如何将列表中的一项与此列表中的所有其他项进行比较,python
我有这样一份清单:如何将列表中的一项与此列表中的所有其他项进行比较,python,python,Python,我有这样一份清单: all = [[a,b,c,d],[r,d,g,s],[e,r,a,b],[p,o,i,u]....(more similar items)] 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]) 我想知道其
all = [[a,b,c,d],[r,d,g,s],[e,r,a,b],[p,o,i,u]....(more similar items)]
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])
我想知道其中有多少项是相同的,所以我需要将all[0]
与all[1]、all[2]…all[(len(all)-1)]
进行比较,然后使用all[1]
与all[2]、all[3]…all[(len(all)-1)]
进行比较,然后all[2]、all[2]、[4]、…all[(len 1
我试过这样的方法:
all = [[a,b,c,d],[r,d,g,s],[e,r,a,b],[p,o,i,u]....(more similar items)]
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])
但我不知道如何继续,我想得到的结果是:
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
这里最简单的算法是n^2。只需在列表上循环两次:
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)
基本上,您要做的是计算每个列表中元素集与其他列表的交点长度。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,但您已经有了想法。设置就是一种方法。
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
all=[[1,2,3,4]、[1,2,5,6]、[4,5,7,8]、[1,8,3,4]]
set_all=[设置(i)为i的集合]
对于范围内的i(len(all)):
对于范围内的j(len(all)):
如果i==j:
持续
ncom=len(集合所有[i]。交点(集合所有[j]))
打印“列表集%s与集%s共有%s个元素”%(i,ncom,j)
列表集0有2个元素与集1相同
列表集0有1个与集2相同的元素
列表集0有3个与集3相同的元素
列表集1有2个元素与集0相同
列表集1有1个元素与集2相同
列表集1有1个元素与集合3相同
列表集2有1个与集0相同的元素
列表集2有1个元素与集合1相同
列表集2有2个元素与集合3相同
列表集3有3个元素与集0相同
列表集3有1个元素与集合1相同
列表集3有2个元素与集合2相同
如果你在寻找最短的答案,因为你和我一样懒惰:)
是否有子列表包含相同的值两次?如果是这样,您关心多重性吗?@DSM子列表不包含重复项:)顺便说一句,您确实不应该使用名称all
,因为您正在覆盖