CODEWITHSUNDEEP
python
obamaslayer89
obamaslayer89

Reputation: 3

Python - TypeError: string indices mustr be integers

I've tried looking at other solutions for this issue, but none of them work or make sense to me.

relevant code:

alphabet = 'abcdefghijklmnopqrstuvwxyz'
options = ["option 1", "option 2", "option 3", "option 4"]
choices = alphabet[0:len(options)]

for i in options:
    print(choices[i], options[i])

Output: print(choices[i], options[i]) TypeError: string indices must be integers

I'm looking to get the following output:

a option 1

b option 2

c option 3

d option 4

Anyone know what I'm screwing up here? Thanks in advance

Upvotes: 0

Views: 69

Answers (4)

a1cd
a1cd

Reputation: 25126

try this:

alphabet = 'abcdefghijklmnopqrstuvwxyz'
options = ["option 1", "option 2", "option 3", "option 4"]

for i in range(len(options)):
    print(alphabet[i] + " " + options[i])

it prints

a option 1
b option 2
c option 3
d option 4

Upvotes: 0

seqre
seqre

Reputation: 178

Instead of trying to slice the alphabet, you can just zip those iterables. It will only zip so many entries how many are in the smaller iterable.

alphabet = 'abcdefghijklmnopqrstuvwxyz'
options = ["option 1", "option 2", "option 3", "option 4"]

for (letter, opt) in zip(alphabet, options):
    print(letter, opt)

Note: that approach will work only for options iterables being smaller or equal to 26 entries. If you had more choices, you could use itertools.cycle, but then you would have multiple options with the same letter, so just adding more unique characters to alphabet or changing it to numbers, could solve that problem.

Upvotes: 0

evan
evan

Reputation: 21

You're iterating through the items of the list, not an index.

You can try this:

for i, element in enumerate(options):
    print(choices[i], element)

Upvotes: 2

user8563312
user8563312

Reputation:

The issue is, that you're iterating over the elements of options which are strings (option1, ...). You have to iterate over the length of your list.

for i in range(len(options)):
    print(choices[i], options[i])

Upvotes: 2

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