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How to make a dictionary using two list but i want both two as a value And Key will be name and sys-id?

I have two values number and sys_id. I have made seperate list for both of the values. How can save it in any other data structure like dictionary, or something else because i those list of number and sys_id are related. I am doing it in Python
Below Is the code what i have done
ticket_num.append(resp['result'][idx]['number'])
sys_id.append(resp['result'][idx]['sys_id']) ```
This is making two list one for ticket_num and sys_id. As Ticket number and sys_id are related for example ticket_num = ['INC00012','INC00013','INC00014' ] ,
sys_id = ['644323432sfasesdf213', '644323432dfgdfkdskrwwr', 'f283044423fdldsf09']
As this list are related like ticket_num[0] is directly link with sys_id[0]
So can i make a dictionary that contains ticket_num, sys_id directly without creating lists(for e.g. : {ticket_num : '...' , sys_id = '....' , ..... }

Use zip with dict
Ex:
ticket_num = ['INC00012','INC00013','INC00014' ]
sys_id = ['644323432sfasesdf213', '644323432dfgdfkdskrwwr', 'f283044423fdldsf09']
print(dict(zip(ticket_num, sys_id)))
Output:
{'INC00012': '644323432sfasesdf213',
'INC00013': '644323432dfgdfkdskrwwr',
'INC00014': 'f283044423fdldsf09'}

Welcome to Stackoverflow.
Do you actually need the lists of ticket numbers and IDs? If not that you could instead consider building the structure you need instead of the lists.
You don't say whether you want to be able to look up IDs from ticket numbers or vice versa. This solution allows you to do either:
idx_from_ticket = {}
ticket_from_idx = {}
# In the loop that produces the values, instead of the current appends ...
temp = resp['result'][idx]
idx = temp['sys_id]
number = temp['number']
idx_from_ticket[number] = idx
ticket_from_idx[idx] = number
The two dictionaries can then be used to correlate the IDs and ticket numbers. If you want to actually do something else then I hope this code gives you enough clues.
If you do already have the lists and want to retain them then the zip function is your friend.
idx_from_ticket = dict(zip(ticket_num, sys_id))
ticket_from_idx = dict(zip(sys_id, ticket_num))
zip, when called with two argument, yields a sequence of two-element tuples, which the
dict function assumes are key/value pairs.

How to dynamically append to array in dict?

This has taken me over a day of trial and error. I am trying to keep a dictionary of queries and their respective matches in a search. My problem is that there can be one or more matches. My current solution is:
match5[query_site] will already have the first match but if it finds another match it will append it using the code below.
temp5=[] #temporary variable to create array
if isinstance(match5[query_site],list): #check if already a list
temp5.extend(match5[query_site])
temp5.append(match_site)
else:
temp5.append(match5[query_site])
match5[query_site]=temp5 #add new location
That if statement is literally to prevent extend converting my str element into an array of letters. If I try to initialize the first match as a single element array I get None if I try to directly append. I feel like there should be a more pythonic method to achieve this without a temporary variable and conditional statement.
Update: Here is an example of my output when it works
5'flank: ['8_73793824', '6_133347883', '4_167491131', '18_535703', '14_48370386']
3'flank: X_11731384
There's 5 matches for my "5'flank" and only 1 match for my "3'flank".

So what about this:
if query_site not in match5: # here for the first time
match5[query_site] = [match_site]
elif isinstance(match5[query_site], str): # was already here, a single occurrence
match5[query_site] = [match5[query_site], match_site] # make it a list of strings
else: # already a list, so just append
match5[query_site].append(match_site)

I like using setdefault() for cases like this.
temp5 = match5.setdefault(query_site, [])
temp5.append(match_site)
It's sort of like get() in that it returns an existing value if the key exists but you can provide a default value. The difference is that if the key doesn't exist already setdefault inserts the default value into the dict.

This is all you need to do
if query_site not in match5:
match5[query_site] = []
temp5 = match5[query_site]
temp5.append(match_site)
You could also do
temp5 = match5.setdefault(query_site, [])
temp5.append(match_site)

Assuming match5 is a dictionary, what about this:
if query_site not in match5: # first match ever
match5[query_site] = [match_site]
else: # entry already there, just append
match5[query_site].append(temp5)
Make the entries of the dictionary to be always a list, and just append to it.

