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How to iterate over a string and for each letter, if it is in a dictionary (key), add values to a new list?

What would I write to iterate over a string and for each letter, if that letter is in a given dictionary (as a key), then add that key's value to an empty list?
For example the string 'cat', I want to search a given dictionary for c, a, and t. Then for each value of each letter, add them to a new list that is a sum of points (values).
I keep getting: TypeError: 'NoneType' object is not iterable
Here's what I have so far:
user_word = cat
values_dict = {'A' : 1, 'E' : 1, 'I' : 1, 'O' : 1, 'U' : 1, 'L' : 1,\
'N' : 1, 'N' : 1, 'S' : 1, 'T' : 1, 'R' : 1,\
'D' : 2, 'G' : 2, 'B' : 3, 'C' : 3, 'M' : 3, 'P' : 3,\
'F' : 4, 'H' : 4, 'V' : 4, 'W' : 4, 'Y' : 4,\
'K' : 5, 'J' : 8, 'X' : 8, 'Q' : 10, 'Z' : 10,}
def get_word_value(user_word, values_dict):
word_pts = 0
for letter in user_word.upper():
for key, value in values_dict.items:
for letter in range(value):
word_pts = word_pts + values_dict[letter]
print(word_pts, 'TEST GOOD')
return word_pts
# expected result is 5 (3+1+1= 5)

Here's a one liner:
sum([values_dict[letter] for letter in user_word.upper()])

I'd suggest you to use
sum([values_dict.get(letter, 0) for letter in user_word.upper()])
This will avoid error if the key is not found.

The TypeError: 'NoneType' object is not iterable is due to line
for key, value in values_dict.items:.
dict.items is a function and not an iterable. To access the actual values, use the parenthesis to iter over the key and values: for key, value in values_dict.items():

Here is your code corrected :
user_word = "cat"
values_dict = {'A' : 1, 'E' : 1, 'I' : 1, 'O' : 1, 'U' : 1, 'L': 1,
'N' : 1, 'S' : 1, 'T' : 1, 'R' : 1,
'D' : 2, 'G' : 2, 'B' : 3, 'C' : 3, 'M' : 3, 'P': 3,
'F' : 4, 'H' : 4, 'V' : 4, 'W' : 4, 'Y' : 4,
'K' : 5, 'J' : 8, 'X' : 8, 'Q' : 10, 'Z' : 10}
def get_word_value(user_word, values_dict):
word_pts = 0
for letter in user_word.upper():
for key, value in zip(values_dict.keys(), values_dict.values()):
if letter == key:
word_pts += value
print('Word pts : {} , TEST GOOD'.format(word_pts))
return word_pts

You can use the operator itemgetter(). It‘s faster than a list comprehension:
from operator import itemgetter
d = {'A': 1, 'T': 1, 'C': 3, 'E': 1}
w = 'cat'
sum(itemgetter(*w.upper())(d))
# 5

user_input = raw_input("Enter Your Word: ")
user_input = user_input.upper()
wordArray = list(user_input)
values_dict = {'A' : 1, 'E' : 1, 'I' : 1, 'O' : 1, 'U' : 1, 'L' : 1,
'N' : 1, 'N' : 1, 'S' : 1, 'T' : 1, 'R' : 1,
'D' : 2, 'G' : 2, 'B' : 3, 'C' : 3, 'M' : 3, 'P' : 3,
'F' : 4, 'H' : 4, 'V' : 4, 'W' : 4, 'Y' : 4,
'K' : 5, 'J' : 8, 'X' : 8, 'Q' : 10, 'Z' : 10,}
def get_word_value(wordArray, values_dict):
word_pts = 0
for letter in wordArray:
word_pts = word_pts + values_dict[letter]
print "Word Points: %d, Test Good." % word_pts
get_word_value(wordArray, values_dict)

Dictionary of punctuation counts for list of strings

How can I use dict comprehension to build a dictionary of punctuation counts for a list of strings? I was able to do it for a single string like this:
import string
test_string = "1990; and 1989', \ '1975/97', '618-907 CE"
counts = {p:test_string.count(p) for p in string.punctuation}
Edit: For anyone who may want this in the future, here's Patrick Artner's answer copied from below, with a very small modification to keep only punctuation counts:
# return punctuation Counter dict for string/list/pd.Series
import string
from collections import Counter
from itertools import chain
def count_punctuation(str_series_or_list):
c = Counter(chain(*str_series_or_list))
unwanted = set(c) - set(string.punctuation)
for unwanted_key in unwanted: del c[unwanted_key]
return c

