CODEWITHSUNDEEP
python
oneman
oneman

Reputation: 811

How to use multiple 'if' statements nested inside an enumerator?

I have a massive string of letters all jumbled up, 1.2k lines long. I'm trying to find a lowercase letter that has EXACTLY three capital letters on either side of it.

This is what I have so far

def scramble(sentence):
    try:
        for i,v in enumerate(sentence):
            if v.islower():
                if sentence[i-4].islower() and sentence[i+4].islower():
    ....
    ....
    except IndexError:
        print() #Trying to deal with the problem of reaching the end of the list


#This section is checking if 
the fourth letters before 
and after i are lowercase to ensure the central lower case letter has
exactly three upper case letters around it

But now I am stuck with the next step. What I would like to achieve is create a for-loop in range of (-3,4) and check that each of these letters is uppercase. If in fact there are three uppercase letters either side of the lowercase letter then print this out.

For example

for j in range(-3,4):
    if j != 0:
        #Some code to check if the letters in this range are uppercase
        #if j != 0 is there because we already know it is lowercase
        #because of the previous if v.islower(): statement.

If this doesn't make sense, this would be an example output if the code worked as expected

scramble("abcdEFGhIJKlmnop")

OUTPUT

EFGhIJK

One lowercase letter with three uppercase letters either side of it.

Upvotes: 0

Views: 87

Answers (3)

derelict
derelict

Reputation: 2082

regex is probably the easiest, using a modified version of @Israel Unterman's answer to account for the outside edges and non-upper surroundings the full regex might be:

s = 'abcdEFGhIJKlmnopABCdEFGGIddFFansTBDgRRQ'

import re
words = re.findall(r'(?:^|[^A-Z])([A-Z]{3}[a-z][A-Z]{3})(?:[^A-Z]|$)', s)

# words is ['EFGhIJK', 'TBDgRRQ']

using (?:.) groups keeps the search for beginning of line or non-upper from being included in match groups, leaving only the desired tokens in the result list. This should account for all conditions listed by OP.

(removed all my prior code as it was generally *bad*)

Upvotes: 1

Paltoquet
Paltoquet

Reputation: 1234

if you can't use regular expression

maybe this for loop can do the trick 

if v.islower():
    if sentence[i-4].islower() and sentence[i+4].islower():
         for k in range(1,4):
            if sentence[i-k].islower() or sentence[i+k].islower():
                break
            if k == 3:
               return i

Upvotes: 1

Israel Unterman
Israel Unterman

Reputation: 13510

Here is a way to do it "Pythonically" without
regular expressions:

s = 'abcdEFGhIJKlmnop'

words = [s[i:i+7] for i in range(len(s) - 7) if s[i:i+3].isupper() and s[i+3].islower() and s[i+4:i+7].isupper()]
print(words)

And the output is:

['EFGhIJK']

And here is a way to do it with regular expressions,
which is, well, also Pythonic :-)

import re
words = re.findall(r'[A-Z]{3}[a-z][A-Z]{3}', s)

Upvotes: 1

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