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Python3-嵌套dict到JSON

python-3.x python

oliverbj · 技术社区 · 7 年前

我想把多个 .txt 文件到“类似表”的数据(包括列和行)。每个 .txt文件 文件应视为新列。

考虑以下内容 .txt文件

File1.txt

Hi there
How are you doing?

What is your name?

File2.txt

Hi
Great!

Oliver, what's yours?

我创建了一个简单的方法,它接受file和integer(来自另一个方法的文件号):

def txtFileToJson(text_file, column):

    data = defaultdict(list)

    i = int(1)
    with open(text_file) as f:

        data[column].append(column)

        for line in f:
            i = i + 1
            for line in re.split(r'[\n\r]+', line):
                data[column] = line

    with open("output.txt", 'a+') as f:
        f.write(json.dumps(data))

所以上面的方法将运行两次(每个文件一次,并附加数据)。

这是 output.txt 运行脚本后的文件:

{"1": "What is your name?"}{"2": "Oliver, what's yours?"}

如您所见,我只能让它为我的每个文件创建一个新的,然后添加整行。

[{
   "1": [{
       "1": "Hi there",
       "2": "How are you doing?",
       "3": "\n"
       "4": "What is your name?"
   },
   "2": [{
       "1": "Hi"
       "2": "Great!",
       "3": "\n",
       "4": "Oliver, what's yours?"
   },
}]

好吧,我玩了一会儿,离得更近了:

myDict = {str(column): []}
i = int(1)
with open(text_file) as f:
    for line in f:
        # data[column].append(column)
        match = re.split(r'[\n\r]+', line)
        if match:
            myDict[str(column)].append({str(i): line})
            i = i + 1

with open(out_file, 'a+') as f:
    f.write(json.dumps(myDict[str(column)]))

下面是输出:

[{"1": "Hi there\n"}, {"2": "How are you doing?\n"}, {"3": "\n"}, {"4": "What is your name?"}]
[{"1": "Hi\n"}, {"2": "Great!\n"}, {"3": "\n"}, {"4": "Oliver, what's yours?"}]

但如您所见,现在我有多个JSON根元素。

解决方案

多亏了乔尼弗里斯,我做到了:

 data = defaultdict(list)

 for path in images.values():
     column = column + 1

     data[str(column)] = txtFileToJson(path, column)

 saveJsonFile(path, data)

def saveJsonFile(text_file, data):

    basename = os.path.splitext(os.path.basename(text_file))
    dir_name = os.path.dirname(text_file) + "/"
    text_file = dir_name + basename[0] + "1.txt"

    out_file = dir_name + 'table_data.txt'

    with open(out_file, 'a+') as f:
        f.write(json.dumps(data))

0 回复 | 直到 6 年前

jonyfries 7 年前

您正在函数本身内创建一个新字典。所以每次你传入一个文本文件时,它都会创建一个新的字典。

def txtFileToJson(text_file, column):
    myDict = {str(column): []}
    i = int(1)
    with open(text_file) as f:
        for line in f:
            # data[column].append(column)
            match = re.split(r'[\n\r]+', line)
            if match:
                myDict[str(column)].append({str(i): line})
                i = i + 1

    with open(out_file, 'a+') as f:
        f.write(json.dumps(myDict[str(column)]))

    return myDict

data = defaultdict(list)

data["1"] = txtFileToJson(text_file, column)
data["2"] = txtFileToJson(other_text_file, other_column)

igrinis 7 年前

def read(text_file):
    data, i = {}, 0
    with open(text_file) as f:
        for line in f:
            i = i + 1
            data['row_%d'%i] = line.rstrip('\n')
    return data

res = {}
for i, fname in enumerate([r'File1.txt', r'File2.txt']):
    res[i] = read(fname)
with open(out_file, 'w') as f:
    json.dump(res, f)

Booboo 7 年前

首先,如果我理解你试图得到一个字典字典的输出,那么让我观察一下,我所理解的你想要的输出似乎是把整个东西都包含在一个列表中,而且,你在字典中有不平衡的开和闭列表方括号,我将忽略这一点,因为我将忽略随附列表。

#!python3

import json
import re

def processTxtFile(text_file, n, data):
    d = {}
    with open(text_file) as f:
        i = 0
        for line in f:
            for line in re.split(r'[\n\r]+', line):
                i = i + 1
                d[str(i)] = line
    data[str(n)] = d


data = dict()
processTxtFile('File1.txt', 1, data)
processTxtFile('File2.txt', 2, data)
with open("output.txt", 'wt') as f:
    f.write(json.dumps(data))

如果你真的需要嵌套在一个列表中,那么你需要替换一个字典

data[str(n)] = d

data[str(n)] = [d]