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The dictionary values created in the loop are overwritten

I am trying to write a function in python which counts the occurrences of letters in the text shifted by x characters. The function works and saves the counted results in the dictionary, but when I want to save this dictionary to another one, the values already saved in it are written by the new dictionary that I add.
Here's my code:
def calcLetter(message):
messageWithoutSpace = message.replace(" ", "")
shift = 6
alphabet = list(string.ascii_lowercase)
position = {}
tempList = {}
for pos in range(0, 6):
for letter in alphabet:
tempList[letter] = {}
textLenght = len(messageWithoutSpace)
tempCount = 0
letterIndex = 0
while textLenght > 0:
if len(messageWithoutSpace) - pos > letterIndex:
if messageWithoutSpace[pos + letterIndex] == letter:
tempCount += 1
letterIndex = letterIndex + shift
textLenght = textLenght - shift
tempList[letter] = tempCount
print("Start position:", pos)
print(tempList)
position[pos] = tempList
print(position)
When I check it for each index, the letters are well counted, but only the last calculations are saved in the final dictionary.
Result:
Start position: 0
{'a': 16, 'b': 1, 'c': 3, 'd': 1, 'e': 10, 'f': 2, 'g': 5, 'h': 5, 'i': 0, 'j': 11, 'k': 7, 'l': 4, 'm': 0, 'n': 9, 'o': 8, 'p': 7, 'q': 3, 'r': 0, 's': 4, 't': 0, 'u': 6, 'v': 7, 'w': 5, 'x': 2, 'y': 5, 'z': 3}
Start position: 1
{'a': 1, 'b': 6, 'c': 1, 'd': 0, 'e': 7, 'f': 20, 'g': 3, 'h': 0, 'i': 7, 'j': 4, 'k': 8, 'l': 2, 'm': 0, 'n': 4, 'o': 0, 'p': 6, 'q': 7, 'r': 10, 's': 3, 't': 5, 'u': 4, 'v': 17, 'w': 1, 'x': 0, 'y': 3, 'z': 4}
Start position: 2
{'a': 4, 'b': 6, 'c': 12, 'd': 6, 'e': 0, 'f': 9, 'g': 4, 'h': 3, 'i': 2, 'j': 0, 'k': 2, 'l': 0, 'm': 4, 'n': 9, 'o': 12, 'p': 2, 'q': 2, 'r': 8, 's': 6, 't': 0, 'u': 5, 'v': 1, 'w': 11, 'x': 2, 'y': 7, 'z': 6}
Start position: 3
