Extracting elements from two lists of dicts if other elements match using comprehension












1














How can I iterate over two lists of dicts, match dicts between the lists by a key, and if there is a match then append a specific key from each dict into a key value pair in a new dictionary. Let me clarify with an example:



l1 = [{'id': 52, 'email': 'someemail@yahoo.com', 'anotherfield': 'some value'},

       .....

      {'id': 98, 'email': 'anotheremail@yahoo.com', 'anotherfield': 'another value'}]



l2 = [{'id': 93, 'email': 'someemail@yahoo.com', 'another key': 'seventeen'},

       .....

      {'id': 101, 'email': 'anotheremail@yahoo.com', 'another key': 'twenty'}]





# match the 'email' keys between each list, and if match, create k, v pair from id's



desired_output = {'52': 93.....'98': 101}


I can achieve this quite easily by simply iterating over each list as follows:



lookup = dict()

for l in l1:

    for p in l2:

        if l['email']==p['email']:

            lookup[l['id']]=p['id']

            break


However this is a bit clunky and I'd prefer some kind of comprehension. My attempt:



lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}





python dictionary list-comprehension







share|improve this question




















  • 1




    Your for-loop is fine. But if you are going to use a comprehension, you'd need {... for l in l1 for p in l2 if l['email']==p['email']} note how it aligns with your for-loop. However, you won't be able to break, making this less ideal. In any event, there are better algorithms than a brute-force check, so, I'd rather change that than convert from a perfectly fine for-loop to comprehension.
    – juanpa.arrivillaga
    Nov 21 '18 at 14:03












  • Interesting, could you provide the full line with your suggested edit?
    – Laurie Bamber
    Nov 21 '18 at 14:04






  • 1




    {l['id']:p['id'] for l in l1 for p in l2 if l['email']==p['email']}
    – juanpa.arrivillaga
    Nov 21 '18 at 14:05












  • This worked really nicely, thanks for the help, and I'll take a look into other methods that might be more robust. Thanks again.
    – Laurie Bamber
    Nov 21 '18 at 14:09






  • 1




    The order of elements in lists is consistency? If not, algorithm is not working. if k['email'] == v['email'] will works then k[N] == v[N]. You need consist of lists and when run your solution. Or, you can create one dict from your lists: d = defaultdict(list), d[email].append(id)
    – Maxim Panfilov
    Nov 21 '18 at 14:12
















1














How can I iterate over two lists of dicts, match dicts between the lists by a key, and if there is a match then append a specific key from each dict into a key value pair in a new dictionary. Let me clarify with an example:



l1 = [{'id': 52, 'email': 'someemail@yahoo.com', 'anotherfield': 'some value'},

       .....

      {'id': 98, 'email': 'anotheremail@yahoo.com', 'anotherfield': 'another value'}]



l2 = [{'id': 93, 'email': 'someemail@yahoo.com', 'another key': 'seventeen'},

       .....

      {'id': 101, 'email': 'anotheremail@yahoo.com', 'another key': 'twenty'}]





# match the 'email' keys between each list, and if match, create k, v pair from id's



desired_output = {'52': 93.....'98': 101}


I can achieve this quite easily by simply iterating over each list as follows:



lookup = dict()

for l in l1:

    for p in l2:

        if l['email']==p['email']:

            lookup[l['id']]=p['id']

            break


However this is a bit clunky and I'd prefer some kind of comprehension. My attempt:



lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}





python dictionary list-comprehension







share|improve this question




















  • 1




    Your for-loop is fine. But if you are going to use a comprehension, you'd need {... for l in l1 for p in l2 if l['email']==p['email']} note how it aligns with your for-loop. However, you won't be able to break, making this less ideal. In any event, there are better algorithms than a brute-force check, so, I'd rather change that than convert from a perfectly fine for-loop to comprehension.
    – juanpa.arrivillaga
    Nov 21 '18 at 14:03












  • Interesting, could you provide the full line with your suggested edit?
    – Laurie Bamber
    Nov 21 '18 at 14:04






