1, the current directory: the os.listdir ( ".") F1 = Open ( 'xuefeng.txt', 'W')
2, the parent directory: the os.listdir ( "..") F1 = Open ( '../ xuefeng.txt', 'W')
3, a root directory written: the os.listdir ( '/') = F1 Open ( '/ xuefeng.txt', 'W')
4, the root directory writing two: the os.listdir ( '\\') = F1 Open ( 'xuefeng.txt \\', 'W')
5, subdirectories: the os.listdir ( 'mytext') = F1 Open ( 'mytext / xuefeng.txt', 'W') # current directory is assumed to have access to a subdirectory named mytext
os.path.isdir () and os.path.isfile () use errors:
#判断是否存在以及为txt文本，后续处理
for file in os.listdir(path):
if os.path.isfile(file) and os.path.splitext(file)[1] == '.txt':
结果if一直为假，
需要了解：
os.path.isfile(path)中path是路径.....
而上面传的是一个文件名.
解决方法：
>>> os.path.isfile(os.path.join(path,files[1]))
TrueSecond, manipulating files and directories
# 查看当前目录的绝对路径:
>>> os.path.abspath('.')
'/Users/michael'
# 在某个目录下创建一个新目录，首先把新目录的完整路径表示出来:
>>> os.path.join('/Users/michael', 'testdir')
'/Users/michael/testdir'
# 然后创建一个目录:
>>> os.mkdir('/Users/michael/testdir')
# 删掉一个目录:
>>> os.rmdir('/Users/michael/testdir')
#拆分路径
>>> os.path.split('/Users/michael/testdir/file.txt')
('/Users/michael/testdir', 'file.txt')
#获得扩展名
>>> os.path.splitext('/path/to/file.txt')
('/path/to/file', '.txt')
以上只对字符串操作，并不要求含有真实文件
以上只对字符串操作，并不要求含有真实文件
以上只对字符串操作，并不要求含有真实文件
以下要求含有test.txt文件
# 对文件重命名:
>>> os.rename('test.txt', 'test.py')
# 删掉文件:
>>> os.remove('test.py')
#复制可以使用 shutil模块提供了copyfile()的函数
#过滤文件（如列出所有目录）
>>> [x for x in os.listdir('.') if os.path.isdir(x)]
['.lein', '.local', '.m2', '.npm', '.ssh', '.Trash', '.vim', 'Applications', 'Desktop', ...]
#过滤文件，如列出.py文件
>>> [x for x in os.listdir('.') if os.path.isfile(x) and os.path.splitext(x)[1]=='.py']
['apis.py', 'config.py', 'models.py', 'pymonitor.py', 'test_db.py', 'urls.py', 'wsgiapp.py']python Traverse Folder
import os
import os.path
rootdir = “d:\data” # 指明被遍历的文件夹
for parent,dirnames,filenames in os.walk(rootdir):
#三个参数：分别返回1.父目录 2.所有文件夹名字（不含路径） 3.所有文件名字
for dirname in dirnames: #输出文件夹信息
print "parent is:" + parent
print "dirname is" + dirname
for filename in filenames: #输出文件信息
print "parent is": + parent
print "filename is:" + filename
print "the full name of the file is:" + os.path.join(parent,filename) #输出文件路径信息
#windows下为：d:\data\query_text\EL_00154Look under the current directory and all subdirectories of the current directory file name contains the specified string, and prints out the relative path
#listdir方法
def searchfile(path,k):
import os
filelist=os.listdir(path)
for x in filelist:
if k in x:
print(os.path.join(path,x))#这边是把x添加到路径后面，不能直接打印x
path= input('Directory: ')
k= input('search: ')
searchfile(path,k)
#遍历方法1
[filename for t in os.walk(search_dir) for filename in t[2] if search_str in os.path.splitext(filename)[0]]
#遍历方法2
import os, logging
def search(s):
rootdir = '.' # 指明被遍历的文件夹
for parent,dirnames,filenames in os.walk(rootdir): #三个参数：分别返回1.父目录 2.所有文件夹名字（不含路径） 3.所有文件名字
for filename in filenames: #输出文件信息
#print "filename is:" + filename
if filename.find(s) != -1:
print "the full path of the file is:" + os.path.abspath(os.path.join(parent,filename)) #输出文件路径信息
if __name__ == '__main__':
search('input')#使用调试logging方法
import os, pdb, logging
logging.basicConfig(level=logging.DEBUG)
def search_dir(path, L):
current_dir = os.listdir(path)
pdb.set_trace()
for n in current_dir:
pdb.set_trace()
new_path = os.path.join(path, n)
if os.path.isfile(new_path): # 需传入路径而非仅仅文件名，否则是FALSE
logging.debug('%s is a file.' % n)
L.append(new_path)
else:
search_dir(new_path, L)
return L
def search(s):
L = search_dir('.', [])
# pdb.set_trace()
for file in L:
pdb.set_trace()
if file.find(s) != -1:
logging.info('找到包含%s的文件路径：%s' % (s, os.path.abspath(file)))
# os.path.abspath(url) 并非返回url真正完整的绝对路径，只是将当前目录与url进行join操作
# 例如，当前目录为 D:/workplace
# url是 test.txt，实际是在 ./aaa/test.txt
# 但该函数返回的是 D:/workplace/test.txt
if __name__ == '__main__':
search('test')Realization mailbox format
^表示行的开头，^\d表示必须以数字开头。
$表示行的结束，\d$表示必须以数字结束。
import re
def regex(s, t):
# re_mail=re.compile(t)
for x in s:
m = re.match(t, x)
n = m.groups()
if n[2]=='com':
print('%s is mail' % x)
else:
print('%s is failed' % x)
ke = r'^([0-9a-zA-Z\_\.]*)\@([a-zA-Z\_\.]*)\.([0-9a-zA-Z\_\.]{3})$'
s1 = ['[email protected]', '[email protected]']
regex(s1, ke)
三、遇到BUG（IOError: [Errno 22] invalid mode ('r') or filename: 'E:\\python_script\x08.txt'）
#错误行
with open('E:\python_script\10.txt','r') as f：This error occurs when using built-in function file () or open () time.
First, because the file does not open mode, the second is the name of the file in question.
The former, then only need to pay attention to whether the file is readable or writable on it.
The latter is associated with the file path problems, and needs r or R escape before the file name, such as: file. (R "e: \ Test.txt", 'r') or a backslash \ into two a, such as file ( "e: \\ Test.txt", 'r').
