Python Challenge level 3 - Solving with string slicing

Question

apologies to posting the question, as has been answered in other questions as well. However, couldn't figure out what's wrong with this solution. The question requires to find the lower-cased characters bordered by 3 upper-cased characters on each side. The code i've writting:

q = ''
for i in range(4,len(x)-4):
    if x[i].islower() and x[i-4].islower() and x[i+4].islower() and x[i-3:i].isupper() and x[i+1:i+4].isupper()  :
        q+=x[i]


print(q)

The string i'm getting is

'lgvcaaginbkvsoezhtlnldslyitlooqfgiksudtm' vs 'linkedlist'

Thanks for the help.

Edit: for some reason the following code seems to work:

q = ''
for i in range(4,len(x)-4):
    if x[i].islower() and x[i-4].islower() and  x[i-3:i].isupper() and x[i+1:i+4].isupper() and x[i+4].islower():
        q+=x[i]

print(q)

Answer 1

This would be a slicing solution:

t = "This is some TEXtTHAt will result in one true result"

def notupper(x): 
    return not x.isupper()

yep = []
# lookup function to execute inside all()
terms = {0:notupper, 8:notupper, 4:notupper}

for part in (t[i:i+9] for i in range(len(t)-7)):
    if len(part)<9:  # fix for XXXxXXX at end of string
      part = (part+"  ")[:9]
    if all(terms.get(index,str.isupper)(ch) for index,ch in enumerate(part)):
        print("Statisfies all terms: ", part)

It slices the text into parts of 9 characters (you need 9 to assure not to have XXXXyXXXX - you need exactly three uppers around your lowers.

It checks the sliced text for notupper on positions 0,8 and 4:

XXXXxXXXX
012345678
lUUUlUUUl   # must statisfy: l=notupper and U=isupper

by pulling this function from the terms-dictionary. All indexes of part that have no matching key in terms are tested with str.isupper using the default param of terms.get(index,str.isupper)

all() makes sure all tests need to evaluate to true to enter the if-condition.

Spaces evaluate to False for isupper and islower().

Readup:

• all()
• dict.get(key,default) and Why dict.get(key) instead of dict[key]?
• enumerate(iterable)

---

You can replace the dict/all part by:

# the ( .. ) are needed to enable line breaking - you could use \ as well
if notupper(part[0]) and notupper(part[8]) and notupper(part[4]) and (
    part[1].isupper() and part[2].isupper() and part[3].isupper() and
    part[5].isupper() and part[6].isupper() and part[7].isupper()):
    print("Statisfies all terms: ", part)

if that is too complicated for your current level of python.

Answer 2

What you are trying to do is match a pattern : Not Upper, Upper, Upper, Upper, Not Upper, Upper, Upper, Upper, Not Upper. This is easier to catch if you use a signature for your string:

>>> t = "This is some TEXtTHAt will result in one true result"
>>> sig = "".join("U" if c.isupper() else "l" for c in t)
>>> sig
'UllllllllllllUUUlUUUllllllllllllllllllllllllllllllll'

You are looking for the lUUUlUUUl substrings in the sig string. Python has no builtin for findall, but you can to iterate over the results of find:

>>> result = ""
>>> i = sig.find("lUUUlUUUl")
>>> while i != -1: # find != -1 means that the substring wasn't found
...     result += t[i+4] # the center `l` 
...     i = sig.find("lUUUlUUUl", i+4) # we can start the next search at the center `l`
... 
>>> print( result )
t

You may also use re.finditer with a regex pattern, but is more complicated.

Answer 3

The text has the new line character "\n" at the end of each line. You need to replace all the "\n" characters with an empty string. This can be done using the RegEx (Regular Expression) module. Add these lines of code and it will work:

import re

x = re.sub(r"\n","", x)