Django turn lists into one list

I want to put this data into one list so i can sort it by timestamp. I tried itertools chain but that didn't really work.
Thank you for your help :)
I'm very bad at making clear what i want to do so im sorry upfront if this takes some explaning.
If i try a chain i get the value back like this.
I want to display it on the html page like this :
date, name , rating, text (newline)
likes comments
Which would work the way i did it but if i want to sort it by time, it wouldn't work so i tried to think of a way to make it into a sortable list. Which can be displayed. Is that understandable ?
['Eva Simia', 'Peter Alexander', {'scale': 5, 'value': 5}, {'scale': 5, 'value': 5}, 1, 0, 1, 0]
it should look like this:
['Peter Alexander, scale:5, value:5, 1,0]
['Eva Simia, scale:5, value:5, 1,0]
for i in user:
name.append(i['name'])
for i in next_level:
rating_values.append(i['rating'])
for i in comment_values:
comments_count.append(i['count'])
for i in likes_values:
likes_count.append(i['count'])
for s in rating_values:
ratings.append(s['value'])
for s in date:
ratings.append(s['date'])
ab = itertools.chain([name], [rating_values],
[comments_count], [likes_values],
[comment_values], [date])
list(ab)

Updated after clarification:
The problem as I understand it:
You have a dataset that is split into several lists, one list per field.
Every list has the records in the same order. That is, user[x]'s rating value is necessarily rating_values[x].
You need to merge that information into a single list of composite items. You'd use zip() for that:
merged = zip(user, next_level, comment_values, likes_values, rating_values, date)
# merged is now [(user[0], next_level[0], comment_values[0], ...),
# (user[1], next_level[1], comment_values[1], ...),
# ...]
From there, you can simply sort your list using sorted():
result = sorted(merged, key=lambda i: (i[5], i[0]))
The key argument must be a function. It is given each item in the list once, and must return the key that will be used to compare items. Here, we build a short function on the fly, that returns the date and username, effectively telling things will be sorted, first by date and if dates are equal, then by username.
[Past answer about itertools.chain, before the clarification]
ab = list(itertools.chain(
(i['name'] for i in user),
(i['rating'] for i in next_level),
(i['count'] for i in comment_values),
(i['count'] for i in likes_values),
(i['value'] for i in rating_values),
(i['date'] for i in date),
))
The point of using itertools.chain is usually to avoid needless copies and intermediary objects. To do that, you want to pass it iterators.
chain will return an iterator that will iterate through each of the given iterators, one at a time, moving to the next iterator when current stops.
Note every iterator has to be wrapped in parentheses, else python will complain. Do not make it square brackets, at it would cause an intermediary list to be built.

You can join list by simply using +.
l = name + rating_values + comments_count + ...

date, rating_values,likes_values,comment_values,next_level,user = (list(t) for t in zip(*sorted(zip(date, rating_values,likes_values,comment_values,next_level,user))))

Python: How to traverse a List[Dict{List[Dict{}]}]

I was just wondering if there is a simple way to do this. I have a particular structure that is parsed from a file and the output is a list of a dict of a list of a dict. Currently, I just have a bit of code that looks something like this:
for i in xrange(len(data)):
for j, k in data[i].iteritems():
for l in xrange(len(data[i]['data'])):
for m, n in data[i]['data'][l].iteritems():
dostuff()
I just wanted to know if there was a function that would traverse a structure and internally figure out whether each entry was a list or a dict and if it is a dict, traverse into that dict and so on. I've only been using Python for about a month or so, so I am by no means an expert or even an intermediate user of the language. Thanks in advance for the answers.
EDIT: Even if it's possible to simplify my code at all, it would help.

You never need to iterate through xrange(len(data)). You iterate either through data (for a list) or data.items() (or values()) (for a dict).
Your code should look like this:
for elem in data:
for val in elem.itervalues():
for item in val['data']:
which is quite a bit shorter.

Will, if you're looking to decend an arbitrary structure of array/hash thingies then you can create a function to do that based on the type() function.
def traverse_it(it):
if (isinstance(it, list)):
for item in it:
traverse_it(item)
elif (isinstance(it, dict)):
for key in it.keys():
traverse_it(it[key])
else:
do_something_with_real_value(it)
Note that the average object oriented guru will tell you not to do this, and instead create a class tree where one is based on an array, another on a dict and then have a single function to process each with the same function name (ie, a virtual function) and to call that within each class function. IE, if/else trees based on types are "bad". Functions that can be called on an object to deal with its contents in its own way "good".

I think this is what you're trying to do. There is no need to use xrange() to pull out the index from the list since for iterates over each value of the list. In my example below d1 is therefore a reference to the current data[i].
for d1 in data: # iterate over outer list, d1 is a dictionary
for x in d1: # iterate over keys in d1 (the x var is unused)
for d2 in d1['data']: # iterate over the list
# iterate over (key,value) pairs in inner most dict
for k,v in d2.iteritems():
dostuff()
You're also using the name l twice (intentionally or not), but beware of how the scoping works.

well, question is quite old. however, out of my curiosity, I would like to respond to your question for much better answer which I just tried.
Suppose, dictionary looks like: dict1 = { 'a':5,'b': [1,2,{'a':100,'b':100}], 'dict 2' : {'a':3,'b':5}}
Solution:
dict1 = { 'a':5,'b': [1,2,{'a':100,'b':100}], 'dict 2' : {'a':3,'b':5}}
def recurse(dict):
if type(dict) == type({}):
for key in dict:
recurse(dict[key])
elif type(dict) == type([]):
for element in dict:
if type(element) == type({}):
recurse(element)
else:
print element
else:
print dict
recurse(dict1)

