Question

这非常基本：几个月前我成功使用python在单个目录中解析json文件。但现在我无法弄清楚我是如何调整它的（队友想出了代码），以便我可以将数据转换为更有用的csv格式。

目前，在使用Python Launcher或Terminal时，我正处于zilch状态。

解析器的样子：

import codecs
import json
import os
import sys
 try:
    import unicodecsv as csv
    except ImportError:
    import csv

    OUTPUT_FILE = 'output.csv'


def process_file(infile, writer):
    print('Processing file: %s' % infile)
    with codecs.open(infile, encoding='utf-8') as infile:
        data = json.load(infile)
            for item in data:
                _id = item['id']
                description =  item['description']
                for gov in item['source']:
                    gov_id = gov['name']
                    for source in item['secondarySource']:
                        source_id = source['sourceId']
                            name = source['name']
                            party = source['party']
                            writer.writerow([_id, description, gov_id, source_id, name, party])


def process_files_in_directory(directory, outfile):
     with codecs.open(outfile, 'w') as outfile:
         writer = csv.writer(outfile)
         writer.writerow(["id", "description", "branch", "sourceID", "name", "party"])
         for f in os.listdir(path):
             if f.endswith('.json'):
                 process_file(f, writer)

USAGE = """
Usage:

            python json_parser.py <source_directory> [<output_file>]

            Where source_directory is path to directory with input JSON files.
            output_file is optional -- defaults to %s
            File names must end with .json
            """ % OUTPUT_FILE


if __name__=='__main__':
    try:
        directory = sys.argv[1]
    except IndexError:
        print(USAGE)
        sys.exit(0)
    if len(sys.argv) > 2:
        outfile = sys.argv[2]
    else:
        outfile = OUTPUT_FILE
    process_files_in_directory(directory, outfile)

Answer 1

您的脚本存在一些格式问题。我不确定它们是否与问题有关...这是你脚本的新版本。基本思想有效，但您可能希望格式化CSV输出以使其更具可读性。为了证明这是有效的，我从命令行运行：

python stackoverflow\junk.py stackoverflow\mydir

stackoverflow\mydir有两个文件：one.json和two.json。

以下代码包含上述评论中的修复

import codecs
import json
import os
import sys

try:
    import unicodecsv as csv
except ImportError:
    import csv

OUTPUT_FILE = 'output.csv'


def process_file(infile, writer):
    print('Processing file: %s' % infile)
    with codecs.open(infile, encoding='utf-8') as infile:
        data = json.load(infile)
        for item in data:
            _id = item['id']
            description = item['description']
            for gov in item['source']:
                gov_id = gov['name']
                for source in item['secondarySource']:
                    source_id = source['sourceId']
                    name = source['name']
                    party = source['party']
                    writer.writerow([_id, description, gov_id, source_id, name, party])


def process_files_in_directory(directory, outfile):
    with codecs.open(outfile, 'w') as outfile:
        writer = csv.writer(outfile)
        writer.writerow(["id", "description", "branch", "sourceID", "name", "party"])
        for f in os.listdir(directory):
            if f.endswith('.json'):
                process_file(os.path.join(directory, f), writer)

USAGE = """
Usage:

            python json_parser.py <source_directory> [<output_file>]

            Where source_directory is path to directory with input JSON files.
            output_file is optional -- defaults to %s
            File names must end with .json
            """ % OUTPUT_FILE

if __name__ == '__main__':
    try:
        directory = sys.argv[1]
    except IndexError:
        print(USAGE)
        sys.exit(0)
    if len(sys.argv) > 2:
        outfile = sys.argv[2]
    else:
        outfile = OUTPUT_FILE
    process_files_in_directory(directory, outfile)

没有从python中的json-parser接收输出

1 个答案: