For Class 11 & above only.
Predict the output of following Co de fragment .
fruit = { }
f1 = ['Apple', 'Banana', 'apple', 'Banana']
for index in fl:
if index in fruit:
fruit[index] += 1
else:
fruit[index] = 1
print(fruit)
print (len(fruit))
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Answers
Corrected Code.
fruit = {}
f1 = ['Apple', 'Banana', 'apple', 'Banana']
for index in f1:
if index in fruit:
fruit[index] += 1
else:
fruit[index] = 1
print(fruit)
print (len(fruit))
Output.
{'Apple': 1, 'Banana': 2, 'apple': 1}
3
Explanation.
We are given with a dictionary named fruit and a list named f1.
After first iteration of loop -
> index = 'Apple' (The first element of list)
Now, check the condition given -
> index in fruit - False as dictionary is empty.
As condition is false, the statements inside else block will be executed.
> fruit[index] = 1
Dictionary becomes - {'Apple' : 1}
After second iteration of loop -
> index = 'Banana'
Now, check the condition given -
> index in fruit - False, its not present.
As condition is false, the statements inside the else block will be executed.
> fruit[index] = 1
Dictionary becomes - {'Apple' : 1, 'Banana' : 1}
After third iteration of loop -
> index = 'apple'
Now, check the condition given -
> index in fruit = False, not present (Note that Python is case-sensitive)
As condition is false, the statements inside the else block will be executed.
> fruit[index] = 1
Dictionary becomes - {'Apple' : 1, 'Banana' : 1, 'apple' : 1}
After fourth iteration of loop -
> index = 'Banana'
Now, check the condition given -
> index in fruit - True, its present.
As condition is true, the statements inside the if block will be executed.
> fruit[index] += 1
Value of 'key' is not changed but 'value' is incremented by 1.
Dictionary becomes - {'Apple': 1, 'Banana': 2, 'apple': 1}
Loop is now terminated.
Now, the dictionary fruit as well as its length is printed. So, output -
{'Apple': 1, 'Banana': 2, 'apple': 1}
3
Answer:
tup1=(1,2,3,4,5,”a”,”b”)
tup1=[1:4]
- Output for the following code will be: