Checking for letter matches

Question

I have a sample string "this string gets stuck". I am trying write a loop that will compare the last last letter of a word with the first letter of the next word to see if the match. So, with this four word string I would check the last letter of "this" ("s") and see if it matches the fist letter off "string" ("s"). I know how to check this for the first and second word with string[0][-1] for the last letter, first word and string[1][0] for the first letter, second word. I am trying to figure out how to loop through each word pairs in the string to check for the same. My thought is to maybe have the first loop = string[0][-1]==string[1][0] and then have the loop advance to each successive pair of words by string[0+1][-1] and string[1+1][0] adding one to each word in the string to compare the first and last letters throughout, but I am not sure how to phrase this.

I know that I could probably use regex for matches, but I am not exactly sure how I loop through with that method.

Answer 1

Robert Haugland
See if this little snippet helps you

lst = ["yellow","white","ergon","night"]

for i in range(1, len(lst)):
if lst[i-1][-1] == lst[i][0]:
print("last letter = first letter")

Answer 2

Prabhas Koya
Just keep practicing and learning

Answer 3

Could you give an example in full of what you want the code to do? Like the end goal of the code.

Not sure if you check index 0 and 1, then 1 and 2 OR after 0 and 1, it's then 2 and 3.

Answer 4

S='this string gets stuck'

print(['match' for i in range(len(S)) if S[i] == ' ' and S[i-1] == S[i+1]])

Answer 5

there is a gap between every word so I used that as reference for checking before letter and after letter if they are equal

Answer 6

Rik Wittkopp
can you say how to write any code efficiently, I am beginner

Answer 7

Rik Whittkopp,
The code worked. I tweeked it to fit the perameters of this problem:
words = input()
wordsp = words.split()
sum = 0
for i in range(1, len(wordsp)):
if wordsp[i-1][-1]==wordsp[i][0]:
sum += 1
else:
print("false")
exit(0)
print("true")
So, I had to break down what you wrote so I could understand and learn from it. In "for i in range(1, len(lst)", I figured the range would have to go the length of the string -1 so as not to run the last word last letter, etc. In my mind I was thinking more like "len(lst)-1" which I believe, also works. Then "if lst[i-1][-1]" I believe this is pointing to the last letter of the word to the left of (i) and "lst[i][0]" points to the first letter of (i). The "exit(0)" I learned from a previous problem where it refers to the "if" block being "true", thus allowing me to move the print statement to the far left to output only one reponse versus a response for each loop. Thanks again, you're a true hero!

Answer 8

Rik Whitkopp,
Okay, I ran in the loop "print([i-1])" and it output "this", "string", "gets". Then "print([i-1][-1]" output the last letters of these three words "s","g","s". "Print([i])" output "string", "gets","stuck" and "print([i][0])" output the fist letter of three words "s","g","s". So, [i-1] is like slicing "string[:-1]" and [i] is "string[1:]". Interesting!

Answer 9

in the if statement:

"if i == ' ' ..." <-- what is that for?

Answer 10

Robert Haugland
👍