Better way to create json file from multiple lists?

Question

I have three lists like below and I want to create a JSON file from them:

devices = ['iphone', 'ipad', 'ipod', 'watch'],
cities = ['NY', 'SFO', 'LA', 'NJ'],
companies = ['Apple', 'Samsung', 'Walmart']

I have done like below.

First manually create a dictionary:

data = {
    'devices': ['iphone', 'ipad', 'ipod', 'watch'],
    'cities': ['NY', 'SFO', 'LA', 'NJ'],
    'companies': ['Apple', 'Samsung', 'Walmart']
}

Then convert it to JSON format like this:

import json

with open('abc.json', 'w') as outfile:
    json.dump(data, outfile, indent=4)

Is there a better way of doing this when we have more number of lists.

Ideally if I have N number of lists, I want to create a JSON formatted file a minimal amount of manual work.

What is the manual work here? Where do those list come from? — Selcuk
– Selcuk, Commented May 25, 2018 at 0:17
@Selcuk I will be provided with these lists in a python file — User12345
– User12345, Commented May 25, 2018 at 0:18
You can import that Python file (say, import input_lists if the file name is input_lists.py) and get a list of attribute names using dir(input_lists). Then it becomes a matter of iterating through that list and eliminating built-in attributes (those that start and end with a __). — Selcuk
– Selcuk, Commented May 25, 2018 at 0:22

martineau · Accepted Answer · 2018-05-25 01:33:22Z

2

Your question doesn't show getting the lists from an external source like another .py file, so here's how to do it given their variable names when they've been defined in-line as shown in it:

import json

devices = ['iphone', 'ipad', 'ipod', 'watch']
cities = ['NY', 'SFO', 'LA', 'NJ']
companies = ['Apple', 'Samsung', 'Walmart']

lists = ['devices', 'cities', 'companies']

data = {listname: globals()[listname] for listname in lists}
with open('abc.json', 'w') as outfile:
    json.dump(data, outfile, indent=4)

Contents of the abc.json file it creates:

{
    "devices": [
        "iphone",
        "ipad",
        "ipod",
        "watch"
    ],
    "cities": [
        "NY",
        "SFO",
        "LA",
        "NJ"
    ],
    "companies": [
        "Apple",
        "Samsung",
        "Walmart"
    ]
}

answered May 25, 2018 at 1:33

martineau

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2 Comments

User12345 Over a year ago

I am getting syntaxError invalid syntax near data = {listname: globals()[listname] for listname in lists} in python 2.6 From Python 2.7 and above there is no error

martineau Over a year ago

@User12345: If you wanted something for other than the current version of Python (which is v3.6.5 I believe), you should have mentioned it in and tagged your question for that version. It may be due to the use of a dictionary display (commonly also called a dictionary comprehension), which is what's going on in that line)—and I'm not sure when they were added but they've been in 2.7.x for a while. Regardless, you should be able to replace it with a few lines of code that uses a regular for loop to populate (build) that dictionary, which will give you something that will work an all of them.

Worm · Accepted Answer · 2018-05-25 01:08:16Z

This method will work for any number of lists providing they have the same format as the ones provided in your question. Hope this helps.

# define the list vars
devices = ['iphone', 'ipad', 'ipod', 'watch'],
cities = ['NY', 'SFO', 'LA', 'NJ'],
companies = ['Apple', 'Samsung', 'Walmart']

# get the variables into a big list
v = locals()['In'][2]

output = {}

#break down the lists and turn them into dict entries
v1 = v.split(',\n')
for each in v1:
    #print(each)
    name = each.split(' = ')[0]
    data = each.split(' = ')[1]
    data = data[2:-2]
    datalist = data.split("', '")
    output[name] = datalist

#show the output
output

#export as JSON
import json

with open('C:\\abc.json', 'w') as outfile:
    json.dump(output, outfile, indent=4)

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