8

我有一个这样的列表：

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

我想得到其中有多少项目是相同的，所以我需要比较all[0]，all[1],all[2]...all[(len(all)-1)]然后用all[1]比较all[2],all[3]...all[(len(all)-1)]，然后all[2]比较all[3],all[4],...all[(len(all)-1)]

我试过这样的事情：

 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])

但不知道如何继续，我想要得到的结果是：

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
4

4 回答 4

4

这里最简单的算法是an^2。只需循环两次您的列表：

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)
于 2012-08-10T13:32:57.303 回答
2

套装是要走的路。. . 

 
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

于 2012-08-10T13:41:08.657 回答
1

基本上你想要做的是计算每个列表中的元素集与其他列表的交集的长度。试试这个：

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 )

输出格式并不完全是您想要的，但您明白了。

于 2012-08-10T13:34:31.307 回答
0

如果您正在寻找最短的答案，因为您像我一样是懒惰的打字员:)

>>> 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]
于 2012-08-10T13:58:02.530 回答