  • 1




    {l['id']:p['id'] for l in l1 for p in l2 if l['email']==p['email']}
    – juanpa.arrivillaga
    Nov 21 '18 at 14:05












  • This worked really nicely, thanks for the help, and I'll take a look into other methods that might be more robust. Thanks again.
    – Laurie Bamber
    Nov 21 '18 at 14:09






  • 1




    The order of elements in lists is consistency? If not, algorithm is not working. if k['email'] == v['email'] will works then k[N] == v[N]. You need consist of lists and when run your solution. Or, you can create one dict from your lists: d = defaultdict(list), d[email].append(id)
    – Maxim Panfilov
    Nov 21 '18 at 14:12














1












1








1







How can I iterate over two lists of dicts, match dicts between the lists by a key, and if there is a match then append a specific key from each dict into a key value pair in a new dictionary. Let me clarify with an example:



l1 = [{'id': 52, 'email': 'someemail@yahoo.com', 'anotherfield': 'some value'},

       .....

      {'id': 98, 'email': 'anotheremail@yahoo.com', 'anotherfield': 'another value'}]



l2 = [{'id': 93, 'email': 'someemail@yahoo.com', 'another key': 'seventeen'},

       .....

      {'id': 101, 'email': 'anotheremail@yahoo.com', 'another key': 'twenty'}]





# match the 'email' keys between each list, and if match, create k, v pair from id's



desired_output = {'52': 93.....'98': 101}


I can achieve this quite easily by simply iterating over each list as follows:



lookup = dict()

for l in l1:

    for p in l2:

        if l['email']==p['email']:

            lookup[l['id']]=p['id']

            break


However this is a bit clunky and I'd prefer some kind of comprehension. My attempt:



lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}





python dictionary list-comprehension







share|improve this question















How can I iterate over two lists of dicts, match dicts between the lists by a key, and if there is a match then append a specific key from each dict into a key value pair in a new dictionary. Let me clarify with an example:



l1 = [{'id': 52, 'email': 'someemail@yahoo.com', 'anotherfield': 'some value'},

       .....

      {'id': 98, 'email': 'anotheremail@yahoo.com', 'anotherfield': 'another value'}]



l2 = [{'id': 93, 'email': 'someemail@yahoo.com', 'another key': 'seventeen'},

       .....

      {'id': 101, 'email': 'anotheremail@yahoo.com', 'another key': 'twenty'}]





# match the 'email' keys between each list, and if match, create k, v pair from id's



desired_output = {'52': 93.....'98': 101}


I can achieve this quite easily by simply iterating over each list as follows:



lookup = dict()

for l in l1:

    for p in l2:

        if l['email']==p['email']:

            lookup[l['id']]=p['id']

            break


However this is a bit clunky and I'd prefer some kind of comprehension. My attempt:



lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}





python dictionary list-comprehension




python dictionary list-comprehension






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 '18 at 13:59

























asked Nov 21 '18 at 13:52









Laurie Bamber

431114




431114








  • 1




    Your for-loop is fine. But if you are going to use a comprehension, you'd need {... for l in l1 for p in l2 if l['email']==p['email']} note how it aligns with your for-loop. However, you won't be able to break, making this less ideal. In any event, there are better algorithms than a brute-force check, so, I'd rather change that than convert from a perfectly fine for-loop to comprehension.
    – juanpa.arrivillaga
    Nov 21 '18 at 14:03












  • Interesting, could you provide the full line with your suggested edit?
    – Laurie Bamber
    Nov 21 '18 at 14:04






  • 1




    {l['id']:p['id'] for l in l1 for p in l2 if l['email']==p['email']}
    – juanpa.arrivillaga
    Nov 21 '18 at 14:05












  • This worked really nicely, thanks for the help, and I'll take a look into other methods that might be more robust. Thanks again.
    – Laurie Bamber
    Nov 21 '18 at 14:09






  • 1




    The order of elements in lists is consistency? If not, algorithm is not working. if k['email'] == v['email'] will works then k[N] == v[N]. You need consist of lists and when run your solution. Or, you can create one dict from your lists: d = defaultdict(list), d[email].append(id)
    – Maxim Panfilov
    Nov 21 '18 at 14:12














