Using 'in' To Test For Part Of One Sublist In Another In Python

Newbie here trying to search for part of one sublist within another sublist. list_1 = [[1, 2, 9], [4, 5, 8]] list_2 = [[1, 2, 3], [4, 5, 6], [1, 2, 5]] for item in list_1: fo

Solution 1:

in looks to see if one sublist is an element (not sublist) of another list:

[1,2] in[[1,2],[3,4]]

would be True.

[1,2] in [1,2,3]

would be False as would:

[1,2] in [1,2]

However:

[1,2] == [1,2]

would be True. Depending on what you're actually trying to do, set objects might be useful.

a = [1,2]
b = [1,2,3]
c = [3,2,1]
d = [1,1,1]
e = set(a)
len(e.intersection(b)) == len(a)  #Truelen(e.intersection(c)) == len(a)  #True -- Order of elements does not matterlen(e.intersection(d)) == len(a)  #False

Solution 2:

Given your example lists:

list_1 = [[1, 2, 9], [4, 5, 8]]
list_2 = [[1, 2, 3], [4, 5, 6], [1, 2, 5]]

This works:

print [this[0:2]==that[0:2] for this in list_1 for that in list_2]
[True, False, True, False, True, False]

Or, use a set:

print [thisforthisin list_1 for that in list_2 ifset(this[0:2])<set(that)]
[[1, 2, 9], [1, 2, 9], [4, 5, 8]]

Be aware that a set is without order, so:

>>> set([1,2])==set([2,1])
True

A typical use of in is with a string:

>>> 'ab'in'cbcbab'True

Or a single element in a sequence:

>>> 100inrange(1000)
True

Or an atomic element in a sequence:

>>> (3,3,3) inzip(*[range(10)]*3)
True

But over lapping list element do not work:

>>> [1,2] in [0,1,2,3]
False

Unless the elements are the same atomic size:

>>> [1,2] in [0,[1,2],3]
True

But you CAN use a string to compare list a being 'in' list b like so:

>>>defstringy(li): return''.join(map(str,li))...>>>stringy([1,2,9][0:2])
'12'
>>>stringy([1,2,9][0:2]) in stringy([1,2,5])
True

So your original intent MAY be to check to see of item[0:2] appears anywhere in otherItem but in the order of 'item' in your loop. You can use a string like so:

>>> print [thisforthisin list_1 for that in list_2 if stringy(this[0:2]) in stringy(that)]
[[1, 2, 9], [1, 2, 9], [4, 5, 8]]

This is different than the set version since '12'!='21' and '12' not in '21' So if you changed the order of the elements of list_2:

list_1 = [[1, 2, 9], [4, 5, 8]]
list_2 = [[1, 2, 3], [4, 5, 6], [1, 5, 2]]print [this for this in list_1 for that in list_2 if set(this[0:2])<set(that)]
[[1, 2, 9], [1, 2, 9], [4, 5, 8]]   # same answer since sets are unordered
print [this for this in list_1 for that in list_2 if stringy(this[0:2]) in stringy(that)]
[[1, 2, 9], [4, 5, 8]]              # different answer...

Solution 3:

printset([1,2]).intersection([1,2,3])==set([1,2])

would be True

using set intersection I think you can get what you want

It is important to note that sets are un-ordered collections unique elements

thus set([1,1,2]) == set([1,2]) and so this may not necessarily work for you for all instances