blocks|key|427080|text|UNIX命令有什么问题？(假设您不是在Windows上工作)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|427081|ls+%7C+xargs+cat+%7C+tee+output.txt可以完成这项工作(如果需要，可以通过子进程从python中调用它)|offset|length|style|CODE|427082|entityMap^0|0|0|V|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

What's wrong with UNIX commands ? (given you're not working on Windows) : 

<code>ls | xargs cat | tee output.txt</code> does the job ( you can call it from python with subprocess if you want)

blocks|key|427101|text|查看文件对象的.read()方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|427102|http://docs.python.org/2/tutorial/inputoutput.html#methods-of-file-objects|offset|length|427103|你可以这样做：|427104|concat+=+""
for+file+in+files:
++++concat+%2B=+open(file).read()|code-block|syntax|javascript|427105|或者一种更“优雅”的python方式：|427106|concat+=+''.join([open(f).read()+for+f+in+files])|427107|根据这篇文章，http://www.skymind.com/~ocrow/python_string/也是最快的。|427108|entityMap|0|LINK|mutability|MUTABLE|url|1|http://www.skymind.com/~ocrow/python_string/^0|0|0|22|0|0|0|0|0|0|7|18|1|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@]|9|@$D|13|E|14|1|15]]|A|$]]|$1|F|3|G|5|6|7|16|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|17|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|19|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@$D|1B|E|1C|1|1D]]|A|$]]|$1|S|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|C]]|Z|$5|V|W|X|A|$Y|10]]]]

Check out the .read() method of the File object:

<a href="http://docs.python.org/2/tutorial/inputoutput.html#methods-of-file-objects" rel="nofollow">http://docs.python.org/2/tutorial/inputoutput.html#methods-of-file-objects</a>

You could do something like:

<pre><code>concat = ""
for file in files:
 concat += open(file).read()
</code></pre>

or a more 'elegant' python-way:

<pre><code>concat = ''.join([open(f).read() for f in files])
</code></pre>

which, according to this article: <a href="http://www.skymind.com/~ocrow/python_string/" rel="nofollow">http://www.skymind.com/~ocrow/python_string/</a> would also be the fastest.

blocks|key|427125|text|如果文件不是巨大的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|427126|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|code-block|syntax|javascript|427127|如果文件太大而无法完全读取并保存在内存中，则算法必须稍有不同，以便通过固定长度的块在循环中读取每个要复制的文件，例如使用read(10000)。|offset|length|style|CODE|427128|entityMap^0|0|0|1O|B|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If the files are not gigantic:

<pre><code>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
</code></pre>

If the files are too big to be entirely read and held in RAM, the algorithm must be a little different to read each file to be copied in a loop by chunks of fixed length, using <code>read(10000)</code> for example.

blocks|key|208981|text|def+concatFiles():
++++path+=+'input/'
++++files+=+os.listdir(path)
++++for+idx,+infile+in+enumerate(files):
++++++++print+("File+#"+%2B+str(idx)+%2B+"++"+%2B+infile)
++++concat+=+''.join([open(path+%2B+f).read()+for+f+in+files])
++++with+open("output_concatFile.txt",+"w")+as+fo:
++++++++fo.write(path+%2B+concat)

if+__name__+==+"__main__":
++++concatFiles()|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|208982|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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()
</code></pre>

blocks|key|208996|text|我不知道是否优雅，但这是可行的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|208997|++++import+glob
++++import+os
++++for+f+in+glob.glob("file*.txt"):
+++++++++os.system("cat+"%2Bf%2B"+>>+OutFile.txt")|code-block|syntax|javascript|208998|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I don't know about elegance, but this works:

<pre><code> import glob
 import os
 for f in glob.glob("file*.txt"):
 os.system("cat "+f+" &gt;&gt; OutFile.txt")
</code></pre>

blocks|key|427195|text|使用shutil.copyfileobj。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|427196|它会自动逐块读取输入文件，这会更高效地读取输入文件，即使某些输入文件太大而无法放入内存，它也会正常工作：|427197|import+shutil

with+open('output_file.txt','wb')+as+wfd:
++++for+f+in+['seg1.txt','seg2.txt','seg3.txt']:
++++++++with+open(f,'rb')+as+fd:
++++++++++++shutil.copyfileobj(fd,+wfd)|code-block|syntax|javascript|427198|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/shutil.html#shutil.copyfileobj^0|2|I|2|I|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Use <a href="https://docs.python.org/3/library/shutil.html#shutil.copyfileobj" rel="nofollow noreferrer"><code>shutil.copyfileobj</code></a>.
It automatically reads the input files chunk by chunk for you, which is more more efficient and reading the input files in and will work even if some of the input files are too large to fit into memory:
<pre><code>import shutil

