Python – Remove Dictionary Key Words
Sometimes, while working with Python strings, we can have a problem in which we need to remove all the words from a string which are a part of key of dictionary. This problem can have application in domains such as web development and day-day programming. Let’s discuss certain ways in which this task can be performed.
Method #1 : Using split() + loop + replace()
The combination of above functions can be used to solve this problem. In this, we perform the task of converting string to list of words using split(). Then we perform a replace of word present in string with empty string using replace().
# Python3 code to demonstrate working of # Remove Dictionary Key Words# Using split() + loop + replace() # initializing stringtest_str = 'gfg is best for geeks' # printing original stringprint("The original string is : " + str(test_str)) # initializing Dictionarytest_dict = {'geeks' : 1, 'best': 6} # Remove Dictionary Key Words# Using split() + loop + replace()for key in test_dict: if key in test_str.split(' '): test_str = test_str.replace(key, "") # printing result print("The string after replace : " + str(test_str)) |
The original string is : gfg is best for geeks The string after replace : gfg is for
Method #2 : Using join() + split()
This is yet another way in which this task can be performed. In this, we reconstruct new string using join(), performing join by the empty string after split.
# Python3 code to demonstrate working of # Remove Dictionary Key Words# Using join() + split() # initializing stringtest_str = 'gfg is best for geeks' # printing original stringprint("The original string is : " + str(test_str)) # initializing Dictionarytest_dict = {'geeks' : 1, 'best': 6} # Remove Dictionary Key Words# Using join() + split()temp = test_str.split(' ')temp1 = [word for word in temp if word.lower() not in test_dict]res = ' '.join(temp1) # printing result print("The string after replace : " + str(res)) |
The original string is : gfg is best for geeks The string after replace : gfg is for
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