  • 1




    Your for-loop is fine. But if you are going to use a comprehension, you'd need {... for l in l1 for p in l2 if l['email']==p['email']} note how it aligns with your for-loop. However, you won't be able to break, making this less ideal. In any event, there are better algorithms than a brute-force check, so, I'd rather change that than convert from a perfectly fine for-loop to comprehension.
    – juanpa.arrivillaga
    Nov 21 '18 at 14:03












  • Interesting, could you provide the full line with your suggested edit?
    – Laurie Bamber
    Nov 21 '18 at 14:04






  • 1




    {l['id']:p['id'] for l in l1 for p in l2 if l['email']==p['email']}
    – juanpa.arrivillaga
    Nov 21 '18 at 14:05












  • This worked really nicely, thanks for the help, and I'll take a look into other methods that might be more robust. Thanks again.
    – Laurie Bamber
    Nov 21 '18 at 14:09






  • 1




    The order of elements in lists is consistency? If not, algorithm is not working. if k['email'] == v['email'] will works then k[N] == v[N]. You need consist of lists and when run your solution. Or, you can create one dict from your lists: d = defaultdict(list), d[email].append(id)
    – Maxim Panfilov
    Nov 21 '18 at 14:12








1




1




Your for-loop is fine. But if you are going to use a comprehension, you'd need {... for l in l1 for p in l2 if l['email']==p['email']} note how it aligns with your for-loop. However, you won't be able to break, making this less ideal. In any event, there are better algorithms than a brute-force check, so, I'd rather change that than convert from a perfectly fine for-loop to comprehension.
– juanpa.arrivillaga
Nov 21 '18 at 14:03






Your for-loop is fine. But if you are going to use a comprehension, you'd need {... for l in l1 for p in l2 if l['email']==p['email']} note how it aligns with your for-loop. However, you won't be able to break, making this less ideal. In any event, there are better algorithms than a brute-force check, so, I'd rather change that than convert from a perfectly fine for-loop to comprehension.
– juanpa.arrivillaga
Nov 21 '18 at 14:03














Interesting, could you provide the full line with your suggested edit?
– Laurie Bamber
Nov 21 '18 at 14:04




Interesting, could you provide the full line with your suggested edit?
– Laurie Bamber
Nov 21 '18 at 14:04




1




1




{l['id']:p['id'] for l in l1 for p in l2 if l['email']==p['email']}
– juanpa.arrivillaga
Nov 21 '18 at 14:05






{l['id']:p['id'] for l in l1 for p in l2 if l['email']==p['email']}
– juanpa.arrivillaga
Nov 21 '18 at 14:05














This worked really nicely, thanks for the help, and I'll take a look into other methods that might be more robust. Thanks again.
– Laurie Bamber
Nov 21 '18 at 14:09




This worked really nicely, thanks for the help, and I'll take a look into other methods that might be more robust. Thanks again.
– Laurie Bamber
Nov 21 '18 at 14:09




1




1




The order of elements in lists is consistency? If not, algorithm is not working. if k['email'] == v['email'] will works then k[N] == v[N]. You need consist of lists and when run your solution. Or, you can create one dict from your lists: d = defaultdict(list), d[email].append(id)
– Maxim Panfilov
Nov 21 '18 at 14:12




The order of elements in lists is consistency? If not, algorithm is not working. if k['email'] == v['email'] will works then k[N] == v[N]. You need consist of lists and when run your solution. Or, you can create one dict from your lists: d = defaultdict(list), d[email].append(id)
– Maxim Panfilov
Nov 21 '18 at 14:12












2 Answers
2






active

oldest

votes


















1














Try this:



from itertools import product

lookup = {k['id']: v['id'] for k, v in product(l1, l2) if k['email'] == v['email']}





share|improve this answer





















  • This is definitely working so thanks for that. Is there any solution which wouldn't involve importing an additional module?
    – Laurie Bamber
    Nov 21 '18 at 14:04










  • importing itertools does not have a lot of costs. but to answer to your question is should say yes. but you should implement product your self(to match all of items from both list together)
    – mehrdad-pedramfar
    Nov 21 '18 at 14:15



















