Question

EncryptedA.txt:

vcbdwvcbwqarr, c xdwzavwqarr u juwqx uwrr cjcm vcbdgzum bkxgmq. yw rrcbtjaqurr qazamgcr u'x wml iwgxgcmb bkxgmq, za'm yykxygujg'x dam vcw fayxgygcrkxum am qcjjk qlmwkr. am azbur ax cgj xaywj tar, xuwrr am qlwgbdgu av vdcxf tjwbfdjwa, a qcmujycm txarwgmgq cx qaywx buxxg'x furck c rrwymarrglar cx qaywx faycbdxwtk vgjlxuj. raywgzgurr zclj bwfdmwq u ruxxg furck cjvcwmgq, qcm qammlaz iwgxgcmb u'x wml a tuvtw c uwrr am fcmyur rwlgzgcrck iwgxgcmb wmgqvc. cx uj a xdaywj, tk'm qlwgbdgu am a jctuxra yygzwquj fwmwrjcwbduj, cf amc av vdxgyazquj vcmfwgmgum jjw'x uwrr am qlwgbdgu cx ywrrcjlwrr cx qaywx vcxf g vcmfwgmgum, km u'x qlgx-qayxgygcrkxum fambcy am a tar. yw'g qcylar am wkuq u "lwgbdxwruwrr cmlwrrkz raqm" lwrg grru qaycrrwy fcwj iwxbdamcz xdalguj qarc ram av vcmfwgmgum. yw'g xdurrlar cx txuygcmmcwbd ixcly c quxyurlar g rrwxtam bxgmgcwbd duxvumcgrr. rcxicxurr a ixgy lwgmgruq quxrum txulm avrrgdwkxgcr zlarruquj u yjamarruwrr cx uj vcxlujcwbd bkxgmq, cv a vurr "qlcxbdkz" a fcyurr wg rxgm. ralwrurr txulm wg yur am "damur yjgm" cv a rxgmgcwbd c qcyurr bkxgmq, lwrg grru rrwxtam rwgzwt am quyam cv avrrgdwkxgcr. tk ycxl cx uj tlabc cycj qarc facmgr amrru. batgc jclwx vcg dkmcmjcrrgcr aruwrr.

I am making a dictionary list of alphabets that counts the number of letters. I almost have the code correct however, I am not able to ignore punctuation and give a value of zero to a letter that does not appear in the text

• your output should contain the key:value pairs 'a': 78 and 'b': 31.
• Ignore punctuation. Punctuation counts must not appear in your dict
• If a given letter does not appear in the text, there must be a key:value pair with value 0.

Example of correct code:

a: 78

b : 31

c,88

d,28

e,0

f,18

etc

import string

encryptedA = open("encryptedA.txt")

encryptedA = encryptedA.read()

def abcfreq(lettervalues):

   

    freq = {}

    

    for j in lettervalues:

        if j in string.punctuation:

            pass

        elif j in freq:

            freq[j] += 1

        else:

            freq[j] = 1

    for key, value in sorted(freq.items()):

         print("%s,%d" % (key, value))

    

x = abcfreq(encryptedA)

print(x)

Answer 1

Changed Code in python

import string

encryptedA = open("encryptedA.txt")
encryptedA = encryptedA.read()

def abcfreq(lettervalues):
    freq = {}
    #lets add all the smlaa letters with count 0 in our dictionary
    for x in string.ascii_lowercase:
        freq[x]=0

    #If the character exists in dictionary then only we will increment the count
    for j in lettervalues:
        if j in freq:
            freq[j] += 1
    # we dont even need to sort now
    for key, value in freq.items():
         print("%s,%d" % (key, value))

x = abcfreq(encryptedA)

New Output

a,78
b,31
c,88
d,28
e,0
f,18
g,78
h,0
i,7
j,36
k,22
l,32
m,76
n,0
o,0
p,0
q,41
r,114
s,0
t,19
u,70
v,27
w,76
x,72
y,40
z,16

Explanation

• Using string.ascii_lowercase we can fill in our dictionary with all small letters with value 0 initially.
• After this while reading character by character we only increment the count if that character is in out dictionary. Hence we ignore all non-characters.

Let me know in comments if you have any doubts. Do leave a thumbs up if this was helpful.