Whenever User Inputs, hits an Out Of Range Error

Question

userInput = input();

for x in range(0,len(userInput)):
    if(userInput[x] == " "):
        spaceValue = x;
rawWord = userInput[0:spaceValue];
secWord = userInput[spaceValue+1:];
wordArray = [];
repeats = len(rawWord); #NEW VAR

if (len(rawWord) == len(secWord)):
    for x in range(0,len(rawWord)):
        wordArray.append(secWord[x]);
    for x in range(0,len(rawWord)):
        for z in range(0,len(rawWord)):
            if((rawWord[x] == wordArray[z])): #Line 15 #repeats insted of wordArray[z]
                wordArray.remove(rawWord[x]);
                repeats = repeats - 1;
                break;

    if(len(wordArray) == 0):
        print("YES");
    else:
        print("NO");
else:
    print("NO");

The code supposed to print YES if the 2 words are same length and have the same letters and NO if they don't.

Error hits at line 15: if((rawWord[x] == wordArray[z])):
IndexError: list index out of range

It works when

Words are same length and same letters.
Words are different length.
Words are same length and all the letters are different.

It doesn't work when

Words are same length and different letters but at least one letter is the same

It must be `len(...) - 1`. If it is `len(...)` it will go over. — Andrew Li, Dec 22 '16 at 13:41
By the way, `range(0, len(userInput))` == `range(len(userInput))` — iFlo, Dec 22 '16 at 13:43
Which `len(...)` are you speaking of, there are a lot. All of them? — stathis21098, Dec 22 '16 at 13:43
You are looping over `for z in range(0,len(rawWord)):` and indexing into `wordArray` which you remove things from in the loop. — doctorlove, Dec 22 '16 at 13:43
@doctorlove so whenever I remove a value, I need to loop 1 less time? — stathis21098, Dec 22 '16 at 13:45
@doctorlove I edited the post, it worked the way you told me, I had to remove 1 loop every time I removed 1 letter because the Array was getting smaller and the loop wasn't (compare letters) — stathis21098, Dec 22 '16 at 13:50

score 0 · Accepted Answer · answered Dec 22 '16 at 14:04

As I noted in a comment, you are looping over for z in range(0,len(rawWord)): but indexing into wordArray which you remove things from in the loop.

Always be worried if you are removing things you are looping over.

Can I suggest a better/more pythonic solution?

from itertools import permutations

userInput = input()
words = userInput.split(' ')
rawWord = words[0]
otherWords = [''.join(p) for p in permutations(words[1])]
if rawWord in otherWords:
    print("YES")
else:
    print("NO")

We can make yours work - don't forget to use your repeats count when you index into the wordArray on line 15. Or what was line 15. I have also removed some un-needed punctuation.

userInput = input()

for x in range(0,len(userInput)):
    if userInput[x] == " ":
        spaceValue = x
rawWord = userInput[0:spaceValue]
secWord = userInput[spaceValue+1:]
wordArray = [];
repeats = len(rawWord)

if len(rawWord) == len(secWord):
    for x in range(0,len(rawWord)):
        wordArray.append(secWord[x])
    for x in range(0,len(rawWord)):
        for z in range(0,len(rawWord)):
            if (rawWord[x] == wordArray[z-repeats]): #Line 15
                wordArray.remove(rawWord[x])
                repeats = repeats - 1
                break;

    if(len(wordArray) == 0):
        print("YES")
    else:
        print("NO")
else:
    print("NO")

Thanks the first one works flawless (ofc and the second), but I will keep the first — stathis21098, Dec 22 '16 at 14:27

Whenever User Inputs, hits an Out Of Range Error

The code supposed to print YES if the 2 words are same length and have the same letters and NO if they don't.

1 Answers1