Question

我需要一些帮助列出目录中的文件并使用Python读取每个文件。我知道如何使用shell命令执行此操作，但有一种Pythonic方法吗？

我想：

1。）列出目录中的所有文件。

2。）抓取最后10个修改/最新文件（最好使用通配符）

3.）阅读所有10个文件的每一行

使用shell命令我可以：

Linux_System# ls -ltr | tail -n 10 
-rw-rw-rw- 1 root root  999934 Jul 26 01:06 data_log.569
-rw-rw-rw- 1 root root  999960 Jul 26 02:05 data_log.570
-rw-rw-rw- 1 root root  999968 Jul 26 03:13 data_log.571
-rw-rw-rw- 1 root root  999741 Jul 26 04:20 data_log.572
-rw-rw-rw- 1 root root  999928 Jul 26 05:31 data_log.573
-rw-rw-rw- 1 root root  999942 Jul 26 06:45 data_log.574
-rw-rw-rw- 1 root root  999916 Jul 26 07:46 data_log.575
-rw-rw-rw- 1 root root  999862 Jul 26 08:59 data_log.576
-rw-rw-rw- 1 root root  999685 Jul 26 10:15 data_log.577
-rw-rw-rw- 1 root root  999633 Jul 26 11:26 data_log.578

Linux_System# cat data_log.{569..578}

使用glob我可以列出文件并打开一个特定文件，但不知道如何只列出最后10个修改过的文件并将通配符文件列表提供给open函数。

import os, fnmatch, glob

files = glob.glob("data_event_log.*")
files.sort(key=os.path.getmtime)
print("\n".join(files))

data_event_log.569
data_event_log.570
data_event_log.571
data_event_log.572
data_event_log.573
data_event_log.574
data_event_log.575
data_event_log.576
data_event_log.577
data_event_log.578

with open(data_event_log.560, 'r') as f:
    output_list = []
    for line in f.readlines():
        if line.startswith('Time'):
            lineRegex = re.compile(r'\d{4}-\d{2}-\d{2}')
            a = (lineRegex.findall(line))

Answer 1

看起来很像你几乎已经做了所有事情

import os.path, glob

files = glob.glob("data_event_log.*")
files.sort(key=os.path.getmtime)
latest=files[-10:] # last 10 entries
print("\n".join(latest))
lineRegex = re.compile(r'\d{4}-\d{2}-\d{2}')
for fn in latest:
    with open(fn) as f:
        for line in f:
            if line.startswith('Time'):          
                a = lineRegex.findall(line)

编辑：

特别是如果你有很多文件，那么

会有更好更简单的解决方案

import os.path, glob, heapq

files = glob.iglob("data_event_log.*")
latest=heapq.nlargest(10, files, key=os.path.getmtime) # last 10 entries
print("\n".join(latest))
lineRegex = re.compile(r'\d{4}-\d{2}-\d{2}')
for fn in latest:
    with open(fn) as f:
        for line in f:
            if line.startswith('Time'):          
                a = lineRegex.findall(line)

Answer 2

您正在寻找的是固定大小的排序缓冲区。 collections.deque这样做，虽然没有排序。所以，这是一个可以满足您需求的缓冲区，main向您展示如何使用它

import bisect
import glob
import operator
import os


class Buffer:
    def __init__(self, maxlen, minmax=1, key=None):
        if key is None: key = lambda x: x
        self.key = key
        self.maxlen = maxlen
        self.buffer = []
        self.keys = []
        self.minmax = minmax  # 1 to track max values, -1 to track min values

        # iterator variables
        self.curr = 0

    def __iter__(self): return self

    def __next__(self):
        if self.curr >= len(self.buffer): raise StopIteration
        self.curr += 1
        return self.buffer[self.curr-1]

    def insert(self, x):
        key = self.key(x)
        idx = bisect.bisect_left(self.keys, key)
        self.keys.insert(idx, key)
        self.buffer.insert(idx, x)
        if len(self.buffer) > self.maxlen:
            if self.minmax>0:
                self.buffer = self.buffer[-1 * self.maxlen :]
                self.keys = self.keys[-1 * self.maxlen :]
            elif self.minmax<0:
                self.buffer = self.buffer[: self.maxlen]
                self.keys = self.keys[: self.maxlen]


def main():
    dirpath = "/path/to/directory"
    modtime = lambda fpath: os.stat(fpath).st_mtime
    buffer = Buffer(10, 1, modtime)
    for fpath in glob.glob(os.path.join(dirpath, "*data_event_log.*")):
        buffer.insert(fpath)

    for fpath in buffer:
        # open the file path and print whatever

Answer 3

pythonic回答：

使用带有lambda函数的sorted()，然后使用列表切片来获得最早的10个或最新的10个或者你有什么。

from glob import glob
from os import stat

files = glob("*")
sorted_list = sorted(files, key=lambda x: stat(x).st_mtime)

truncated_list = sorted_list[-10:]

Python列出了最后10个修改过的文件并读取了所有10个文件的每一行

3 个答案: