list index error python (numeric words program) - python-3.x

Question

I'm trying to write a program in Python 3.
This is how it works:
The input is a word.
And the program has to look if the word contains a dutch numeric word.
A word cannot contain more than 1 numeric word and if a word doesn't contain a numeric word than it has to print 'geen' (none in dutch).
example:
Input = BETWEEN
Output = TWEE (two in dutch)
Input = ZEEVERS
Output = ZES (six in dutch)
Here is my code:
import sys
invoer = input()
letterwoorden = [['T','W','E','E'], ['D','R','I','E'], ['V','I','E','R'],
['V','I','J','F'], ['Z','E','S'], ['Z','E','V','E','N'], ['A','C','H','T'],
['N','E','G','E','N']]
antwoord = []
woord = [str(a) for a in str(invoer)]
L = len(woord)
a = 0
b = 0
c = 0
for i in range(0, 8):
for j in range(0, len(letterwoorden[a])):
if letterwoorden[a][b] == woord[c]:
antwoord.append(letterwoorden[a][b])
b = b + 1
c = c + 1
else:
c = c + 1
if antwoord == letterwoorden[a]:
print(antwoord)
break
else:
a = a + 1
antwoord.clear()
if antwoord != letterwoorden[a]:
print('geen')
Could someone help me with the error on line 21? (list index out of range)
Thanks!
The code is not fully done but when the input is TWEET the output is TWEE,
when the input is BETWEEN it gives the error.

Answer 1

Even though usr2564301's first solution looks good, I wanted to add a couple things. You'll find this in the comments in the code.
Here's how I'd modify your code:
## use strings instead of list of strings
## ['T', 'W', 'E', 'E'] ==> 'TWEE'
## they are also iterable (you can loop through them)
## as well as indexable (['T', 'W', 'E', 'E'][0] == 'TWEE'[0])
letterwoorden = ['TWEE', 'DRIE', 'VIER', 'VIJF', 'ZES', 'ZEVEN', 'ACHT', 'NEGEN']
## keep the string
woord = input()
print('Input:', woord)
## answer should be string as well
antwoord = ''
## name your variables something meaningful
## how does one differentiate between b and c? smh
letterWoordenIndex = 0
woordenIndex = 0
## for-loops can be powerful in Python
## you might've been used to index-based loops from other languages :|
## this way, tempWord can replace all occurrences of letterwoorden[a]
for tempWord in letterwoorden:
## reset your variables at the beginning of the loop
letterWoordenIndex = woordenIndex = 0
# """ ## debug output printing word
print('Word:', tempWord, '?')
# """
## loop through the length of word
## use _ to indicate that the variable won't be used
for _ in range(len(woord)):
# """ ## debug output comparing letters/indices
print(tempWord[letterWoordenIndex],
'({})'.format(letterWoordenIndex),
'<=>', woord[woordenIndex],
'({})'.format(woordenIndex))
# """
## check current indices equals match
if tempWord[letterWoordenIndex] == woord[woordenIndex]:
antwoord += tempWord[letterWoordenIndex] ## append char to string using +
## increment indices
letterWoordenIndex += 1
woordenIndex += 1
## check if index is filled
if letterWoordenIndex == len(tempWord):
break
else:
woordenIndex += 1
# """ ## debug output comparing final results
print(antwoord, '>==<', tempWord)
# """
## assert that the accumulated result (antwoord)
## equates to the tempWord
if antwoord == tempWord:
# """ ## debug assert true
print('Yep\n')
# """
print(antwoord)
break
## no need to use else here
## if the code inside the above if executes, then it's already game-over
antwoord = '' ## empty the string
# """ ## debug assert false
print('Nope\n')
# """
## antwoord would be empty if everything failed
if antwoord == '':
print('GEEN')

Answer 2

You are iterating over the wrong list with your line
for j in range(0, len(letterwoorden[a])):
as this increases a for all letterwoordenagain – you already iterate over letterwoorden with the first loop, for i in range(0, 8):. Changing that to iterate over the word in question led to another error if you run out of 'source' characters (the length of invoer) OR of 'compare' characters (the word woord[c]). You must also take care to reset your b and c counters when restarting a compare.
At the very end, you test antwoord[a] but a may be out of range
The following code works
import sys
invoer = input()
letterwoorden = [['T','W','E','E'], ['D','R','I','E'], ['V','I','E','R'],
['V','I','J','F'], ['Z','E','S'], ['Z','E','V','E','N'], ['A','C','H','T'],
['N','E','G','E','N']]
antwoord = []
woord = [str(a) for a in str(invoer)]
L = len(woord)
a = 0
b = 0
c = 0
for i in range(len(letterwoorden)):
b = 0
c = 0
for j in range(len(woord)):
print ('test', a,b,c, letterwoorden[a][b], woord[c])
if letterwoorden[a][b] == woord[c]:
antwoord.append(letterwoorden[a][b])
b = b + 1
if b >= len(letterwoorden[a]):
break
c = c + 1
else:
c = c + 1
if antwoord == letterwoorden[a]:
print(antwoord)
break
else:
a = a + 1
antwoord = []
if antwoord == []:
print('geen')
A few additional notes: you don't need to maintain separate variables if you already have a loop variable. Where you use a, for example, you can also use the loop i; and the same with j.
About all of your comparing can be done much more efficiently, since Python can immediately check if a phrase contains another phrase: if straw in haystack. So, basically,
invoer = 'ZEEVERS'
letterwoorden = ['TWEE', 'DRIE', 'VIER', 'VIJF', 'ZES', 'ZEVEN', 'ACHT', 'NEGEN']
for word in letterwoorden:
if word in invoer:
print (word)
break
else:
print ('geen')
where the else is linked to the for and not to the else, to check if the loop stopped for a result.