在单个表达式中读取文件，正确关闭它们

Question

我想读取 python 中文件列表的内容。我的第一个想法是

contents = [open(f).read() for f in files]

但这会使文件保持打开状态，直到对象被垃圾回收，并显示 ResourceWarning。

关闭文件需要多重理解：

fds = [open(f) for f in files]
contents = [fd.read() for fd in fds]
[fd.close() for fd in fds]

...这是不自然的。

或一个循环：

contents = []
for f in files:
    with open(f) as fd:
        contents.append(f.read())

...非常冗长，读起来很长。

还有其他选择吗？

Answer 1

你可以使用一个函数：

def read_contents(filename):
    with open(filename) as file:
        return file.read()

contents = [read_contents(f) for f in files]

Answer 2

您可以使用 pathlib。

from pathlib import Path
contents_text = [Path(f).read_text() for f in files]
contents_bytes = [Path(f).read_bytes() for f in files]

里面只有：

class Path:

    # ....

    def read_bytes(self):
        """
        Open the file in bytes mode, read it, and close the file.
        """
        with self.open(mode='rb') as f:
            return f.read()

    def read_text(self, encoding=None, errors=None):
        """
        Open the file in text mode, read it, and close the file.
        """
        with self.open(mode='r', encoding=encoding, errors=errors) as f:
            return f.read()

Answer 3

您可以使用 ExitStack 上下文管理器。您的用例与文档中显示的示例略有不同。

from contextlib import ExitStack

with ExitStack() as es:
    contents = [es.enter_context(open(f)).read() for f in files]

在单个表达式中读取文件，正确关闭它们

Read files in a single expression, properly closing them

python

expression

list-comprehension

file

contextmanager