{'a': 13, 'b': 6, 'c': 2, 'd': 5, 'e': 2, 'f': 11, 'g': 0, 'h': 2, 'i': 0, 'j': 11, 'k': 1, 'l': 1, 'm': 3, 'n': 2, 'o': 7, 'p': 12, 'q': 2, 'r': 0, 's': 8, 't': 11, 'u': 5, 'v': 3, 'w': 0, 'x': 8, 'y': 0, 'z': 8}
Start position: 4
{'a': 7, 'b': 0, 'c': 6, 'd': 8, 'e': 14, 'f': 0, 'g': 4, 'h': 3, 'i': 9, 'j': 2, 'k': 3, 'l': 1, 'm': 10, 'n': 0, 'o': 5, 'p': 4, 'q': 10, 'r': 7, 's': 9, 't': 0, 'u': 0, 'v': 6, 'w': 6, 'x': 9, 'y': 0, 'z': 0}
Start position: 5
{'a': 2, 'b': 0, 'c': 5, 'd': 4, 'e': 5, 'f': 5, 'g': 0, 'h': 4, 'i': 0, 'j': 7, 'k': 5, 'l': 9, 'm': 0, 'n': 7, 'o': 6, 'p': 16, 'q': 0, 'r': 2, 's': 3, 't': 14, 'u': 2, 'v': 3, 'w': 4, 'x': 1, 'y': 10, 'z': 9}
{0: {'a': 2, 'b': 0, 'c': 5, 'd': 4, 'e': 5, 'f': 5, 'g': 0, 'h': 4, 'i': 0, 'j': 7, 'k': 5, 'l': 9, 'm': 0, 'n': 7, 'o': 6, 'p': 16, 'q': 0, 'r': 2, 's': 3, 't': 14, 'u': 2, 'v': 3, 'w': 4, 'x': 1, 'y': 10, 'z': 9}, 1: {'a': 2, 'b': 0, 'c': 5, 'd': 4, 'e': 5, 'f': 5, 'g': 0, 'h': 4, 'i': 0, 'j': 7, 'k': 5, 'l': 9, 'm': 0, 'n': 7, 'o': 6, 'p': 16, 'q': 0, 'r': 2, 's': 3, 't': 14, 'u': 2, 'v': 3, 'w': 4, 'x': 1, 'y': 10, 'z': 9}, 2: {'a': 2, 'b': 0, 'c': 5, 'd': 4, 'e': 5, 'f': 5, 'g': 0, 'h': 4, 'i': 0, 'j': 7, 'k': 5, 'l': 9, 'm': 0, 'n': 7, 'o': 6, 'p': 16, 'q': 0, 'r': 2, 's': 3, 't': 14, 'u': 2, 'v': 3, 'w': 4, 'x': 1, 'y': 10, 'z': 9}, 3: {'a': 2, 'b': 0, 'c': 5, 'd': 4, 'e': 5, 'f': 5, 'g': 0, 'h': 4, 'i': 0, 'j': 7, 'k': 5, 'l': 9, 'm': 0, 'n': 7, 'o': 6, 'p': 16, 'q': 0, 'r': 2, 's': 3, 't': 14, 'u': 2, 'v': 3, 'w': 4, 'x': 1, 'y': 10, 'z': 9}, 4: {'a': 2, 'b': 0, 'c': 5, 'd': 4, 'e': 5, 'f': 5, 'g': 0, 'h': 4, 'i': 0, 'j': 7, 'k': 5, 'l': 9, 'm': 0, 'n': 7, 'o': 6, 'p': 16, 'q': 0, 'r': 2, 's': 3, 't': 14, 'u': 2, 'v': 3, 'w': 4, 'x': 1, 'y': 10, 'z': 9}, 5: {'a': 2, 'b': 0, 'c': 5, 'd': 4, 'e': 5, 'f': 5, 'g': 0, 'h': 4, 'i': 0, 'j': 7, 'k': 5, 'l': 9, 'm': 0, 'n': 7, 'o': 6, 'p': 16, 'q': 0, 'r': 2, 's': 3, 't': 14, 'u': 2, 'v': 3, 'w': 4, 'x': 1, 'y': 10, 'z': 9}}

