Search in lists of lists by given index

Question

I have a list of two-item lists and need to search for things in it  If the list is  list      a   b      a   c      b   d     I can search for a pair easily by doing   a   b   in list  Now  is there a way to see if I have a pair in which a string is present in just the second position  I can do this  for i in range  0  len list        if list i  1   search         found 1  But is there a  better  way without the for loop  I don t need to know i or keep the loop going after it s found

User · Answer

I was searching for a deep find for dictionaries and didn't find one. Based on this article I was able to create the following. Thanks and Enjoy!!

def deapFind( theList, key, value ):
    result = False
    for x in theList:
        if( value == x[key] ):
            return True
    return result

theList = [{ "n": "aaa", "d": "bbb" }, { "n": "ccc", "d": "ddd" }]
print 'Result: ' + str (deapFind( theList, 'n', 'aaa'))

I'm using == instead of the in operator since in returns true for partial matches. IOW: searching aa on the n key returns true. I don't think that would be desired.

HTH

User · Answer

Here s the Pythonic way to do it   data      a   b      a   c      b   d    search    c  any e 1     search for e in data    Or    well  I m not going to claim this is the  one true Pythonic way  to do it because at some point it becomes a little subjective what is Pythonic and what isn t  or which method is more Pythonic than another  But using any   is definitely more typical Python style than a for loop as in e g  RichieHindle s answer    Of course there is a hidden loop in the implementation of any  although it breaks out of the loop as soon as it finds a match     Since I was bored I made a timing script to compare performance of the different suggestions  modifying some of them as necessary to make the API the same  Now  we should bear in mind that fastest is not always best  and being fast is definitely not the same thing as being Pythonic  That being said  the results are    strange  Apparently for loops are very fast  which is not what I expected  so I d take these with a grain of salt without understanding why they ve come out the way they do   Anyway  when I used the list defined in the question with three sublists of two elements each  from fastest to slowest I get these results    RichieHindle s answer with the for loop  clocking in at 0 22   s Terence Honles  first suggestion which creates a list  at 0 36   s Pierre-Luc Bedard s answer  last code block   at 0 43   s Essentially tied between Markus s answer and the for loop from the original question  at 0 48   s Coady s answer using operator itemgetter    at 0 53   s Close enough to count as a tie between Alex Martelli s answer with ifilter   and Anon s answer  at 0 67   s  Alex s is consistently about half a microsecond faster  Another close-enough tie between jojo s answer  mine  Brandon E Taylor s  which is identical to mine   and Terence Honles  second suggestion using any    all coming in at 0 81-0 82   s And then user27221 s answer using nested list comprehensions  at 0 95   s   Obviously the actual timings are not meaningful on anyone else s hardware  but the differences between them should give some idea of how close the different methods are   When I use a longer list  things change a bit  I started with the list in the question  with three sublists  and appended another 197 sublists  for a total of 200 sublists each of length two  Using this longer list  here are the results    RichieHindle s answer  at the same 0 22   s as with the shorter list Coady s answer using operator itemgetter    again at 0 53   s Terence Honles  first suggestion which creates a list  at 0 36   s Another virtual tie between Alex Martelli s answer with ifilter   and Anon s answer  at 0 67   s Again a close-enough tie between my answer  Brandon E Taylor s identical method  and Terence Honles  second suggestion using any    all coming in at 0 81-0 82   s   Those are the ones that keep their original timing when the list is extended  The rest  which don t  are   The for loop from the original question  at 1 24   s Terence Honles  first suggestion which creates a list  at 7 49   s Pierre-Luc Bedard s answer  last code block   at 8 12   s Markus s answer  at 10 27   s jojo s answer  at 19 87   s And finally user27221 s answer using nested list comprehensions  at 60 59   s

User · Answer

k old post but no one use list expression to answer  P  list      a   b      a   c      b   d     Search    c     return if it find in either item 0 or item 1 print  x for x y in list if x    Search or y    Search     return if it find in item 1 print  x for x y in list if y    Search

User · Answer

gt  gt  gt  the list      a   b      a   c      b   d      gt  gt  gt   b  in zip  the list  1  True   zip   takes a bunch of lists and groups elements together by index  effectively transposing the list-of-lists matrix  The asterisk takes the contents of the list and sends it to zip as arguments  so you re effectively passing the three lists separately  which is what zip wants  All that remains is to check if  b   or whatever  is in the list made up of elements with the index you re interested in

User · Answer

gt  gt  gt  my list      a    b      a    c      b    d      gt  gt  gt   d  in  x 1  for x in my list  True   Editing to add   Both David s answer using any and mine using in will end when they find a match since we re using generator expressions   Here is a test using an infinite generator to show that   def mygen            Infinite generator         while True          yield  xxx     Just to include a non-match in the generator         yield  d   print  d  in  x for x in mygen          True print any  d     x for x in mygen       True   print  q  in  x for x in mygen          Never ends if uncommented   print any  q     x for x in mygen       Never ends if uncommented   I just like simply using in instead of both    and any

User · Answer

Nothing wrong with using a gen exp  but if the goal is to inline the loop      gt  gt  gt  import itertools  operator  gt  gt  gt   b  in itertools imap operator itemgetter 1   the list  True   Should be the fastest as well

User · Answer

Given below is a simple way to find exactly where in the list the item is   for i in range  0 len a    sublist a i  for i in range 0 len sublist        if search  sublist i           print  found in sublist     a  str i

User · Answer

You re always going to have a loop - someone might come along with a clever one-liner that hides the loop within a call to map   or similar  but it s always going to be there   My preference would always be to have clean and simple code  unless performance is a major factor   Here s perhaps a more Pythonic version of your code   data      a   b      a   c      b   d    search    c  for sublist in data      if sublist 1     search          print  Found it    sublist         break   Prints  Found it    a    c     It breaks out of the loop as soon as it finds a match    You have a typo  by the way  in   b  d

User · Answer

gt  gt  gt  the list      a   b      a   c      b  d      gt  gt  gt  any  c     x 1  for x in the list  True

User · Answer

What about   list      a   b      a   c      b   d     search    b   filter lambda x x 1   search list    This will return each list in the list of lists with the second element being equal to search

User · Answer

the above all look good  but do you want to keep the result   if so     you can use the following  result    element for element in data if element 1     search    then a simple  len result    lets you know if anything was found  and now you can do stuff with the results   of course this does not handle elements which are length less than one  which you should be checking unless you know they always are greater than length 1  and in that case should you be using a tuple   tuples are immutable    if you know all items are a set length you can also do   any second    search for    second in data    or for len data 0      4   any second    search for    second       in data       and I would recommend using  for element in data           instead of  for i in range len data              for future uses  unless you want to save or use  i   and just so you know the  0  is not required  you only need use the full syntax if you are starting at a non zero value

User · Answer

I think using nested list comprehensions is the most elegant way to solve this  because the intermediate result is the position where the element is  An implementation would be   list      a   b      a   c      b   d     search    c  any    list index x  x index y   for x in list for y in x if y    search

User · Answer

Markus has one way to avoid using the word for -- here s another  which should have much better performance for long the lists      import itertools found   any itertools ifilter lambda x x 1    b   the list

[python] Search in lists of lists by given index

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