How to iterate through a nested list in python?

Question

I want to iterate through a list that has a lot of dictionaries inside it. The json response I'm trying to iterate looks something like this:

user 1 JSON response:
[
 {
 "id": "333",
 "name": "hello"
 },
 {
 "id": "999",
 "name": "hi"
 },
 {
 "id": "666",
 "name": "abc"
 },
]

user 2 JSON response:
[
 {
 "id": "555",
 "name": "hello"
 },
 {
 "id": "1001",
 "name": "hi"
 },
 {
 "id": "26236",
 "name": "abc"
 },
]

This is not the actual JSON response but it is structured the same way. What I'm trying to do is to find a specific id and store it in a variable. The JSON response I'm trying to iterate is not organized and changes every time depending on the user. So I need to find the specific id which would be easy but there are many dictionaries inside the list. I tried iterating like this:

    for guild_info in guilds:
        for guild_ids in guild_info:

This returns the first dictionary which is id: 333. For example, I want to find the value 666 and store it in a variable. How would I do that?

You say you need to "find a specific id and store it in a variable". Presumably you already know the id, which is what you're searching for. Are you looking to assign a variable to the corresponding name? Also, do you need to loop through multiple user lists, or only one at a time? (In other words, each user is a list of dictionaries. Do you in fact have a list of lists of dictionaries?) — CrazyChucky
– CrazyChucky, Commented Apr 11, 2021 at 16:03

tlouarn · Accepted Answer · 2021-04-11 15:56:44Z

1

What you have is a list of dictionaries.

When you run for guild_info in guilds: you will iterate through dictionaries, so here each guild_info will be a dictionary. Therefore simply take the key id like so: guild_info['id'].

If what you want to do is find the name corresponding to a specific id, you can use list comprehension and take its first element, as follows:

name = [x['name'] for x in guilds if x['id'] == '666'][0]

answered Apr 11, 2021 at 15:56

tlouarn

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3 Comments

CrazyChucky Over a year ago

Only downside of this list comprehension is that it will keep searching through all the user's guilds even after it's found the match. Probably nothing to worry about though, performance-wise, unless you've got a whole bunch of users and/or guilds to process.

CrazyChucky Over a year ago

It will also throw an IndexError in the event that there was no match.

CrazyChucky Over a year ago

Based on this answer, your solution can be slightly modified to avoid both issues: name = next((x['name'] for x in guilds if x['id'] == '666'), None). Or if it's certain there will always be a match, that can be simplified to name = next(x['name'] for x in guilds if x['id'] == '666').

CrazyChucky · Accepted Answer · 2021-04-11 16:24:01Z

Here's a function that will search only until it finds the matching id and then return, which avoids checking further entries unnecessarily.

def get_name_for_id(user, id_to_find):
    # user is a list, and each guild in it is a dictionary.
    for guild in user:
        if guild['id'] == id_to_find:
            # Once the matching id is found, we're done.
            return guild['name']

    # If the loop completes without returning, then there was no match.
    return None

user = [
    {
        "id": "333",
        "name": "hello"
    },
    {
        "id": "999",
        "name": "hi"
    },
    {
        "id": "666",
        "name": "abc"
    },
]

name = get_name_for_id(user, '666')
print(name)
name2 = get_name_for_id(user, '10000')
print(name2)

Output:

abc
None

Sanket Wagh · Accepted Answer · 2021-04-11 16:00:23Z

0

This will create a loop which will iterate to the list of dictionaries.If you are looking for simple approach

for every_dictionary in List_of_dictionary:
    for every_dictionary_item in every_dictionary.keys():
        print(every_dictionary[every_dictionary_item])

answered Apr 11, 2021 at 16:00

Sanket Wagh

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