Why count yourself?
import string
from collections import Counter
test_string = "1990; and 1989', \ '1975/97', '618-907 CE"
c = Counter(test_string) # counts all occurences
for p in string.punctuation: # prints the one in string.punctuation
print(p , c[p]) # access like dictionary (its a subclass of dict)
print(c)
Output:
! 0
" 0
# 0
$ 0
% 0
& 0
' 4
( 0
) 0
* 0
+ 0
, 2
- 1
. 0
/ 1
: 0
; 1
< 0
= 0
> 0
? 0
# 0
[ 0
\ 1
] 0
^ 0
_ 0
` 0
{ 0
| 0
} 0
~ 0
Counter({'9': 7, ' ': 6, '1': 4, "'": 4, '7': 3, '0': 2, '8': 2, ',': 2, ';': 1, 'a': 1, 'n': 1, 'd': 1, '\\': 1, '5': 1, '/': 1, '6': 1, '-': 1, 'C': 1, 'E': 1})
Counter is dictionary-like: see https://docs.python.org/2/library/collections.html#collections.Counter
Edit: multiple strings in a list:
import string
from collections import Counter
from itertools import chain
test_strings = [ "1990; and 1989', \ '1975/97', '618-907 CE" , "someone... or no one? that's the question!", "No I am not!"]
c = Counter(chain(*test_strings))
for p in string.punctuation:
print(p , c[p])
print(c)
Output: (removed 0-Entries)
! 2
' 5
, 2
- 1
. 3
/ 1
; 1
? 1
\ 1
Counter({' ': 15, 'o': 8, '9': 7, 'n': 6, "'": 5, 'e': 5, 't': 5, '1': 4, 'a': 3, '7': 3, 's': 3, '.': 3, '0': 2, '8': 2, ',': 2, 'm': 2, 'h': 2, '!': 2, ';': 1, 'd': 1, '\\': 1, '5': 1, '/': 1, '6': 1, '-': 1, 'C': 1, 'E': 1, 'r': 1, '?': 1, 'q': 1, 'u': 1, 'i': 1, 'N': 1, 'I': 1})

counting individual characters in a large list of passwords without killing loop

I'm trying to get a count on, for example, how many a's are in a massive list of passwords out of curiosity. but i think when i'm trying to add a count for a character it's killing the loop that is going through all the characters.
#Examine passwords.txt
file = open('passwords.txt','r')
a = 0
b = 0
c = 0
d = 0
e = 0
f = 0
g = 0
h = 0
i = 0
j = 0
k = 0
l = 0
m = 0
n = 0
o = 0
p = 0
q = 0
r = 0
s = 0
t = 0
u = 0
v = 0
w = 0
x = 0
y = 0
z = 0
with open('passwords.txt','r') as fileobj:
for line in fileobj:
for char in line:
if char == a:
a += 1
elif char == b:
b += 1
print(a)
print(b)
print(c)
print(d)
print(e)
print(f)

You should use a dictionary (or a list which has an order) to store how many counts of each letter there are. This is much better than using 26 one letter variables which is ridiculous!
To create a dictionary, you can use a dictionary-comprehension with str.count on the entire contents of the file.
with open('passwords.txt','r') as fileobj:
text = fileobj.read()
letterCounts = {c: text.count(c) for c in "abcdefghijklmnopqrstuvwxyz"}
which would give letterCounts as something like:
{'s': 0, 'a': 4, 'o': 0, 'i': 0, 'm': 0, 'k': 0, 'q': 0, 'y': 0, 'c': 1, 'j': 0, 'b': 4, 'g': 0, 'd': 1, 'h': 0, 'e': 0, 'f': 0, 'u': 0, 'n': 0, 'w': 0, 't': 0, 'x': 0, 'p': 0, 'l': 0, 'r': 0, 'z': 0, 'v': 0}

from collections import Counter
with open('passwords.txt','r') as fileobj:
counts = Counter()
for line in fileobj:
counts.update(line)
counts is a Counter keeping track of the counts of all the characters that appear in the file. YOu would access the number of as with counts['a']

How to separate upper and lower case with counter?

