How can I get dictionary of lists in one line

Question

I have below lines of code. I want to get the stack name in info (marked as stack). I am still in learning phase. I only want first 2 loops (for and if) and in the info I need a separate loop where I loop through dictionary whose values are in a list. I compare these values to volume id (the volume.id that I get from 1st for loop) and when I get a match, I put the key of that dictionary to stack. What I tried is below but this is not right because I am unnecessary looping and I am getting long list of repeated volumes from the first loop.
Can anyone please let me know how can I loop through the elements of list values of dictionary in one line under stack? Please let me know if I need to give more information.

for volume in volumes:
        if volume.snapshot_id == snapshot_id:
            for key,value in tempDict.items():
                for elem in value:
                    if volume.id == elem:
                       stack = key
                       info={'type':'volume',
                             'id':volume.id,
                             'display_name':volume.display_name,
                             'snapshot_id':volume.snapshot_id,
                             'created_at':volume.created_at,
                             'snapshots':[], 
                             'stack': stack}# This is where I need to work
                       vol_list.append(info)}

I also tried below for stack but I also need to add if volume.id == elem and if match then key goes in stack)

stack: {k:[elem for elem in v] for k,v in tempDict.iteritems()}}

My dictionary looks like below

{
    "deployCI2": [
        "094fd196-20f0-4e8d-b946-f74a56d2f319",
        "6a1ce382-98c6-4058-a929-95a7d2415fd0",
        "156a1409-c89d-48fe-af6a-cad12d985b7a",
        "7376f485-6f70-4e35-a16e-b3136523206b"
    ],
    "deployCI3": [
        "c8fff661-4482-4908-b722-4fac0227a8b0",
        "929cf1fa-3fa6-4f95-8464-d58e5490f4cf",
        "4c7a7755-8576-43d0-9e7a-ae7319f40f6d",
        "a8460ed5-913b-4670-a1d4-43af508c8187"
    ],
    "Lomka": [
        "17efb1f1-3bf2-44f0-97a1-ebca4b6a2a30"
    ],
    "HeenaStackXYZ": [],
    "deployCI4": [
        "9f8ffa3c-460d-43a9-8113-58e891340e1b",
        "6e535e92-4da2-4228-a6ab-c8fc8d31adcd",
        "8e26a35e-7fb9-43b3-8026-d1283f7b678c",
        "f40e5c29-b4df-4cfb-9d7f-3bcc9c4dcf9f"
    ],
    "HeenaStackXYZ-VM1-ne4rpss4bnft": [],
    "ci_a3077f8b": [
        "0956d6a3-87f4-43c3-a2f7-8a963b2d00c3"
    ]
}

What should output look like? Actually It is going to html. My major concern is getting the stack name. What I have done

1. Fetch the stack names and the ID of the volumes associated with the stack. You see that in my dictionary. 
2. I compare all the volumes in my environment and see if any volume id matches with the stack's volume ID. If yes, I need to grab that stack name and put that in the info section (That's where stack comes)

A first thing: don't ever call a variable `dict` in Python, this a reserved name for dictionary type. Now, how do you initialise your dictionary in the first place? — Arthur Spoon, Aug 04 '17 at 09:07
In fact it's not a built-in function, [`dict()`](https://docs.python.org/2/library/stdtypes.html#dict) is a built-in class — Arount, Aug 04 '17 at 09:10
@GhilasBELHADJ True, it's not reserved (otherwise it would be impossible to use it as a variable name), but it's not a function: it's a type (aka class). — PM 2Ring, Aug 04 '17 at 09:10
@UbdusSamad `dict` is not reserved, you can override it at will. — Ghilas BELHADJ, Aug 04 '17 at 09:11
Reserved words are things like `for`, `in`, `while`, etc. Try using them as variable names. ;) — PM 2Ring, Aug 04 '17 at 09:12
Think about reversing a dictionary. See this: https://stackoverflow.com/questions/35945473/how-to-reverse-a-dictionary-in-python — damisan, Aug 04 '17 at 09:15
Can you override dict? I never knew! Thanks, I'll dig through it! — Ubdus Samad, Aug 04 '17 at 09:15
@HeenashreeKhandelwal there's a closing bracket at the end of your last line of code that doesn't match any opening bracket. Can you fix that? Can you also tell us what the ouput you are giving comes from in your code? (like, what object you are printing to get this output) — Arthur Spoon, Aug 04 '17 at 09:17
But anyway, this discussion about built-in names isn't helping the OP solve her main problem... — PM 2Ring, Aug 04 '17 at 09:17
@Arount: Since you're nitpicking: `dict()` isn't a built-in class. `dict()` is just an empty dict. `dict` is the built-in class ;) — Eric Duminil, Aug 04 '17 at 09:18
@HeenashreeKhandelwal You should explain that data a bit. Are the stack names the keys like `"deployCI2"` and `"Lomka"`? And are the volume IDs the strings like "094fd196-20f0-4e8d-b946-f74a56d2f319"? — PM 2Ring, Aug 04 '17 at 09:19
@PM2Ring, yes, the key is the name of the stack like deployCI2 and the value is the volume IDs — Heenashree Khandelwal, Aug 04 '17 at 09:21
@Eric Duminil I was wondering that as well but felt it would be stupid to ask! — Ubdus Samad, Aug 04 '17 at 09:23
@HeenashreeKhandelwal if you post the output directly and not a description of it, it will be easier for us to help you. — Ghilas BELHADJ, Aug 04 '17 at 09:29
@GhilasBELHADJ, I have put code at http://paste.openstack.org/show/617518/ — Heenashree Khandelwal, Aug 04 '17 at 09:44
@HeenashreeKhandelwal it does not help a lot. I'll ask the question differently. what is the value you're waiting for in your `output` variable? — Ghilas BELHADJ, Aug 04 '17 at 10:16

score 0 · Accepted Answer · answered Aug 04 '17 at 12:58

0

I tried reversing the dictionary and for the stack I added - 'stack':{value for key, value in reversedDict.items() if volume.id == key}} This worked for me.

answered Aug 04 '17 at 12:58

Heenashree Khandelwal

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How can I get dictionary of lists in one line

1 Answers1