count characters

Question

Yes it is my homework.
I have to write a program the coounts the occurency of each character of a given word.
word = input("please enter a word")
Georg
for c in word:
#now lets say c is G
G=1
... E=1 O=1 R=1..
but now G=1 is wrong and how can I check that I already have a variable G?

Answer 1

Using dictionary to keep count of how many times a variable appeared in a word

Answer 2

To get the appropriate result, string method “upper()” should be used to get G = 2 and not G = 1 , g = 1

Answer 3

https://code.sololearn.com/cxcK635842F2/?ref=app

Answer 4

wait... my effort... but it is exactly that

Answer 5

Abhay I needed the hint to do it with dictionary.
With a cheatsheet on my desktop I can code simple programs. If one tells me to do with a list, I can often do it.
But list,set,dictionary where to use when is a problem to me.
Anyways I can do it now.
And also that other 4 homeworks.
If I have to find all successors characters of a character in a text than it is again a dictionary with key character and value a list..,,right?

Answer 6

yes

Answer 7

Abhay look it did it.

Answer 8

loop through a set of the word.

Answer 9

GeorgeT Nice one ,I had one solution but it has way to many lines than that ,ty for showing me other way to do it. :-)

Answer 10

Use for loop and dict

Answer 11

GeorgeT that's not a try ,you are doing nothing in that code ,it's not that difficult if you know which letter is going to appear in the word ,like
count1=0
count2=0
for i in word
If i==o:
coun1+=1
Elif i==r:
count2+=1
but then you have to print seperate variables and know which variable is for which letter ,best is to use dictionary

Answer 12

rodwynnejones oh ...but why does it not show an error?
I have G two times.
https://code.sololearn.com/cxcK635842F2/?ref=app

Answer 13

Abhay lol... you are welcome.
i tried count[i] +=1 because I didn't understand fully your code but got a key error.
That was my idea for that.

Answer 14

As i mentioned previously...
word = input("please enter a word")
for x in set(word):
print(x, '=', word.count(x), end=', ')

Answer 15

Dr. Exceptional yes good point.