1














Solution for unconsistency lists:



l1 = [{"email": "email1", "id": 1}, {"email": "email2", "id": 2}, {"email": "email3", "id": 3}]

l2 = [{"email": "email2", "id": 22}, {"email": "email4", "id": 4}, {"email": "email1", "id": 11}, ]



emails = {}

lookup = {}



for el in l1:

    emails[el["email"]] = el["id"]



for el in l2:

    email = el["email"]

    if email in emails:

        lookup[emails[email]] = el["id"]



# {1: 11, 2: 22}

print(lookup)



# bad solution from question

lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}



# {} - empty

print(lookup)


If you need more lists - extend solution, update emails dictionary on all loops before finally loop






share|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Try this:



    from itertools import product
    
lookup = {k['id']: v['id'] for k, v in product(l1, l2) if k['email'] == v['email']}





    share|improve this answer





















    • This is definitely working so thanks for that. Is there any solution which wouldn't involve importing an additional module?
      – Laurie Bamber
      Nov 21 '18 at 14:04










    • importing itertools does not have a lot of costs. but to answer to your question is should say yes. but you should implement product your self(to match all of items from both list together)
      – mehrdad-pedramfar
      Nov 21 '18 at 14:15
















    1














    Try this:



    from itertools import product
    
lookup = {k['id']: v['id'] for k, v in product(l1, l2) if k['email'] == v['email']}





    share|improve this answer





















    • This is definitely working so thanks for that. Is there any solution which wouldn't involve importing an additional module?
      – Laurie Bamber
      Nov 21 '18 at 14:04










    • importing itertools does not have a lot of costs. but to answer to your question is should say yes. but you should implement product your self(to match all of items from both list together)
      – mehrdad-pedramfar
      Nov 21 '18 at 14:15














    1












    1








    1






    Try this:



    from itertools import product
    
lookup = {k['id']: v['id'] for k, v in product(l1, l2) if k['email'] == v['email']}





    share|improve this answer












    Try this:



    from itertools import product
    
lookup = {k['id']: v['id'] for k, v in product(l1, l2) if k['email'] == v['email']}






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 21 '18 at 13:56









    mehrdad-pedramfar

    4,66711236




    4,66711236












    • This is definitely working so thanks for that. Is there any solution which wouldn't involve importing an additional module?
      – Laurie Bamber
      Nov 21 '18 at 14:04










    • importing itertools does not have a lot of costs. but to answer to your question is should say yes. but you should implement product your self(to match all of items from both list together)
      – mehrdad-pedramfar
      Nov 21 '18 at 14:15


















    • This is definitely working so thanks for that. Is there any solution which wouldn't involve importing an additional module?
      – Laurie Bamber
      Nov 21 '18 at 14:04










    • importing itertools does not have a lot of costs. but to answer to your question is should say yes. but you should implement product your self(to match all of items from both list together)
      – mehrdad-pedramfar
      Nov 21 '18 at 14:15
















    This is definitely working so thanks for that. Is there any solution which wouldn't involve importing an additional module?
    – Laurie Bamber
    Nov 21 '18 at 14:04




    This is definitely working so thanks for that. Is there any solution which wouldn't involve importing an additional module?
    – Laurie Bamber
    Nov 21 '18 at 14:04












    importing itertools does not have a lot of costs. but to answer to your question is should say yes. but you should implement product your self(to match all of items from both list together)
    – mehrdad-pedramfar
    Nov 21 '18 at 14:15




    importing itertools does not have a lot of costs. but to answer to your question is should say yes. but you should implement product your self(to match all of items from both list together)
    – mehrdad-pedramfar
    Nov 21 '18 at 14:15













    1














    Solution for unconsistency lists:



    l1 = [{"email": "email1", "id": 1}, {"email": "email2", "id": 2}, {"email": "email3", "id": 3}]
    
l2 = [{"email": "email2", "id": 22}, {"email": "email4", "id": 4}, {"email": "email1", "id": 11}, ]
    

    
emails = {}
    
lookup = {}
    

    
for el in l1:
    
    emails[el["email"]] = el["id"]
    

    
for el in l2:
    
    email = el["email"]
    
    if email in emails:
    
        lookup[emails[email]] = el["id"]
    

    
# {1: 11, 2: 22}
    
print(lookup)
    

    
# bad solution from question
    
lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}
    

    
# {} - empty
    
print(lookup)


    If you need more lists - extend solution, update emails dictionary on all loops before finally loop






    share|improve this answer


























      1














      Solution for unconsistency lists:



      l1 = [{"email": "email1", "id": 1}, {"email": "email2", "id": 2}, {"email": "email3", "id": 3}]
      
l2 = [{"email": "email2", "id": 22}, {"email": "email4", "id": 4}, {"email": "email1", "id": 11}, ]
      

      
emails = {}
      
lookup = {}
      

      
for el in l1:
      
    emails[el["email"]] = el["id"]
      

      
for el in l2:
      
    email = el["email"]
      
    if email in emails:
      
        lookup[emails[email]] = el["id"]
      

      
# {1: 11, 2: 22}
      
print(lookup)
      

      
# bad solution from question
      
lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}
      

      
# {} - empty
      
print(lookup)


      If you need more lists - extend solution, update emails dictionary on all loops before finally loop






      share|improve this answer
























        1












        1








        1






        Solution for unconsistency lists:



        l1 = [{"email": "email1", "id": 1}, {"email": "email2", "id": 2}, {"email": "email3", "id": 3}]
        
l2 = [{"email": "email2", "id": 22}, {"email": "email4", "id": 4}, {"email": "email1", "id": 11}, ]
        

        
emails = {}
        
lookup = {}
        

        
for el in l1:
        
    emails[el["email"]] = el["id"]
        

        
for el in l2:
        
    email = el["email"]
        
    if email in emails:
        
        lookup[emails[email]] = el["id"]
        

        
# {1: 11, 2: 22}
        
print(lookup)
        

        
# bad solution from question
        
lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}
        

        
# {} - empty
        
print(lookup)


        If you need more lists - extend solution, update emails dictionary on all loops before finally loop






        share|improve this answer












        Solution for unconsistency lists:



        l1 = [{"email": "email1", "id": 1}, {"email": "email2", "id": 2}, {"email": "email3", "id": 3}]
        
l2 = [{"email": "email2", "id": 22}, {"email": "email4", "id": 4}, {"email": "email1", "id": 11}, ]
        

        
emails = {}
        
lookup = {}
        

        
for el in l1:
        
    emails[el["email"]] = el["id"]
        

        
for el in l2:
        
    email = el["email"]
        
    if email in emails:
        
        lookup[emails[email]] = el["id"]
        

        
# {1: 11, 2: 22}
        
print(lookup)
        

        
# bad solution from question
        
lookup = {k['id']: v['id'] for k, v in zip(l1, l2) if k['email'] == v['email']}
        

        
# {} - empty
        
print(lookup)


        If you need more lists - extend solution, update emails dictionary on all loops before finally loop







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 21 '18 at 14:39









        Maxim Panfilov

        150111




        150111






























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            0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43
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            How to know if a Active Directory user can login interactively

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            1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor
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            TypeError: fit_transform() missing 1 required positional argument: 'X'

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            0 I am trying to do Feature Scaling in a dataset, but I get an error and have no idea how to proceed: > Traceback (most recent call last): > > File "<ipython-input-10-71bea414b4d0>", line 22, in <module> > x_train = sc_X.fit_transform(x_train) > > TypeError: fit_transform() missing 1 required positional argument: 'X' and here is my code: import pandas as pd # Importing the dataset dataset = pd.read_csv('Data.csv') X = dataset.iloc[:, :-1].values y = dataset.iloc[:, 3].values # Taking care of missing data from sklearn.preprocessing import Imputer imputer = Imputer(missing_values="NaN", strategy="mean", axis=0) imputer = Imputer.fit(imputer,X[:,1:3]) X[:, 1:3] = Imputer.transform(imputer,X[
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