I want to do the following in Python:
A = [1, 2, 3, 4, 5, 6, 7, 7, 7]
C = A - [3, 4] # Should be [1, 2, 5, 6, 7, 7, 7]
C = A - [4, 3] # Should not be removing anything, because sequence 4, 3 is not found
So, I simply want to remove the first appearance of a sublist (as a sequence) from another list. How can I do that?
Edit: I am talking about lists, not sets. Which implies that ordering (sequence) of items matter (both in A and B), as well as duplicates.
Use sets:
C = list(set(A) - set(B))
In case you want to mantain duplicates and/or oder:
filter_set = set(B)
C = [x for x in A if x not in filter_set]
If you want to remove exact sequences, here is one way:
Find the bad indices by checking to see if the sublist matches the desired sequence:
bad_ind = [range(i,i+len(B)) for i,x in enumerate(A) if A[i:i+len(B)] == B]
print(bad_ind)
#[[2, 3]]
Since this returns a list of lists, flatten it and turn it into a set:
bad_ind_set = set([item for sublist in bad_ind for item in sublist])
print(bad_ind_set)
#set([2, 3])
Now use this set to filter your original list, by index:
C = [x for i,x in enumerate(A) if i not in bad_ind_set]
print(C)
#[1, 2, 5, 6, 7, 7, 7]
The above bad_ind_set will remove all matches of the sequence. If you only want to remove the first match, it's even simpler. You just need the first element of bad_ind (no need to flatten the list):
bad_ind_set = set(bad_ind[0])
Update: Here is a way to find and remove the first matching sub-sequence using a short circuiting for loop. This will be faster because it will break out once the first match is found.
start_ind = None
for i in range(len(A)):
if A[i:i+len(B)] == B:
start_ind = i
break
C = [x for i, x in enumerate(A)
if start_ind is None or not(start_ind <= i < (start_ind + len(B)))]
print(C)
#[1, 2, 5, 6, 7, 7, 7]
I considered this question was like one substring search, so KMP, BM etc sub-string search algorithm could be applied at here. Even you'd like support multiple patterns, there are some multiple pattern algorithms like Aho-Corasick, Wu-Manber etc.
Below is KMP algorithm implemented by Python which is from GitHub Gist. PS: the author is not me. I just want to share my idea.
class KMP:
def partial(self, pattern):
""" Calculate partial match table: String -> [Int]"""
ret = [0]
for i in range(1, len(pattern)):
j = ret[i - 1]
while j > 0 and pattern[j] != pattern[i]:
j = ret[j - 1]
ret.append(j + 1 if pattern[j] == pattern[i] else j)
return ret
def search(self, T, P):
"""
KMP search main algorithm: String -> String -> [Int]
Return all the matching position of pattern string P in S
"""
partial, ret, j = self.partial(P), [], 0
for i in range(len(T)):
while j > 0 and T[i] != P[j]:
j = partial[j - 1]
if T[i] == P[j]: j += 1
if j == len(P):
ret.append(i - (j - 1))
j = 0
return ret
Then use it to calcuate out the matched position, finally remove the match:
A = [1, 2, 3, 4, 5, 6, 7, 7, 7, 3, 4]
B = [3, 4]
result = KMP().search(A, B)
print(result)
#assuming at least one match is found
print(A[:result[0]:] + A[result[0]+len(B):])
Output:
[2, 9]
[1, 2, 5, 6, 7, 7, 7, 3, 4]
[Finished in 0.201s]
PS: You can try other algorithms also. And @Pault 's answers is good enough unless you care about the performance a lot.
Here is another approach:
# Returns that starting and ending point (index) of the sublist, if it exists, otherwise 'None'.
def findSublist(subList, inList):
subListLength = len(subList)
for i in range(len(inList)-subListLength):
if subList == inList[i:i+subListLength]:
return (i, i+subListLength)
return None
# Removes the sublist, if it exists and returns a new list, otherwise returns the old list.
def removeSublistFromList(subList, inList):
indices = findSublist(subList, inList)
if not indices is None:
return inList[0:indices[0]] + inList[indices[1]:]
else:
return inList
A = [1, 2, 3, 4, 5, 6, 7, 7, 7]
s1 = [3,4]
B = removeSublistFromList(s1, A)
print(B)
s2 = [4,3]
C = removeSublistFromList(s2, A)
print(C)
Using numpy:
import numpy as np
A = [1, 2, 3, 4, 5, 6, 7, 7, 7]
B = [3,4]
C = [4,3]
list(np.setdiff1d(A,B, assume_unique=True))
output: [1, 2, 5, 6, 7, 7, 7]
list(np.setdiff1d(A,C, assume_unique=True))
output: [1, 2, 5, 6, 7, 7, 7]