I am thinking of something with collections
s = 'Hello Mr. Rogers, how are you this fine Tuesday?'
import collections
c = collections.Counter(s)
As a result I have
Counter({' ': 8,
',': 1,
'.': 1,
'?': 1,
'H': 1,
'M': 1,
'R': 1,
'T': 1,
'a': 2,
'd': 1,
'e': 5,
'f': 1,
'g': 1,
'h': 2,
'i': 2,
'l': 2,
'n': 1,
'o': 4,
'r': 3,
's': 3,
't': 1,
'u': 2,
'w': 1,
'y': 2})
If I try sum I got syntax problem
print sum(1 for i in c if i.isupper())
File "<ipython-input-21-66a7538534ee>", line 4
print sum(1 for i in c if i.isupper())
^
SyntaxError: invalid syntax
How should I count only upper or lower from the counter?

You lack the () in your generator expresion:
sum((1 for x in c if x.isupper()))
4
EDIT: As #Błotosmętek sugest, you lack the () in your print, i guess you are using python3, you should use print()

You can try something like this:
import collections
s = 'Hello Mr. Rogers, how are you this fine Tuesday?'
c = collections.Counter([ch for ch in s if ch.isupper()])
# Change to ch.islower() if you need lower case
# c = collections.Counter([ch for ch in s if ch.islower()])
print(c)

Python - Updating dictionary values after a .count() function while iterating

I'm having a problem with counting letters in a string and then updating this value to a dictionary.
I am iterating through a dictionary of alphabet letters and using the .count() function at each pair to compare to a sentence.
The count is working and returns the number you would expect (in this case xval), however this is not updating the dictionary when I use dic[k] = xval.
Infuriatingly simple. Please offer some advice if you can.
import string
type = ("cat sat on the mat").lower()
dic = dict.fromkeys(string.ascii_lowercase, 0)
print(dic)
for k, v in dic.items():
xval = type.count(k)
print(xval)
dic[k] = xval
print ("%s: %s" % (k, v))
And here is the output for completion. Many thanks.
{'m': 0, 'l': 0, 'h': 0, 'f': 0, 'd': 0, 'x': 0, 'i': 0, 's': 0, 'r': 0, 'u': 0, 'z': 0, 't': 0, 'c': 0, 'a': 0, 'q': 0, 'p': 0, 'j': 0, 'n': 0, 'g': 0, 'w': 0, 'o': 0, 'e': 0, 'k': 0, 'b': 0, 'v': 0, 'y': 0}
0
f: 0
0
q: 0
0
r: 0
1
m: 0
1
c: 0
3
a: 0
0
u: 0
0
g: 0
1
e: 0
0
d: 0
0
v: 0
1
h: 0
0
y: 0
0
k: 0
4
t: 0
0
i: 0
1
o: 0
0
w: 0
0
b: 0
1
s: 0
0
l: 0
0
j: 0
0
x: 0
0
z: 0
1
n: 0
0
p: 0
Process finished with exit code 0

You are printing k and v which do not reflect the updated dictionary.
import string
type = ("cat sat on the mat").lower()
dic = dict.fromkeys(string.ascii_lowercase, 0)
print(dic)
for k in dic:
xval = type.count(k)
print(xval)
dic[k] = xval
print k, ":", dic[k]
outputs
{'a': 0, 'c': 0, 'b': 0, 'e': 0, 'd': 0, 'g': 0, 'f': 0, 'i': 0, 'h': 0, 'k': 0, 'j': 0, 'm': 0, 'l': 0, 'o': 0, 'n': 0, 'q': 0, 'p': 0, 's': 0, 'r': 0, 'u': 0, 't': 0, 'w': 0, 'v': 0, 'y': 0, 'x': 0, 'z': 0}
3
a : 3
1
c : 1
0
b : 0
1
e : 1
0
d : 0
0
g : 0
0
f : 0
0
i : 0
1
h : 1
0
k : 0
0
j : 0
1
m : 1
0
l : 0
1
o : 1
1
n : 1
0
q : 0
0
p : 0
1
s : 1
0
r : 0
0
u : 0
4
t : 4
0
w : 0
0
v : 0
0
y : 0
0
x : 0
0
z : 0