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Python: compare items in a list with each other

I'm currently looking for a way to compare list items with each other going from left to right.

Here is my list:

mylist = [[15], [14, 15], [19, 20], [13], [3], [65, 19], [19, 20, 31]]

I need to compare the first element with all the others and check if any of the values ​​matches, the same for the second and third elements and so on, until it reaches the end of the list. If there is a match, I need to print it. I need to print a value that has a match and the index with which it maps.

For instance.

index 0 matched with index 1 for value 15.
index 2 matched with index 5 for value 19
index 2 matched with index 6 for value 19
index 2 matched with index 6 for value 20
index 5 matched with index 6 for value 19.

How can i do this?

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3 answers
mylist = [[15], [14, 15], [19, 20], [13], [3], [65, 19], [19, 20, 31]]
my_set_list = [set(item) for item in mylist]
for i1, item in enumerate(my_set_list):
    for i2 in xrange(i1 + 1, len(my_set_list)):
        for val in (item & my_set_list[i2]):
            print "Index {} matched with index {} for value {}".format(i1,i2,val)

Output

Index 0 matched with index 1 for value 15.
Index 2 matched with index 5 for value 19.
Index 2 matched with index 6 for value 19.
Index 2 matched with index 6 for value 20.
Index 5 matched with index 6 for value 19.
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The naive solution is to iterate over each pair, which is slow. However, you can do something like this:

  • dict, ints ( ) , .
  • , dict.
  • dict , , "".

:

>>> mylist = [[15], [14, 15], [19, 20], [13], [3], [65, 19], [19, 20, 31]]
>>> 
>>> pairs = dict()
>>> 
>>> 
>>> from collections import defaultdict
>>> 
>>> pairs = defaultdict(list)
>>> 
>>> for idx, sub in enumerate(mylist):
...     for a in sub:
...         pairs[a].append(idx)
... 
>>> pairs
defaultdict(<type 'list'>, {65: [5], 3: [4], 13: [3], 14: [1], 15: [0, 1], 19: [2, 5, 6], 20: [2, 6], 31: [6]})

( , ). , , . , 19 [2, 5, 6], :

  • 2 5 -
  • 2 6 -
  • 5 6 -

. . , O (n) .

:

>>> from itertools import combinations
>>> 
>>> for k,v in pairs.iteritems():
...     if len(v) > 1:
...         for a,b in combinations(v, 2):
...             print "Index {} matched with index {} for value {}".format(a, b, k)
... 
Index 0 matched with index 1 for value 15
Index 2 matched with index 5 for value 19
Index 2 matched with index 6 for value 19
Index 5 matched with index 6 for value 19
Index 2 matched with index 6 for value 20
+3

You can use itertools.combinationsthis to save you a nested loop:

In [770]: l2=[(i,set(mylist[i])) for i in range(len(mylist))]
     ...: for (i, a), (j, b) in combinations(l2,2):
     ...:     for v in a.intersection(b):
     ...:         print "Index {} matched with index {} for value {}".format(i,j,v)
Index 0 matched with index 1 for value 15
Index 2 matched with index 5 for value 19
Index 2 matched with index 6 for value 19
Index 2 matched with index 6 for value 20
Index 5 matched with index 6 for value 19
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