查看File对象的.read()方法:

http://docs.python.org/2/tutorial/inputoutput.html#methods-of-file-objects

你可以这样做:

concat = ""
for file in files:
    concat += open(file).read()

或者更“优雅”的python方式:

concat = ''.join([open(f).read() for f in files])

根据这篇文章，http://www.skymind.com/~ocrow/python_string/也将是最快的。

我不知道什么叫优雅，但这招管用:

    import glob
    import os
    for f in glob.glob("file*.txt"):
         os.system("cat "+f+" >> OutFile.txt")

如果文件不是很大:

with open('newfile.txt','wb') as newf:
    for filename in list_of_files:
        with open(filename,'rb') as hf:
            newf.write(hf.read())
            # newf.write('\n\n\n')   if you want to introduce
            # some blank lines between the contents of the copied files

如果文件太大，不能完全读取并保存在RAM中，则算法必须稍微不同，以固定长度的块读取循环中复制的每个文件，例如使用read(10000)。

查看File对象的.read()方法:

http://docs.python.org/2/tutorial/inputoutput.html#methods-of-file-objects

你可以这样做:

concat = ""
for file in files:
    concat += open(file).read()

或者更“优雅”的python方式:

concat = ''.join([open(f).read() for f in files])

根据这篇文章，http://www.skymind.com/~ocrow/python_string/也将是最快的。

def concatFiles():
    path = 'input/'
    files = os.listdir(path)
    for idx, infile in enumerate(files):
        print ("File #" + str(idx) + "  " + infile)
    concat = ''.join([open(path + f).read() for f in files])
    with open("output_concatFile.txt", "w") as fo:
        fo.write(path + concat)

if __name__ == "__main__":
    concatFiles()

  import os
  files=os.listdir()
  print(files)
  print('#',tuple(files))
  name=input('Enter the inclusive file name: ')
  exten=input('Enter the type(extension): ')
  filename=name+'.'+exten
  output_file=open(filename,'w+')
  for i in files:
    print(i)
    j=files.index(i)
    f_j=open(i,'r')
    print(f_j.read())
    for x in f_j:
      outfile.write(x)