Computer Science
Predict the output:
text = "abracadabraaabbccrr"
counts = {}
ct = 0
lst = []
for word in text:
if word not in lst:
lst.append(word)
counts[word] = 0
ct = ct + 1
counts[word] = counts[word] + 1
print(counts)
print(lst)
Answer
Output
{'a': 7, 'b': 4, 'r': 4, 'c': 3, 'd': 1}
['a', 'b', 'r', 'c', 'd']
Explanation
This python program counts the frequency of each character in a string. Here is a step-by-step explanation of the program:
- Initialize the variables
- The
textvariable stores the input string "abracadabraaabbccrr". - The
countsvariable is a dictionary that stores the frequency of each character in the string. - The
ctvariable is a counter that keeps track of the total number of characters in the string. - The
lstvariable is a list that stores the unique characters in the string.
- The
- Loop through each character
wordin thetextstring- If the character has not been seen before, it is added to the
lstlist and a new key is added to thecountsdictionary with a value of 0. - The
ctvariable is incremented by 1. - The value of the character's key in the
countsdictionary is incremented by 1. This value keeps a count of the number of times this character has appeared in the string so far.
- If the character has not been seen before, it is added to the
- Finally, the
countsdictionary and thelstlist are printed to the console. Thecountsdictionary displays the frequency of each character in the string, and thelstlist displays the unique characters in the string.
Related Questions
Predict the output:
d = dict() d['left'] = '<' d['right'] = '>' print('{left} and {right} or {right} and {left}')Predict the output:
d = dict() d['left'] = '<' d['right'] = '>' d['end'] = ' ' print(d['left'] and d['right'] or d['right'] and d['left']) print(d['left'] and d['right'] or d['right'] and d['left'] and d['end']) print((d['left'] and d['right'] or d['right'] and d['left']) and d['end']) print("end")Predict the output:
list1 = [2, 3, 3, 2, 5,3, 2, 5, 1,1] counts = {} ct = 0 lst = [] for num in list1: if num not in lst: lst.append(num) counts[num] = 0 ct = ct+1 counts[num] = counts[num]+1 print(counts) for key in counts.keys(): counts[key] = key * counts[key] print(counts)Create a dictionary 'ODD' of odd numbers between 1 and 10, where the key is the decimal number and the value is the corresponding number in words.
Perform the following operations on this dictionary :- Display the keys
- Display the values
- Display the items
- Find the length of the dictionary
- Check if 7 is present or not
- Check if 2 is present or not
- Retrieve the value corresponding to the key 9
- Delete the item from the dictionary corresponding to the key 9