How to compare an element of a tuple (int) to determine if it exists in a list

I have the two following lists:
# List of tuples representing the index of resources and their unique properties
# Format of (ID,Name,Prefix)
resource_types=[('0','Group','0'),('1','User','1'),('2','Filter','2'),('3','Agent','3'),('4','Asset','4'),('5','Rule','5'),('6','KBase','6'),('7','Case','7'),('8','Note','8'),('9','Report','9'),('10','ArchivedReport',':'),('11','Scheduled Task',';'),('12','Profile','<'),('13','User Shared Accessible Group','='),('14','User Accessible Group','>'),('15','Database Table Schema','?'),('16','Unassigned Resources Group','#'),('17','File','A'),('18','Snapshot','B'),('19','Data Monitor','C'),('20','Viewer Configuration','D'),('21','Instrument','E'),('22','Dashboard','F'),('23','Destination','G'),('24','Active List','H'),('25','Virtual Root','I'),('26','Vulnerability','J'),('27','Search Group','K'),('28','Pattern','L'),('29','Zone','M'),('30','Asset Range','N'),('31','Asset Category','O'),('32','Partition','P'),('33','Active Channel','Q'),('34','Stage','R'),('35','Customer','S'),('36','Field','T'),('37','Field Set','U'),('38','Scanned Report','V'),('39','Location','W'),('40','Network','X'),('41','Focused Report','Y'),('42','Escalation Level','Z'),('43','Query','['),('44','Report Template ','\\'),('45','Session List',']'),('46','Trend','^'),('47','Package','_'),('48','RESERVED','`'),('49','PROJECT_TEMPLATE','a'),('50','Attachments','b'),('51','Query Viewer','c'),('52','Use Case','d'),('53','Integration Configuration','e'),('54','Integration Command f'),('55','Integration Target','g'),('56','Actor','h'),('57','Category Model','i'),('58','Permission','j')]
# This is a list of resource ID's that we do not want to reference directly, ever.
unwanted_resource_types=[0,1,3,10,11,12,13,14,15,16,18,20,21,23,25,27,28,32,35,38,41,47,48,49,50,57,58]
I'm attempting to compare the two in order to build a third list containing the 'Name' of each unique resource type that currently exists in unwanted_resource_types. e.g. The final result list should be:
result = ['Group','User','Agent','ArchivedReport','ScheduledTask','...','...']
I've tried the following that (I thought) should work:
result = []
for res in resource_types:
if res[0] in unwanted_resource_types:
result.append(res[1])
and when that failed to populate result I also tried:
result = []
for res in resource_types:
for type in unwanted_resource_types:
if res[0] == type:
result.append(res[1])
also to no avail. Is there something i'm missing? I believe this would be the right place to perform list comprehension, but that's still in my grey basket of understanding fully (The Python docs are a bit too succinct for me in this case).
I'm also open to completely rethinking this problem, but I do need to retain the list of tuples as it's used elsewhere in the script. Thank you for any assistance you may provide.

Your resource types are using strings, and your unwanted resources are using ints, so you'll need to do some conversion to make it work.
Try this:
result = []
for res in resource_types:
if int(res[0]) in unwanted_resource_types:
result.append(res[1])
or using a list comprehension:
result = [item[1] for item in resource_types if int(item[0]) in unwanted_resource_types]

The numbers in resource_types are numbers contained within strings, whereas the numbers in unwanted_resource_types are plain numbers, so your comparison is failing. This should work:
result = []
for res in resource_types:
if int( res[0] ) in unwanted_resource_types:
result.append(res[1])

The problem is that your triples contain strings and your unwanted resources contain numbers, change the data to
resource_types=[(0,'Group','0'), ...
or use int() to convert the strings to ints before comparison, and it should work. Your result can be computed with a list comprehension as in
result=[rt[1] for rt in resource_types if int(rt[0]) in unwanted_resource_types]
If you change ('0', ...) into (0, ... you can leave out the int() call.
Additionally, you may change the unwanted_resource_types variable into a set, like
unwanted_resource_types=set([0,1,3, ... ])
to improve speed (if speed is an issue, else it's unimportant).

The one-liner:
result = map(lambda x: dict(map(lambda a: (int(a[0]), a[1]), resource_types))[x], unwanted_resource_types)
without any explicit loop does the job.
Ok - you don't want to use this in production code - but it's fun. ;-)
Comment:
The inner dict(map(lambda a: (int(a[0]), a[1]), resource_types)) creates a dictionary from the input data:
{0: 'Group', 1: 'User', 2: 'Filter', 3: 'Agent', ...
The outer map chooses the names from the dictionary.