CODEWITHSUNDEEP
python
manxing
manxing

Reputation: 3305

how to compare one item in a list with all the other items in this list, python

I have a list like this:

 all = [[a,b,c,d],[r,d,g,s],[e,r,a,b],[p,o,i,u]....(more similar items)]

I want to get how many items are the same among them, so I need to compare all[0] with all[1],all[2]...all[(len(all)-1)], and then use all[1] to compare with all[2],all[3]...all[(len(all)-1)], then all[2] to compare with all[3],all[4],...all[(len(all)-1)]

I tried something like this:

 for i in range(len(all)):
     print len(all[i] & all[i+1]) ##how many identical items shared by all[0] and all[1]
     print len(all[i+1] & all[i+2])

but don't know how to continue, The result I want to get is:

item1 has 3 same values with item2, 
      has 4 same values with item3,
      has 1 same values with item4....

item2 has 3 same values with item1,
      has 2 same values with item3,
      etc

Upvotes: 8

Views: 12179

Answers (4)

Meitham
Meitham

Reputation: 9670

If you're looking for the shortest answer because you're a lazy typer like me :)

>>> my_list = [['a','b','c','d'],['r','d','g','s'],['e','r','a','b'],['p','o','i','u']]
>>> for i, sub_list in enumerate(my_list):
...     print 'item %d shares with %r'%(i, map(lambda a, b: len(set(a) & set(b)), sub_list, my_list))
...
item 0 shares with [1, 0, 0, 0]
item 1 shares with [0, 1, 0, 0]
item 2 shares with [0, 1, 1, 0]
item 3 shares with [0, 0, 0, 1]

Upvotes: 0

reptilicus
reptilicus

Reputation: 10397

Sets are the way to go. . . 

 
all = [[1,2,3,4],[1,2,5,6],[4,5,7,8],[1,8,3,4]]
set_all = [set(i) for i in all]
for i in range(len(all)):
    for j in range(len(all)):
        if i == j: 
            continue
        ncom = len(set_all[i].intersection(set_all[j]))
        print "List set %s has %s elements in common with set %s" % (i, ncom, j)



List set 0 has 2 elements in common with set 1
List set 0 has 1 elements in common with set 2
List set 0 has 3 elements in common with set 3
List set 1 has 2 elements in common with set 0
List set 1 has 1 elements in common with set 2
List set 1 has 1 elements in common with set 3
List set 2 has 1 elements in common with set 0
List set 2 has 1 elements in common with set 1
List set 2 has 2 elements in common with set 3
List set 3 has 3 elements in common with set 0
List set 3 has 1 elements in common with set 1
List set 3 has 2 elements in common with set 2

Upvotes: 3

Gordon Bailey
Gordon Bailey

Reputation: 3911

Basically what you want to do is count the length of the intersections of the set of elements in each list with each other list. Try this:

a = [['a','b','c','d'],['r','d','g','s'],['e','r','a','b'],['p','o','i','u']]

for i in range(len(a)):
   for j in range(len(a)):
      print "item%d has %d same values as item%d" % ( i, len(set(a[i]) & set(a[j])) ,j )

The output format is not exactly what you wanted, but you get the idea.

Upvotes: 1

stderr
stderr

Reputation: 8722

The simplest algorithm here is a n^2. Just loop over your list twice:

for x, left in enumerate(all):
    for y, right in enumerate(all):
        common = len(set(left) & set(right))
        print "item%s has %s values in common with item%s"%(x, common, y)

Upvotes: 5

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