Python supports a list as its list element and hence a matrix can be formed. Sometimes we might have a utility in which we require to perform a search in that list of list i.e matrix and its a very common in all the domains of coding. Let’s discuss certain ways in which this can be performed.
Method #1 : Using any() + list comprehension
The any function can be used to perform the task of if condition and the check for each element in the nested list can be computed using the list comprehension.
# Python3 code to demonstrate # Search in Matrix # using any() + list comprehension # initializing list test_list = [[4, 5, 6], [10, 2, 13], [1, 11, 18]] # printing original list print("The original list : " + str(test_list)) # using any() + list comprehension # to Search in Matrix res = any(13 in sub for sub in test_list) # printing result print("Is 13 present in Matrix ? : " + str(res)) |
The original list : [[4, 5, 6], [10, 2, 13], [1, 11, 18]] Is 13 present in Matrix ? : True
Method #2 : Using set.issubset() + itertools.chain()
The issubset method can be used to check for the membership in sublist and chain function can be used to perform this task for the each element in the Matrix, in a faster way as it works on iterators.
# Python3 code to demonstrate # Search in Matrix # using set.issubset() + itertools.chain() from itertools import chain # initializing list test_list = [[4, 5, 6], [10, 2, 13], [1, 11, 18]] # printing original list print("The original list : " + str(test_list)) # using set.issubset() + itertools.chain() # to Search in Matrix res = {13}.issubset(chain.from_iterable(test_list)) # printing result print("Is 13 present in Matrix ? : " + str(res)) |
The original list : [[4, 5, 6], [10, 2, 13], [1, 11, 18]] Is 13 present in Matrix ? : True
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