How could this be done easier? My output is correct and the code is working fine, but it is not very efficient.

Question

https://code.sololearn.com/cguxGjDsJ96g/?ref=app

Answer 1

Vic ten Bokum
The following code is also a more efficient solution:

text = input()
my_dict = {}

for t in text:
my_dict.update({t:text.count(t)})

print(my_dict)

In the above code, the update function updates 'my_dict' with key:value pairs. As the for loop is iterating through 'text', my_dict is simultaneously being updated with the current value of 't' as the key, and the amount of times 't' occurs in 'text', as the value. As you can see, the update function is used to update a dictionary with key:value pairs.

Answer 2

Python can be super short, and very useful for that are the so-called "comprehensions".

dictio = {letter: text.count(letter) for letter in text}

This line alone creates the full dictionary and thereby replaces lines 6 to 29.

Here you can read more about the topic:

https://code.sololearn.com/ck6HicJ6Z8jG/?ref=app

Answer 3

Use the stored digit as a counter because your program only works if the input has 10 repeated letters. So, what you want to do is:
if you find a new letter, its value must be 1, else, if the letter is repeated, increment the stored value.

That sould work.

Answer 4

HonFu that's amazing, thank you!

Answer 5

What are you trying to do? If you're trying to count the number of characters of a word/words entered by the user, you can use for loop, or len() function, both ways would be correct.

for loop:

https://code.sololearn.com/cTTS7EoNGwaP/?ref=app

len() function:

https://code.sololearn.com/cwFJADdKlxBa/?ref=app

Answer 6

Lamron I am trying to have a dictionary as output that has the letters of the input as keys and the number of times those letters are in the input string as values. The output of my code is exactly how I want it to be, but I was wondering if it coud be done in a different, more efficient, way.

Answer 7

I don't know how to do this, soz 🤷‍♂️

Answer 8

Edgar Sabido Thanks, I'll try that :)

Answer 9

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