Python—并行读取多个大文件并分别生成它们
我有多个大文件,需要一行一行地生成循环式文件。类似于以下伪代码:Python—并行读取多个大文件并分别生成它们,python,file-io,bigdata,Python,File Io,Bigdata,我有多个大文件,需要一行一行地生成循环式文件。类似于以下伪代码: def get(self): with open(file_list, "r") as files: for file in files: yield file.readline() 我该如何做到这一点?使用上下文管理器会很棘手(或需要一些附加的库),但如果没有上下文管理器,应该不会很困难open() def get(files_list): f
def get(self):
with open(file_list, "r") as files:
for file in files:
yield file.readline()
我该如何做到这一点?使用上下文管理器会很棘手(或需要一些附加的库),但如果没有上下文管理器,应该不会很困难open()
def get(files_list):
file_handles = [open(f, 'r') for f in files_list]
while file_handles:
for fd in file_handles:
line = fd.readline()
if line:
yield line
else:
file_handles.remove(fd)
我假设您希望继续运行,直到从每个文件中读取每一行,较短的文件在到达EOF时会掉落。itertools
有几个配方,其中有一个非常简洁的循环配方。我还可以使用多个文件上下文管理器:
from itertools import cycle, islice
from contextlib import ExitStack
# https://docs.python.org/3.8/library/itertools.html#itertools-recipes
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
num_active = len(iterables)
nexts = cycle(iter(it).__next__ for it in iterables)
while num_active:
try:
for next in nexts:
yield next()
except StopIteration:
# Remove the iterator we just exhausted from the cycle.
num_active -= 1
nexts = cycle(islice(nexts, num_active))
...
def get(self):
with open(files_list) as fl:
filenames = [x.strip() for x in fl]
with ExitStack() as stack:
files = [stack.enter_context(open(fname)) for fname in filenames]
yield from roundrobin(*files)
尽管如此,也许最好的设计是使用控制反转,并将文件对象序列作为参数提供给.get
,因此调用代码应注意使用退出堆栈:
class Foo:
...
def get(self, files):
yield from roundrobin(*files)
# calling code:
foo = Foo() # or however it is initialized
with open(files_list) as fl:
filenames = [x.strip() for x in fl]
with ExitStack() as stack:
files = [stack.enter_context(open(fname)) for fname in filenames]
for line in foo.get(files):
do_something_with_line(line)
发现一个可能的重复项:
文件列表
是一个文件名字符串的python列表。open
不接受列表作为输入,是吗?它采用类似于对象的str
,字节
,os.path
。我错过什么了吗?不,你是对的。然而,我的代码只是伪代码。我假设存在这样一个“开放”命令。