Python – Remove Record if Nth Column is K
Sometimes while working with list of records, we can have a problem in which we need to perform removal of records on the basis of presence of certain element at Nth position of record. Lets discuss certain ways in which this task can be performed.
Method #1 : Using loop
This is brute force method in which this task can be performed. In this, we iterate for each element in list and exclude it if its Nth column is equal to K.
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# Python3 code to demonstrate # Remove Record if Nth Column is K# using loop # Initializing listtest_list = [(5, 7), (6, 7, 8), (7, 8, 10), (7, 1)] # printing original listprint("The original list is : " + str(test_list)) # Initializing K K = 7 # Initializing N N = 1 # Remove Record if Nth Column is K# using loopres = []for sub in test_list: if (sub[N] != K): res.append(sub) # printing result print ("List after removal : " + str(res)) |
The original list is : [(5, 7), (6, 7, 8), (7, 8, 10), (7, 1)] List after removal : [(7, 8, 10), (7, 1)]
Method #2 : Using list comprehension
This is yet another way in which this task can be performed. In this, we perform this task in similar way as above but in one liner form.
# Python3 code to demonstrate # Remove Record if Nth Column is K# using list comprehension # Initializing listtest_list = [(5, 7), (6, 7, 8), (7, 8, 10), (7, 1)] # printing original listprint("The original list is : " + str(test_list)) # Initializing K K = 7 # Initializing N N = 1 # Remove Record if Nth Column is K# using list comprehensionres = [sub for sub in test_list if sub[N] != K] # printing result print ("List after removal : " + str(res)) |
The original list is : [(5, 7), (6, 7, 8), (7, 8, 10), (7, 1)] List after removal : [(7, 8, 10), (7, 1)]