This happens because python dictionaries are mutable. You are not making a copy of the dictionary by adding it to your list, you are just referencing the object (and later making changes). Because your list contains repeated references to the same object, the content of all list entries changes.
Try importing copy and changing position[pos] = tempList to position[pos] = copy.copy(tempList).

you need to copy tempList
>>> d = {1: '1', 2: '2'}
>>> l = [d, d.copy()]
>>> d[1] = '3'
>>> l
[{1: '3', 2: '2'}, {1: '1', 2: '2'}]

Vowel count for each line of txt file

Trying to count the number of vowels in each line of a text file. So far I have the following;
lc=0
lst=[]
vowel_count={}
with open('frankenstein.txt', 'r') as f:
for line in f:
no_of_characters=len(line)
l1=line.lower()
lc+=1
if no_of_characters !=1:
for vowel in "aeiou":
count=l1.count(vowel)
vowel_count[vowel]=count
print(lc, vowel_count)
print(lc, vowel_count)
I want the output to display the number of vowels and the corresponding line number, which is the purpose of the "lc" variable.
However when the code runs it gives the following output;
2128 {'a': 2, 'e': 6, 'i': 0, 'o': 0, 'u': 1}
2128 {'a': 2, 'e': 6, 'i': 5, 'o': 0, 'u': 1}
2128 {'a': 2, 'e': 6, 'i': 5, 'o': 4, 'u': 1}
2128 {'a': 2, 'e': 6, 'i': 5, 'o': 4, 'u': 3}
2129 {'a': 3, 'e': 6, 'i': 5, 'o': 4, 'u': 3}
2129 {'a': 3, 'e': 6, 'i': 5, 'o': 4, 'u': 3}
2129 {'a': 3, 'e': 6, 'i': 4, 'o': 4, 'u': 3}
2129 {'a': 3, 'e': 6, 'i': 4, 'o': 6, 'u': 3}
2129 {'a': 3, 'e': 6, 'i': 4, 'o': 6, 'u': 2}
2130 {'a': 3, 'e': 6, 'i': 4, 'o': 6, 'u': 2}
2130 {'a': 3, 'e': 5, 'i': 4, 'o': 6, 'u': 2}
2130 {'a': 3, 'e': 5, 'i': 2, 'o': 6, 'u': 2}
2130 {'a': 3, 'e': 5, 'i': 2, 'o': 10, 'u': 2}
2130 {'a': 3, 'e': 5, 'i': 2, 'o': 10, 'u': 3}
2131 {'a': 3, 'e': 5, 'i': 2, 'o': 10, 'u': 3}
2131 {'a': 3, 'e': 8, 'i': 2, 'o': 10, 'u': 3}
2131 {'a': 3, 'e': 8, 'i': 2, 'o': 10, 'u': 3}
2131 {'a': 3, 'e': 8, 'i': 2, 'o': 6, 'u': 3}
2131 {'a': 3, 'e': 8, 'i': 2, 'o': 6, 'u': 3}
2132 {'a': 3, 'e': 8, 'i': 2, 'o': 6, 'u': 3}
I get get several outputs for each line, how do i stop this?

You're printing status after each new found vowel. Move your print out of the inner for loop.

Adding value depending on dictionary value

Say I have a dictionary like this,
SCRABBLE_LETTER_VALUES = {
'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h': 4, 'i': 1, 'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3, 'q': 10, 'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y': 4, 'z': 10
}
Now, say I have a variable like this.
letter = 'i'
I want to add the corresponding value to the variable score. So since letter is equal to i, then score should equal 1

SCRABBLE_LETTER_VALUES = {'a': 1, 'c': 3, 'b': 3, 'e': 1, 'd': 2, 'g': 2, 'f': 4, 'i': 1, 'h': 4, 'k': 5, 'j': 8, 'm': 3, 'l': 1, 'o': 1, 'n': 1, 'q': 10, 'p': 3, 's': 1, 'r': 1, 'u': 1, 't': 1, 'w': 4, 'v': 4, 'y': 4, 'x': 8, 'z': 10}
>>> SCRABBLE_LETTER_VALUES['i']
1
>>> SCRABBLE_LETTER_VALUES['z']
10

Dictionary values are accessed using dictionary_name[key]. So in this case:
score+=SCRABBLE_LETTER_VALUES[letter]
You also need to assign score to something before you do this:
score=0
You might find it useful to read the documentation on dictionaries: https://docs.python.org/3/tutorial/datastructures.html#dictionaries

Firstly you need to know how python plays with dictionary- the question you asked in very simple thus people are downvoting!
Answering you question:
You have the following code in mind and struggling for the rest. I will be answering them accordingly under your statements.
SCRABBLE_LETTER_VALUES = {
'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h': 4, 'i': 1, 'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3, 'q': 10, 'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y': 4, 'z': 10
}
letter = 'i'
In python to get value for a corresponding key it work like dictionary[key], which gives the output.
So for your problem if we do SCRABBLE_LETTER_VALUES[letter], its the same as doing SCRABBLE_LETTER_VALUES['i'] where 'i' is the key and we will get 1 as output.
Hence for SCRABBLE_LETTER_VALUES[letter] we get 1 as output!
I want to add the corresponding value to the variable score. So since letter is equal to i
Assigning the corresponding value to the variable score is score = SCRABBLE_LETTER_VALUES[letter]
then score should equal 1
and we know now for sure that the value of score is 1. Don't we?
For SCRABBLE_LETTER_VALUES[letter] value is 1 when letter= 'i' thus score is 1

How to sum items in a for loop in python?

