[SOLVED] What am I doing wrong - letter counter exercise in intermediate python

Question

**This is the exercise:**

Given a string as input, you need to output how many times each letter appears in the string.
You decide to store the data in a dictionary, with the letters as the keys, and the corresponding counts as the values.
Create a program to take a string as input and output a dictionary, which represents the letter count.

Sample Input:
hello

Sample Output:
{'h': 1, 'e': 1, 'l': 2, 'o': 1}

**Here's my code:**

text = input()
dict = {}
count = 0
for x in text:
count += 1
dict[x] = count
print(dict)

But it's not counting how many times each letter appears, it's saying where in the string the last instance of each letter appears.

How do I fix this?

(Not looking for other solutions, I've already found some great ones - such as this https://code.sololearn.com/cZw2842qky59/?ref=app - just wanting to work through this approach for my own understanding.)

Answer 1

# without module import and use of implicit nested loops implied by call to count() method:

text = input()
dict = {}
for char in text:
if char in dict:
dict[char] += 1
else:
dict[char] = 1
print(dict)

# shortest and most efficient (most pythonic) solution would be to use built in module collections:

from collections import Counter

print(dict(Counter(input())))

Answer 2

Try my simple solution!

text = input()
dict = {}

for t in text:
dict[t] = text.count(t)

print(dict)

Answer 3

In your code ,count is incremented on every iteration over text , so basically you end up with dictionary as last position of each letter in text .

What you should do is check if letter exists as key in dictionary or not .

=>If it exists update the value for that key by 1 .
=>if it doesn't exists assign 1 to your new key .

eg.
for x in text:
if x in dict:
dict[x]+=1
else:
dict[x]=1

Answer 4

text = input()
dict = {}

for i in text.lower():
dict[i] = text.count(i)
print(dict)

Answer 5

text = input()
dict = {x: text.count(x) for x in text}
print(dict)

Answer 6

text = input("> ")
dict = {}
for letter in text:
dict.update({letter: text.count(letter)})
print(dict)

very similar to jacky lei code

Answer 7

text = input()
dict = {}
count = 0
for x in text:
count += 1
dict[x] = text.count(x)

Print(dict)

Answer 8

def letter_count(text, letter):
counters = 0

for char in text:
if (char == letter):
counters = counters +1

return counters
#your code goes here

text = input()
letter = input()

print(letter_count(text, letter))

Answer 9

Check this out-
This works

text = input()
dict = {}
#your code goes here

for n in text:
keys = dict.keys()
if n in keys:
dict[n] += 1
else:
dict[n] = 1

print(dict)

Answer 10

text = input()
dict = {}

for t in text:
dict[t] = text.count(t)

print(dict)

This is Correct answer

Answer 11

Abhay and visph - thank you!

Answer 12

text = input()
dict = {}
#your code goes here

for i in text:
dict[i] = text.count(i)
print(dict)

Answer 13

Another way to go for it:

text = input()
dict = {}
#your code goes here
for character in text:
if character.isalpha():
dict[character] = text.count(character)
text.replace(character,"")

print(dict)

Answer 14

text = input()
dict = {}
for x in text:
if x in dict:
dict[x] += 1
else:
dict[x] = 1
print(dict)
print(dict(Counter(input())))

#letter_count #python

Answer 15

text = input()
dict = {}

for i in text.lower():
dict[i] = text.count(i)
print(dict)

Answer 16

text = input()
dict = {}
for char in text:
if char in dict:
dict[char] += 1
else:
dict[char] = 1
print(dict)

Answer 17

text = input()
dict = {}

for i in text:
dict[i]=text.count(i)
print(dict)

this is the simplest solution i could think of.

Answer 18

My :

text = input()
dict = {}
#your code goes here
for x in text:
dict[x] = text.count(x)
print(dict)

!!! Try before !!!

Answer 19

correct answer is

def letter_count(text, letter):
counters = 0

for char in text:
if (char == letter):
counters = counters +1

return counters
#your code goes here

text = input()
letter = input()

print(letter_count(text, letter))

Answer 20

Use code:
text = input()
dict = {}
for char in text:
if char in dict:
dict[char] += 1
else:
dict[char] = 1
print(dict)

input:- hello