在多个文件的所有行上进行迭代的最pythonic方法是什么? [英] What's the most pythonic way to iterate over all the lines of multiple files?

问题描述

我想将许多文件都当作一个文件来对待.用[发生器]/不将整个文件读入内存的[文件名] => [文件对象] => [行]的正确pythonic方法是什么?

I want to treat many files as if they were all one file. What's the proper pythonic way to take [filenames] => [file objects] => [lines] with generators/not reading an entire file into memory?

我们都知道打开文件的正确方法:

We all know the proper way to open a file:

with open("auth.log", "rb") as f:
    print sum(f.readlines())

我们知道将多个迭代器/生成器链接为一个长链的正确方法:

And we know the correct way to link several iterators/generators into one long one:

>>> list(itertools.chain(range(3), range(3)))
[0, 1, 2, 0, 1, 2]

但是如何将多个文件链接在一起并保留上下文管理器?

but how do I link multiple files together and preserve the context managers?

with open("auth.log", "rb") as f0:
    with open("auth.log.1", "rb") as f1:
        for line in itertools.chain(f0, f1):
            do_stuff_with(line)

    # f1 is now closed
# f0 is now closed
# gross

我可以忽略上下文管理器并执行类似的操作，但是感觉不正确:

I could ignore the context managers and do something like this, but it doesn't feel right:

files = itertools.chain(*(open(f, "rb") for f in file_names))
for line in files:
    do_stuff_with(line)

或者是异步IO-PEP 3156 的用途，我只是必须稍后再等待优雅的语法?

Or is this kind of what Async IO - PEP 3156 is for and I'll just have to wait for the elegant syntax later?

推荐答案

总是有 fileinput .

for line in fileinput.input(filenames):
    ...

阅读源文件，看来fileinput.FileInput可以可用作上下文管理器 ¹ .要解决此问题，您可以使用 contextlib.closing ，因为FileInput实例具有健全的已实现的close方法:

Reading the source however, it appears that fileinput.FileInput can't be used as a context manager¹. To fix that, you could use contextlib.closing since FileInput instances have a sanely implemented close method:

from contextlib import closing
with closing(fileinput.input(filenames)) as line_iter:
    for line in line_iter:
        ...

---

上下文管理器的另一种选择是编写一个简单的函数，遍历文件并随行生成行:

---

An alternative with the context manager, is to write a simple function looping over the files and yielding lines as you go:

def fileinput(files):
    for f in files:
        with open(f,'r') as fin:
            for line in fin:
                yield line

这里不需要真正的itertools.chain恕我直言...神奇之处在于yield语句，该语句用于将普通函数转换为异常懒惰的生成器.

No real need for itertools.chain here IMHO ... The magic here is in the yield statement which is used to transform an ordinary function into a fantastically lazy generator.

¹ 顺便说一句，从python3.2开始，fileinput.FileInput 被实现为上下文管理器，其功能与我们之前使用contextlib所做的完全一样.现在我们的示例变为:

¹As an aside, starting with python3.2, fileinput.FileInput is implemented as a context manager which does exactly what we did before with contextlib. Now our example becomes:

# Python 3.2+ version
with fileinput.input(filenames) as line_iter:
    for line in line_iter:
        ...

尽管其他示例也适用于python3.2 +.

although the other example will work on python3.2+ as well.

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