How can I test if a list contains another list (ie. it's a contiguous subsequence). Say there was a function called contains:
contains([1,2], [-1, 0, 1, 2]) # Returns [2, 3] (contains returns [start, end])
contains([1,3], [-1, 0, 1, 2]) # Returns False
contains([1, 2], [[1, 2], 3]) # Returns False
contains([[1, 2]], [[1, 2], 3]) # Returns [0, 0]
Edit:
contains([2, 1], [-1, 0, 1, 2]) # Returns False
contains([-1, 1, 2], [-1, 0, 1, 2]) # Returns False
contains([0, 1, 2], [-1, 0, 1, 2]) # Returns [1, 3]
This question is related to
python
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contains
list-comparison
If all items are unique, you can use sets.
>>> items = set([-1, 0, 1, 2])
>>> set([1, 2]).issubset(items)
True
>>> set([1, 3]).issubset(items)
False
Here is my answer. This function will help you to find out whether B is a sub-list of A. Time complexity is O(n).
`def does_A_contain_B(A, B): #remember now A is the larger list
b_size = len(B)
for a_index in range(0, len(A)):
if A[a_index : a_index+b_size]==B:
return True
else:
return False`
Here's a straightforward algorithm that uses list methods:
#!/usr/bin/env python
def list_find(what, where):
"""Find `what` list in the `where` list.
Return index in `where` where `what` starts
or -1 if no such index.
>>> f = list_find
>>> f([2, 1], [-1, 0, 1, 2])
-1
>>> f([-1, 1, 2], [-1, 0, 1, 2])
-1
>>> f([0, 1, 2], [-1, 0, 1, 2])
1
>>> f([1,2], [-1, 0, 1, 2])
2
>>> f([1,3], [-1, 0, 1, 2])
-1
>>> f([1, 2], [[1, 2], 3])
-1
>>> f([[1, 2]], [[1, 2], 3])
0
"""
if not what: # empty list is always found
return 0
try:
index = 0
while True:
index = where.index(what[0], index)
if where[index:index+len(what)] == what:
return index # found
index += 1 # try next position
except ValueError:
return -1 # not found
def contains(what, where):
"""Return [start, end+1] if found else empty list."""
i = list_find(what, where)
return [i, i + len(what)] if i >= 0 else [] #NOTE: bool([]) == False
if __name__=="__main__":
import doctest; doctest.testmod()
May I humbly suggest the Rabin-Karp algorithm if the big list is really big. The link even contains almost-usable code in almost-Python.
After OP's edit:
def contains(small, big):
for i in xrange(1 + len(big) - len(small)):
if small == big[i:i+len(small)]:
return i, i + len(small) - 1
return False
If we refine the problem talking about testing if a list contains another list with as a sequence, the answer could be the next one-liner:
def contains(subseq, inseq):
return any(inseq[pos:pos + len(subseq)] == subseq for pos in range(0, len(inseq) - len(subseq) + 1))
Here unit tests I used to tune up this one-liner:
I think this one is fast...
def issublist(subList, myList, start=0):
if not subList: return 0
lenList, lensubList = len(myList), len(subList)
try:
while lenList - start >= lensubList:
start = myList.index(subList[0], start)
for i in xrange(lensubList):
if myList[start+i] != subList[i]:
break
else:
return start, start + lensubList - 1
start += 1
return False
except:
return False
a=[[1,2] , [3,4] , [0,5,4]]
print(a.__contains__([0,5,4]))
It provides true output.
a=[[1,2] , [3,4] , [0,5,4]]
print(a.__contains__([1,3]))
It provides false output.
Smallest code:
def contains(a,b):
str(a)[1:-1].find(str(b)[1:-1])>=0
I tried to make this as efficient as possible.
It uses a generator; those unfamiliar with these beasts are advised to check out their documentation and that of yield expressions.
Basically it creates a generator of values from the subsequence that can be reset by sending it a true value. If the generator is reset, it starts yielding again from the beginning of sub.
