Django - Get query parameters and put it into list of dictionary

Question

I have a GET query the format is like this:

?title1=a&test1=b&title2=a&test2=b&..&titleN=a&testN=b

The view contains just the code above

def index(request):
    # This is my view
    print request.GET.items():

This is the result returned I got from the view when I run the query :

{ 'title1':'a', 'test1':'b', 'title2':'a', 'test2':'b','titleN':'a', 'testN':'b' }

I want to create a new list of dictionary.

[
    {
        'title1' : 'a' , 
        'test1':'b'
    }, 
    {
        'title2' : 'a' ,
         'test2':'b'
    },
    {
        'titleN' : 'a',
        'testN':'b'
    }
]

As you can notice I want to arrange all the data by number and N here is a Number

The question is how to make a list like I mentioned above , the only convention is the dictionary contains keys with the same last number .

And then I want the list become like this :

[
    {
        'title' : 'a' , 
        'test':'b'
    }, 
    {
        'title' : 'a' ,
         'test':'b'
    },
    {
        'title' : 'a',
        'test':'b'
    }
]

Yes, it ends with a number and it contain a word like the example mentioned above the only important thing it's the number. — Selim Ajimi, Jan 29 '17 at 21:22

DYZ · Answer 1 · 2017-01-29T21:45:43.763

2

One way to solve the problem is to use regulare expressions for extracting the number and then itertools for grouping by that number.

import re, itertools

# This is the grouper function; used first for sorting, 
# then for actual grouping

def grouper(key):
    return re.findall(r'\d+$',key[0])[0]

# Sort the dictionary by the number
sorted_dict = sorted(original_dict.items(), key=grouper)

# Group the sorted dictionary items and convert the groups into dicts
result = [dict(vals) for _,vals in (itertools.groupby(sorted_dict, key=grouper))]
#[{'title1': 'a', 'test1': 'b'}, 
# {'title11': 'a', 'test11': 'b'}, 
# {'title2': 'a', 'test2': 'b'}]

edited Jan 29 '17 at 21:45

answered Jan 29 '17 at 20:55

DYZ

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in the result what does the dict(vals) contain ? – Selim Ajimi Jan 29 '17 at 21:07
1

It converts a list of tuples into a dictionary. – DYZ Jan 29 '17 at 21:10
Sorry i'm new to python what is the role of _ ? – Selim Ajimi Jan 29 '17 at 21:12
2

http://stackoverflow.com/questions/5893163/what-is-the-purpose-of-the-single-underscore-variable-in-python – DYZ Jan 29 '17 at 21:14
2

@SelimAjimi It is convention in python to use `_` for variable which won't be used – Moinuddin Quadri Jan 29 '17 at 21:14
I got an error in result = [dict(vals) for _,vals in (itertools.groupby(sorted_dict), key=grouper)] On key=grouper and also x is not defined in the grouper function – Selim Ajimi Jan 29 '17 at 21:43
There were two minor typos. – DYZ Jan 29 '17 at 21:47
The solution works fine However if the query not passed without any ending number it ends with an error is there any solution . This is only for security reason. – Selim Ajimi Jan 29 '17 at 22:01
Shouldn't that have been checked at the time of costructing the original dictionary. – DYZ Jan 29 '17 at 22:03

score 2 · Accepted Answer · answered Feb 02 '17 at 21:57

This one doesn't depend on the order of the items

import itertools
import re
num = re.compile('[0-9]+')
alpha = re.compile('[a-z]+')
query = { 'title1':'a', 'test1':'b', 'title2':'a', 'test2':'b','titleN':'a', 'testN':'b' }

result = [dict((num.sub('', w[0]), w[1]) for z, w in v) for i, v in itertools.groupby(((alpha.sub('', k), (k, v)) for k, v in query.items()), lambda x: x[0])]

# result:
# [{'test': 'b', 'title': 'a'},
#  {'test': 'b', 'title': 'a'},
#  {'testN': 'b', 'titleN': 'a'}]

Moinuddin Quadri · Answer 3 · 2017-01-29T21:00:38.510

Assuming the test and title will be holding integer suffix in continuous series of numbers starting from 1, below is a possible approach via using list comprehension to achieve this:

>>> my_dict = { 'title1':'a', 'test1':'b', 'title2':'a', 'test2':'b','title3':'a', 'test3':'b' }
>>> n = int(len(my_dict)/2)

>>> [{'title{}'.format(i), my_dict['title{}'.format(i)], 'test{}'.format(i), my_dict['test{}'.format(i)],} for i in range(1, n+1)]
[{'test1', 'b', 'a', 'title1'}, {'b', 'test2', 'a', 'title2'}, {'b', 'a', 'title3', 'test3'}]

I will suggest that instead of passing params to your apis as:

title1=a&test1=b&title2=a&test2=b

you should be passing it as:

title=a1,a2,a3&test=b1,b2,b3

The reason is: all the values should be mapped with single title and test parameter.

In this case, your code should be:

title = request.GET['title']
test = request.GET['test']

my_list = [{'title{}'.format(i): ti, 'test{}'.format(i): te } for i, (ti, te) in enumerate(zip(title, test))]

where value hold by my_list will be:

[{'test0': 'b1', 'title0': 'a1'}, {'test1': 'b2', 'title1': 'a2'}, {'title2': 'a3', 'test2': 'b3'}]

@DYZ Check the start of my answer. I already mentioned that I am assuming that it won't happen that way. If it does, my answer won't work ;) — Moinuddin Quadri, Jan 29 '17 at 21:17

score 1 · Answer 4 · answered Jan 29 '17 at 20:54

Not so familiar with django, so there could be a more efficient way to do this, but... You could do something like the following:

 query_result = request.GET.items()
 list_of_dicts = []
 for i in range(1, N):
     d ={'title'+str(i):d['title'+str(i)],'test'+str(i):d['test'+str(i)]}
     list_of_dicts.append(d)

score 1 · Answer 5 · answered Jan 29 '17 at 21:00

Cumbersome but still oneliner:

result = sorted([{k:v,"title"+k[4:]:d["title"+k[4:]]} for k,v in d.items() if k.startswith("test")],key=lambda x : sorted(x.keys()))

create the sub-dicts with testN & corresponding titleN keys+values (shortcut: 4 is the length of test string)
sort the list according to the sorted list of keys

result:

[{'test1': 'b', 'title1': 'a'}, {'test2': 'b', 'title2': 'a'}, {'titleN': 'a', 'testN': 'b'}]

Django - Get query parameters and put it into list of dictionary

5 Answers5