I have the following code:
dict = {
'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h': 4, 'i': 1,
'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3, 'q': 10,
'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y': 4, 'z': 10
}
word = 'banana'
for letter in word:
print dict[letter]
I get the following output
3
1
1
1
1
How can I add these values?
That is how can I print the output as 8

You can use sum and a generator expression:
>>> # Please don't name a variable `dict` -- it overshadows the built-in
>>> dct = { 'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h': 4, 'i': 1, 'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3, 'q': 10, 'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y': 4, 'z': 10 }
>>> word = 'banana'
>>> print sum(dct[letter] for letter in word)
8
>>>
Note that the above solution assumes that all of the characters in word can be found in dct. If this isn't always the case, then you can use dict.get to avoid a KeyError:
>>> dct = { 'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h': 4, 'i': 1, 'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3, 'q': 10, 'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y': 4, 'z': 10 }
>>> word = '$banana1'
>>> print sum(dct.get(letter, 0) for letter in word)
8
>>>

Here is another way that you can sum the items in the list.
dict = {
'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h': 4, 'i': 1,
'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3, 'q': 10,
'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y': 4, 'z': 10
}
word = 'banana'
sum = 0
for letter in word:
sum+=dict[letter]
print sum

comparing dicts with additional values

I have 2 lists of python dicts: realList and expectedList
I want realList to be considered equal to expectedList if there is one, and only one, "sub-dict" on realList for each dict of expectedList
By sub-dict I mean a dict with at least the same key/values, but which might have additional key/values.
so, for example:
realDict = [{'a': 1, 'b': 2}, {'a': 2, 'b': 3}]
==
expectedDict = [{'a': 1}, {'a': 2, 'b': 3}]
realDict = [{'a': 1, 'b': 2}, {'a': 2, 'b': 3}]
!=
expectedDict = [{'a': 2}, {'a': 2, 'b': 3}]
realDict = [{'a': 1, 'b': 2}, {'a': 1, 'b': 3}]
!=
expectedDict = [{'a': 1}]
Any modules to help? Only idea I can think of is iterating over expectedDict and removing a dict from it and from realDict whenever they match. Then, in the end, both must be [].

This works, but I'm unsure how performant it is:
def complete_subset(real_list, expected_list):
real_set_list = [set(d.items()) for d in real_list]
expected_set_list = [set(d.items()) for d in expected_list]
while len(real_set_list):
real_len = len(real_set_list)
i = 0
for real_set in real_set_list:
for expected_set in expected_set_list:
if not len(expected_set - real_set):
real_set_list.remove(real_set)
expected_set_list.remove(expected_set)
i = i + 1
if i == real_len:
break
return (not len(real_set_list)) and (not len(expected_set_list))
Here are my tests:
print complete_subset([{'a': 1, 'b': 2}, {'a': 2, 'b': 3}], [{'a': 1}, {'a': 2, 'b': 3}]) == True
print complete_subset([{'a': 1, 'b': 2}, {'a': 2, 'b': 3}], [{'a': 2}, {'a': 2, 'b': 3}]) == False
print complete_subset([{'a': 1, 'b': 2}, {'a': 2, 'b': 3}], [{'a': 1}]) == False
print complete_subset([{'a': 1, 'b': 2}, {'a': 1, 'b': 2}], [{'a': 1}, {'b': 2}]) == True
print complete_subset([
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
], [
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8, 'i': 9},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7, 'h': 8},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6, 'g': 7},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6},
{'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5},
{'a': 1, 'b': 2, 'c': 3, 'd': 4},
{'a': 1, 'b': 2, 'c': 3},
{'a': 1, 'b': 2},
{'a': 1},
{'a': 1, 'c': 3, 'd': 4, 'f': 6, 'g': 7, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'c': 3, 'e': 5, 'f': 6, 'h': 8, 'i': 9, 'j': 10},
{'a': 1, 'b': 2, 'd': 4, 'e': 5, 'g': 7, 'h': 8, 'j': 10},
]) == True

In case your two lists are to be compared element by element, you can do it this way.
def compare(realDict, expectedDict):
if len(readDict) != len(expectedDict):
return False
for d1, d2 in zip(realDict, expectedDict):
for key in d2:
if key not in d1 or d1[key] != d2[key]:
return False
return True