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Python: Convert a string to all uppercase if it contains at least 2 uppercase characters in the first 4 characters

Python String: Exercise-21 with Solution

Write a Python function to convert a given string to all uppercase if it contains at least 2 uppercase characters in the first 4 characters.

Python String Exercises: Convert a string to all uppercase if it contains at least 2 uppercase characters in the first 4 characters

Sample Solution:-

Python Code:

def to_uppercase(str1):
    num_upper = 0
    for letter in str1[:4]: 
        if letter.upper() == letter:
            num_upper += 1
    if num_upper >= 2:
        return str1.upper()
    return str1

print(to_uppercase('Python'))
print(to_uppercase('PyThon'))

Sample Output:

Python                                                                                                        
PYTHON

Flowchart:

Flowchart: Convert a string to all uppercase if it contains at least 2 uppercase characters in the first 4 characters

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Python: Tips of the Day

How to make a flat list out of list of lists?

Given a list of lists l

flat_list = [item for sublist in l for item in sublist]

which means:

flat_list = []
for sublist in l:
    for item in sublist:
        flat_list.append(item)

is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:

flatten = lambda l: [item for sublist in l for item in sublist]

As evidence, you can use the timeit module in the standard library:

$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

Ref: https://bit.ly/3dKsNTR