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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            Refactoring coordinates for Minecraft Pi buildings written in Python

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            0 $begingroup$ I'm fairly new to coding and I wrote this script in Python for a house with a roof in Minecraft Pi. I've found different ways to write the coordinates to make the code base shorter, but I'm wondering if it is possible for me to refactor it any further so I don't have to list each coordinate line by line. import mcpi.minecraft as minecraft import mcpi.block as block mc = minecraft.Minecraft.create() p = mc.player.getTilePos() mc.setBlocks(p.x+1, p.y, p.z+1, #Main structure starting coordinates (DO NOT ALTER) p.x+19, p.y+4, p.z+6, #Structure size/coordinates (Alters cube size) block.STONE_BRICK) mc.setBlocks(p.x+2, p.y+1, p.z+2, #Hollowing out main structure interior p.x+18, p.y+3, p.z+5, block.AIR) mc.setBlocks(p.x, p.y...
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