with open('output_file.txt','wb') as wfd:
 for f in ['seg1.txt','seg2.txt','seg3.txt']:
 with open(f,'rb') as fd:
 shutil.copyfileobj(fd, wfd)
</code></pre>

blocks|key|209074|text|@inspectorG4dget答案的替代方案(2016年3月29日的最佳答案)。我测试了3个436MB的文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|209075|@inspectorG4dget解决方案:+162秒|209076|解决方案如下:+125秒|209077|from+subprocess+import+Popen
filenames+=+['file1.txt',+'file2.txt',+'file3.txt']
fbatch+=+open('batch.bat','w')
str+="type+"
for+f+in+filenames:
++++str%2B=+f+%2B+"+"
fbatch.write(str+%2B+"+>+file4results.txt")
fbatch.close()
p+=+Popen("batch.bat",+cwd=r"Drive:\Path\to\folder")
stdout,+stderr+=+p.communicate()|code-block|syntax|javascript|209078|这个想法是创建一个批处理文件并执行它，利用“旧的好技术”。它是半蟒蛇，但运行速度更快。适用于windows。|209079|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

An alternative to @inspectorG4dget answer (best answer to date 29-03-2016). I tested with 3 files of 436MB. 

@inspectorG4dget solution: 162 seconds

The following solution : 125 seconds

<pre><code>from subprocess import Popen
filenames = ['file1.txt', 'file2.txt', 'file3.txt']
fbatch = open('batch.bat','w')
str ="type "
for f in filenames:
 str+= f + " "
fbatch.write(str + " &gt; file4results.txt")
fbatch.close()
p = Popen("batch.bat", cwd=r"Drive:\Path\to\folder")
stdout, stderr = p.communicate()
</code></pre>

The idea is to create a batch file and execute it, taking advantage of "old good technology". Its semi-python but works faster. Works for windows.

blocks|key|209128|text|如果目录中有很多文件，那么glob2可能是生成文件名列表的更好选择，而不是手动编写它们。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|209129|import+glob2

filenames+=+glob2.glob('*.txt')++#+list+of+all+.txt+files+in+the+directory

with+open('outfile.txt',+'w')+as+f:
++++for+file+in+filenames:
++++++++with+open(file)+as+infile:
++++++++++++f.write(infile.read()%2B'\n')|code-block|syntax|javascript|209130|entityMap^0|D|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If you have a lot of files in the directory then <code>glob2</code> might be a better option to generate a list of filenames rather than writing them by hand.

<pre><code>import glob2

filenames = glob2.glob('*.txt') # list of all .txt files in the directory

with open('outfile.txt', 'w') as f:
 for file in filenames:
 with open(file) as infile:
 f.write(infile.read()+'\n')
</code></pre>

blocks|key|427331|text|outfile.write(infile.read())+#+time:+2.1085190773010254s
shutil.copyfileobj(fd,+wfd,+1024*1024*10)+#+time:+0.60599684715271s|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|427332|一个简单的基准测试表明，shutil的性能更好。|unstyled|427333|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>outfile.write(infile.read()) # time: 2.1085190773010254s
shutil.copyfileobj(fd, wfd, 1024*1024*10) # time: 0.60599684715271s
</code></pre>

A simple benchmark shows that the shutil performs better.

blocks|key|427337|text|++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%2B'.'%2Bexten
++output_file=open(filename,'w%2B')
++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)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|427338|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> 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)
</code></pre>

I have a list of 20 file names, like <code>['file1.txt', 'file2.txt', ...]</code>. I want to write a Python script to concatenate these files into a new file. I could open each file by <code>f = open(...)</code>, read line by line by calling <code>f.readline()</code>, and write each line into that new file. It doesn't seem very &quot;elegant&quot; to me, especially the part where I have to read/write line by line.
Is there a more &quot;elegant&quot; way to do this in Python?

How do I concatenate text files in Python?

Windows

Python

我有一个包含20个文件名的列表，比如['file1.txt', 'file2.txt', ...]。我想写一个Python脚本，将这些文件连接成一个新文件。我可以通过f = open(...)打开每个文件，通过调用f.readline()逐行读取，并将每一行写入新文件。这对我来说似乎不是很“优雅”，特别是我必须逐行读/...

问如何在Python中连接文本文件？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在Python中连接文本文件？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在Python中连接文本文件？
EN