OK, I did't know if you were able to re-organize the data. I know nothing
about AWS load balancers, but it's unfortunate that the data is laid out in a
way that makes dealing with it difficult.
But it sounds like you have worked it out. Best of luck.
Irv
> On Apr 9, 2017, at 2:21 PM, Kenton Brede <kbr...@gmail.com> wrote:
>
> Thanks for the response Irv. On one level I'm glad to know that someone
> more knowledgeable than myself sees this data structure as difficult. :) I
> was thinking it was an easy problem to solve. Unfortunately that is the
> structure I have to use.
>
> The data comes from pulling back tag information on AWS load balancers. So
> each list within the overall parent list are tags from one load balancer.
> Each dict pair inside the inner lists, is one tag. To simplify, below is a
> representation of one load balancer and two tags. The parent list contains
> multiple inner lists, each for a single load balancer.
>
> [
> [
> {
> "Value": "20176783-181622543367489",
> "Key": "Resource_group_id"
> },
> {
> "Value": "shibboleth-prd-alb",
> "Key": "Name"
> }
> ],
> ]
>
> I have an ugly solution of 'for loops' and 'if statements' that works,
> which uses a different path to get to the information. I was just hoping
> to be more direct by matching the 'Value' of 'Key: Resource_group_id' and
> pull back the 'Value' of 'Key: Name' and other tags as needed.
>
> Thanks,
>
> Kenton Brede
>
>
>
>
>
> On Sat, Apr 8, 2017 at 11:29 PM, Irv Kalb <i...@furrypants.com> wrote:
>
>> [ Sorry, forgot the important stuff! ]
>>
>> What you want to do is tricky because your data structure is difficult to
>> deal with. My guess is that it has to do with a misconception about how a
>> Python dictionary works. Yes, it is a series of key/value pairs, but not
>> the way you have it. It looks like you put together dictionaries where
>> each dictionary has a 'Value' and a 'Key'.
>>
>> Instead, _each_ item in a dictionary is a key value pair. The key is
>> typically a string, and the value is obviously some value associated with
>> that key. For example, if you have the ability to rebuild your data a
>> different way, it looks like it would be better to deal with it something
>> like this:
>>
>> aList = [
>> {'Name':'shibboleth-prd', 'Billing':'kmvu',
>> 'Resource_group_id': '20179204-181622543367489'},
>> {'Name':'shibboleth-tst', 'Resource_group_id':'20172857-
>> 152037106154311'}
>> ]
>>
>> This is a list of dictionaries. However, I'm not sure what you are trying
>> to do with this data. I'm guessing that you want to match a resource group
>> id, and if you find it, print the name and the billing info if they exist.
>> If so, you may want something like this (untested):
>>
>> def printInfo(thisGroupID):
>> for thisDict in aList: # loop through all dictionaries in the list
>> if thisGroupID == aList['Resource_group_id']:
>> if 'Name' in thisDict: # if thisDict has a key called 'Name'
>> print ('Name is', thisDict['Dict'])
>> if 'Billing' in thisDict: # if thisDict has a key called
>> 'Billing'
>> print ('Billing is', thisDict['Billing'])
>>
>> Hope this helps,
>>
>> Irv
>>> On Apr 8, 2017, at 9:04 PM, Irv Kalb <i...@furrypants.com> wrote:
>>>
>>> What you want to do is tricky because your data structure is difficult
>> to deal with. My guess is that it has to do with a misconception about how
>> a Python dictionary works. Yes, it is a series of key/value pairs, but not
>> the way you have it. It looks like you put together dictionaries where
>> each dictionary has a 'Value' and a 'Key'.
>>>
>>> Instead, _each_ item in a dictionary is a key value pair. The key is
>> typically a string, and the value is obviously some value associated with
>> that key. For example, if you have the ability to rebuild your data a
>> different way, it looks like it would be better to deal with it something
>> like this:
>>>
>>> aList = [
>>> {'Name':'shibboleth-prd', 'Billing':'kmvu',
>> 'Resource_group_id': '20179204-181622543367489'},
>>> {'Name':'shibboleth-tst', 'Resource_group_id':'20172857-
>> 152037106154311'}
>>> ]
>>>
>>> This is a list of dictionaries. However, I'm not sure what you are
>> trying to do with this data. I'm guessing that you want to match a
>> resource group id, and if you find it, print the name and the billing info
>> if they exist. If so, you may want something like this (untested):
>>>
>>> def printInfo(thisGroupID):
>>> for thisDict in aList: # loop through all dictionaries in the list
>>> if thisGroupID == aList['Resource_group_id']:
>>> if 'Name' in thisDict: # if thisDict has a key called 'Name'
>>> print ('Name is', thisDict['Dict'])
>>> if 'Billing' in thisDict: # if thisDict has a key called
>> 'Billing'
>>> print ('Billing is', thisDict['Billing'])
>>>
>>> Hope this helps,
>>>
>>> Irv
>>>
>>>
>>>
>>>> On Apr 8, 2017, at 5:55 PM, Kenton Brede <kbr...@gmail.com> wrote:
>>>>
>>>> This is an example of the data I'm working with. The key/value pairs
>> may
>>>> come in any order. There are some keys like the 'Resource_group_id' key
>> and
>>>> the 'Name' key which will always be present, but other lists may have
>>>> unique keys.
>>>>
>>>> alist = [[{u'Value': 'shibboleth-prd', u'Key': 'Name'}, {u'Value':
>> 'kvmu',
>>>> u'Key': 'Billing'},
>>>> {u'Value': '20179204-181622543367489', u'Key':
>>>> 'Resource_group_id'}],
>>>> [{u'Value': '20172857-152037106154311', u'Key':
>>>> 'Resource_group_id'},
>>>> {u'Value': 'shibboleth-tst', u'Key': 'Name'}]]
>>>>
>>>> What I want to do is something along the lines of:
>>>>
>>>> for a in alist:
>>>> if a['Resource_group_id'] == '01234829-2041523815431':
>>>> print the Value of 'Name'
>>>> print the Value of 'Billing'
>>>>
>>>> I've found I can do the following, to print the value of 'Name' but that
>>>> only works if the 'Resource_group_id' key is the first key in the list
>> and
>>>> the 'Name' key is in the second slot. If each list contained the same
>>>> keys, I could probably sort the keys and use [num] to pull back values,
>> but
>>>> they don't.
>>>>
>>>> for a in alist:
>>>> if a[0]['Key'] == 'Resource_group_id' and a[0]['Value'] ==
>>>> '20172857-152037106154311':
>>>> print a[1]['Value']
>>>>
>>>> There has to be a way to do this but I've been pounding away at this for
>>>> hours. Any help appreciated. I'm new to Python and not a programmer,
>> so
>>>> go easy on me. :)
>>>> --
>>>> https://mail.python.org/mailman/listinfo/python-list
>>>>
>>>
>>
>> --
>> https://mail.python.org/mailman/listinfo/python-list
>>
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