在字符串中查找字母相邻字母（2 种类型：aABClg = ABC，aAbcd = aA）

Question

EDIT: The problem, and the answer lies in using the enumerate function to access the index of iterables, but I'm still working on applying it properly.

I was asked to generate a random word with N length, and to print uppercase alphabet neighbours and lower - upper neighbours. I really don't know how to put this better. The example is in the title, here is my code so far, and I think it works, I just need to fix the error made by the index search in the ascii_uppercase list variable.

Also, please excuse the messy do - while loop at the beginning.

import string
import random
 
letters = list(string.ascii_uppercase + string.ascii_lowercase)
 
enter = int(input('N: '))
def randomised(signs = string.ascii_uppercase + string.ascii_lowercase, X = enter):
        return( ''.join(random.choice(signs) for _ in range(X)))
 
if enter == 1:
    print('END')
   
while enter > 1:
    enter = int(input('N: '))
    word1 = randomised()
    word2 = list(word1)
    neighbour = ''
    same = ''
    for j in word2:
        if word2[j] and word2[j+1] in string.ascii_uppercase and letters.index(j) == word2.index(j) and letters.index(j+1) == word2.index(j+1):
            same += j
            same += j+1
           
    for i in word2:
        if word2[i] == i.upper and word2[i+1] == (i+1).upper:
            neighbour += i
            neighbour += i+1
    print('Created: {}, Neighbour uppercase letters: {}, Neighbour same letters: {}' .format(word1,same,neighbour))

expected behaviour:

N: 7
Created: aaBCDdD, Neighbour uppercase letters: BCD, Neighbour same letters: dD
N: 1
N: END

Answer 1

I am not so sure, but i think your problem might stem from the use of "i+1" and "j+1" without limiting the iterations stop before the last one.

The next thing is that you need to put this code in english for people to be able to provide better answers, i am not native english, but the core of my code is in english so it will be understandable worldwide, there are other general improvements that can be done, but those are learn while coding.

I hope your assignment goes great.

Another recommendation you can use "a".islower() to see if the character is lowercase, isupper() to se if it is uppercase, you don't need to import the whole alphabet. There are many builtin functions to deal with common scenarios and those tend to be more efficient than what most people would do without them.

Answer 2

Here is a simple working code (without formatting of the output). I used a simple pairwise iteration to compare each character with the previous one.

# generation of random word
N = 50
word = ''.join(random.choices(ascii_letters, k=N))

# forcing word for testing
word = 'aaBCDdD'

# test of conditions
cond1 = ''
cond2 = ''
flag1 = False  # flag to add last character of stretch

for a,b in zip('_'+word, word+'_'):
    if a.isupper() and b.isupper() and ord(a) == ord(b)-1:
        cond1 += a
        flag1 = True
    elif flag1:
        flag1 = False
        cond1 += a
    if a.islower() and b.isupper() and a.lower() == b.lower():
        cond2 += a+b

print(cond1, cond2, sep='\n')
# BCD
# dD

NB. In case the conditions are met several times in the word, the identified patterns will just be concatenated

Example on random word of 500 characters:

IJRSOPHILMMNVW
kKdDyY