Not sure if i am using dictionaries properly

Question

I was trying to create a text analyzer with both dictionary and string formatting, but i seem to have gotten an error at the second last statement. It says:

line 25 in <module>
msg = ("{0} = {1}%".format(letters, dict(letters)))
ValueError: dictionary update sequence element #0 has length 1; 2 is required.

How do i solve this:

#text analyzer
#starting of the program
textfile = input("Enter text document:")
with open(textfile) as f:
text = f.read()
print(text)

def count_char(text, char):
count = 0
dict = {}
for c in text:
if c == char:
count += 1
dict[char] = count*100/len(text)

characters = ["abcdefghijklmnopqrstuvwxyz"]
text = text.lower()
for letters in characters:
count_char(text, str(letters))
msg = ("{0} = {1}%".format(letters, dict(letters)))
print(msg)

EDIT:
text = "Pretend that this is text in a file"
print(text)

def count_char(text, char):
count = 0
dict = {}
for c in text:
if c == char:
count += 1
dict[char] = count*100/len(text)

characters = ["abcdefghijklmnopqrstuvwxyz"]
text = text.lower()
for letters in characters:
count_char(text, str(letters))
msg = ('{0} = {1}%'.format(letters, dict[letters]))

Answer 1

# There are several mistakes:

# in count_char() function, don't use a local dict if you want to access it from outside (your latest letters loop)...
# rather than handle a global var, let it return the count:

def count_char(text, char):
count = 0
for c in text:
if c == char:
count += 1
return count*100/len(text)

# definition of list of char doesn't need to be written inside squared brackets, as string are already a kind of list:
characters = "abcdefghijklmnopqrstuvwxyz" # else you later iterate over the unque string in the list ^^

text = text.lower()
for letters in characters:
c = count_char(text, letters) # no need for str(), as letters is already of string type (once char string)
msg = "{0} = {1}%".format(letters, c) # use the returned value of count_char() rather than trying to use a bad used dict ;)
print(msg)

Answer 2

Dont use dict(letters), but dict[letters] (squared brackets, not rounded ones are used to access a key-index value)...

By the way, I cannot run your code, as you use external file(s) ^^

Answer 3

When i tried that out, i got a different error:
"TypeError: 'type' object is not subscriptable"

I don't understand what i am doing wrong
I just edited the code, to run without external files.