Then it just compares successive values of sequence with the generator yields, resetting the generator if they don't match.
When the generator runs out of values, i.e. reaches the end of sub without being reset, that means that we've found our match.
Since it works for any sequence, you can even use it on strings, in which case it behaves similarly to str.find, except that it returns False instead of -1.
As a further note: I think that the second value of the returned tuple should, in keeping with Python standards, normally be one higher. i.e. "string"[0:2] == "st". But the spec says otherwise, so that's how this works.
It depends on if this is meant to be a general-purpose routine or if it's implementing some specific goal; in the latter case it might be better to implement a general-purpose routine and then wrap it in a function which twiddles the return value to suit the spec.
def reiterator(sub):
"""Yield elements of a sequence, resetting if sent ``True``."""
it = iter(sub)
while True:
if (yield it.next()):
it = iter(sub)
def find_in_sequence(sub, sequence):
"""Find a subsequence in a sequence.
>>> find_in_sequence([2, 1], [-1, 0, 1, 2])
False
>>> find_in_sequence([-1, 1, 2], [-1, 0, 1, 2])
False
>>> find_in_sequence([0, 1, 2], [-1, 0, 1, 2])
(1, 3)
>>> find_in_sequence("subsequence",
... "This sequence contains a subsequence.")
(25, 35)
>>> find_in_sequence("subsequence", "This one doesn't.")
False
"""
start = None
sub_items = reiterator(sub)
sub_item = sub_items.next()
for index, item in enumerate(sequence):
if item == sub_item:
if start is None: start = index
else:
start = None
try:
sub_item = sub_items.send(start is None)
except StopIteration:
# If the subsequence is depleted, we win!
return (start, index)
return False
Dave answer is good. But I suggest this implementation which is more efficient and doesn't use nested loops.
def contains(small_list, big_list):
"""
Returns index of start of small_list in big_list if big_list
contains small_list, otherwise -1.
"""
loop = True
i, curr_id_small= 0, 0
while loop and i<len(big_list):
if big_list[i]==small_list[curr_id_small]:
if curr_id_small==len(small_list)-1:
loop = False
else:
curr_id_small += 1
else:
curr_id_small = 0
i=i+1
if not loop:
return i-len(small_list)
else:
return -1
The problem of most of the answers, that they are good for unique items in list. If items are not unique and you still want to know whether there is an intersection, you should count items:
from collections import Counter as count
def listContains(l1, l2):
list1 = count(l1)
list2 = count(l2)
return list1&list2 == list1
print( listContains([1,1,2,5], [1,2,3,5,1,2,1]) ) # Returns True
print( listContains([1,1,2,8], [1,2,3,5,1,2,1]) ) # Returns False
You can also return the intersection by using ''.join(list1&list2)
There's an all() and any() function to do this.
To check if big contains ALL elements in small
result = all(elem in big for elem in small)
To check if small contains ANY elements in big
result = any(elem in big for elem in small)
the variable result would be boolean (TRUE/FALSE).
Here a solution with less line of code and easily understandable (or at least I like to think so).
If you want to keep order (match only if the smaller list is found in the same order on the bigger list):
def is_ordered_subset(l1, l2):
# First check to see if all element of l1 are in l2 (without checking order)
if not set(l1).issubset(l2):
return False
length = len(l1)
# Make sublist of same size than l1
list_of_sublist = [l2[i:i+length] for i, x in enumerate(l2)]
#Check if one of this sublist is l1
return l1 in list_of_sublist
This works and is fairly fast since it does the linear searching using the builtin list.index() method and == operator:
def contains(sub, pri):
M, N = len(pri), len(sub)
i, LAST = 0, M-N+1
while True:
try:
found = pri.index(sub[0], i, LAST) # find first elem in sub
except ValueError:
return False
if pri[found:found+N] == sub:
return [found, found+N-1]
else:
i = found+1
Source: Stackoverflow.com