使用 python 脚本从 hdfs (hadoop) 目录获取文件列表

Question

How to get a list of files from hdfs (hadoop) directory using python script?

I have tried with following line:

dir = sc.textFile("hdfs://127.0.0.1:1900/directory").collect()

The directory have list of files "file1,file2,file3....fileN". By using the line i got all the content list only. But i need to get list of file names.

Can anyone please help me to find out this problem?

Thanks in advance.

Answer 1

Use subprocess

import subprocess
p = subprocess.Popen("hdfs dfs -ls <HDFS Location> |  awk '{print $8}'",
    shell=True,
    stdout=subprocess.PIPE,
    stderr=subprocess.STDOUT)

for line in p.stdout.readlines():
    print line

EDIT: Answer without python. The first option can be used to recursively print all the sub-directories as well. The last redirect statement can be omitted or changed based on your requirement.

hdfs dfs -ls -R <HDFS LOCATION> | awk '{print $8}' > output.txt
hdfs dfs -ls <HDFS LOCATION> | awk '{print $8}' > output.txt

EDIT: Correcting a missing quote in awk command.

Answer 2

import subprocess

path = "/data"
args = "hdfs dfs -ls "+path+" | awk '{print $8}'"
proc = subprocess.Popen(args, stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)

s_output, s_err = proc.communicate()
all_dart_dirs = s_output.split() #stores list of files and sub-directories in 'path'

Answer 3

Why not have the HDFS client do the hard work by using the -C flag instead of relying on awk or python to print the specific columns of interest?

ie Popen(['hdfs', 'dfs', '-ls', '-C', dirname])

Afterwards, split the output on new lines and then you will have your list of paths.

Here's an example along with logging and error handling (including for when the directory/file doesn't exist):

from subprocess import Popen, PIPE
import logging
logger = logging.getLogger(__name__)

FAILED_TO_LIST_DIRECTORY_MSG = 'No such file or directory'

class HdfsException(Exception):
    pass

def hdfs_ls(dirname):
    """Returns list of HDFS directory entries."""
    logger.info('Listing HDFS directory ' + dirname)
    proc = Popen(['hdfs', 'dfs', '-ls', '-C', dirname], stdout=PIPE, stderr=PIPE)
    (out, err) = proc.communicate()
    if out:
        logger.debug('stdout:\n' + out)
    if proc.returncode != 0:
        errmsg = 'Failed to list HDFS directory "' + dirname + '", return code ' + str(proc.returncode)
        logger.error(errmsg)
        logger.error(err)
        if not FAILED_TO_LIST_DIRECTORY_MSG in err:
            raise HdfsException(errmsg)
        return []
    elif err:
        logger.debug('stderr:\n' + err)
    return out.splitlines()

Answer 4

For python 3:

    from subprocess import Popen, PIPE
hdfs_path = '/path/to/the/designated/folder'
process = Popen(f'hdfs dfs -ls -h {hdfs_path}', shell=True, stdout=PIPE, stderr=PIPE)
std_out, std_err = process.communicate()
list_of_file_names = [fn.split(' ')[-1].split('/')[-1] for fn in std_out.decode().readlines()[1:]][:-1]
list_of_file_names_with_full_address = [fn.split(' ')[-1] for fn in std_out.decode().readlines()[1:]][:-1]

Answer 5

use the following:

hdfsdir = r"hdfs://VPS-DATA1:9000/dir/"
filepaths = [ line.rsplit(None,1)[-1] for line in sh.hdfs('dfs','-ls',hdfsdir).split('\n') if len(line.rsplit(None,1))][1:]

for path in filepaths:
    print(path)

Answer 6

to get list of hdfs files in a drectory:

hdfsdir = /path/to/hdfs/directory
    filelist = [ line.rsplit(None,1)[-1] for line in sh.hdfs('dfs','-ls',hdfsdir).split('\n') if len(line.rsplit(None,1))][1:]
    for path in filelist:
        #reading data file from HDFS 
        with hdfs.open(path, "r") as read_file:
            #do what u wanna do
            data = json.load(read_file)

this list is a list of all files in hdfs directory

filelist = [ line.rsplit(None,1)[-1] for line in sh.hdfs('dfs','-ls',hdfsdir).split('\n') if len(line.rsplit(None,1))][1:]

Answer 7

How to get a list of files from hdfs (hadoop) directory using python script?

I have tried with following line:

dir = sc.textFile("hdfs://127.0.0.1:1900/directory").collect()

The directory have list of files "file1,file2,file3....fileN". By using the line i got all the content list only. But i need to get list of file names.

Can anyone please help me to find out this problem?

Thanks in advance.