Python: Check whether a given string is an "isogram" or not

Python Basic - 1: Exercise-85 with Solution

From Wikipedia:
An isogram (also known as a "nonpattern word") is a logological term for a word or phrase without a repeating letter. It is also used by some people to mean a word or phrase in which each letter appears the same number of times, not necessarily just once. Conveniently, the word itself is an isogram in both senses of the word, making it autological.
Write a Python program to check whether a given string is an "isogram" or not.

Sample Solution:

Python Code:

def check_isogram(str1):
    return len(str1) == len(set(str1.lower()))


Sample Output:


Pictorial Presentation:

Python: Check whether a given string is an 'isogram' or not.
Python: Check whether a given string is an 'isogram' or not.
Python: Check whether a given string is an 'isogram' or not.


Flowchart: Python - Check whether a given string is an 'isogram' or not.

Python Code Editor:

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Previous: Write a Python program that accepts a list of numbers. Count the negative numbers and compute the sum of the positive numbers of the said list. Return these values through a list.
Next: Write a Python program to count the number of equal numbers from three given integers.

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Python: Tips of the Day

Iterating over dictionaries using 'for' loops:

I am a bit puzzled by the following code: d = {'x': 1, 'y': 2, 'z': 3} for key in d: print key, 'corresponds to', d[key] What I don't understand is the key portion. How does Python recognize ...

key is just a variable name.

for key in d:

For Python 3.x:

for key, value in d.items():

For Python 2.x:

for key, value in d.iteritems():

To test for yourself, change the word key to poop.

In Python 3.x, iteritems() was replaced with simply items(), which returns a set-like view backed by the dict, like iteritems() but even better. This is also available in 2.7 as viewitems().

The operation items() will work for both 2 and 3, but in 2 it will return a list of the dictionary's (key, value) pairs, which will not reflect changes to the dict that happen after the items() call. If you want the 2.x behavior in 3.x, you can call list(d.items()).

Ref: https://bit.ly/37dm0Qo