Python3: TypeError: sequence item 0: expected str instance, list found

Question

I'm trying to print all of the possible character substitutions for the word "otaku".

#!/usr/bin/python3

import itertools

user_input = "otaku"

dict = {
'c': ['c', 'C'],
'a': ['a', 'A', '@'],
't': ['t', 'T'],
'k': ['k', 'K'],
'u': ['u', 'U'],
'e': ['e', 'E', '3'],
'o': ['o', 'O', '0']
}

output = ""

for i in itertools.product(dict['o'],dict['t'],dict['a'],dict['k'],dict['u']):
    output += ''.join(i) + "\n"

print(output)

The above script works but I need the itertools.product() input (dict['o'],dict['t'],dict['a'],dict['k'],dict['u']) to be dynamic (e.g., new_list):

#!/usr/bin/python3

import itertools

user_input = "otaku"

dict = {
'c': ['c', 'C'],
'a': ['a', 'A', '@'],
't': ['t', 'T'],
'k': ['k', 'K'],
'u': ['u', 'U'],
'e': ['e', 'E', '3'],
'o': ['o', 'O', '0']
}

new_list = []

for i in user_input:
    new_list.append(dict[i])

output = ""

for i in itertools.product(new_list):
    output += ''.join(i) + "\n"

print(output)

This errors with:

TypeError: sequence item 0: expected str instance, list found

I tried the solutions found here, but converting the list to a str breaks the itertools.product line.

How can I pass dynamic input into itertools.product()?

Desired output:

otaku
otakU
otaKu
otaKU
otAku
otAkU
otAKu
otAKU
ot@ku
ot@kU
ot@Ku
ot@KU
oTaku
oTakU
oTaKu
oTaKU
oTAku
oTAkU
oTAKu
oTAKU
oT@ku
oT@kU
oT@Ku
oT@KU
Otaku
OtakU
OtaKu
OtaKU
Ot@KU

MusHusKat · Accepted Answer · 2021-02-02 04:38:36Z

1

try for i in itertools.product(*new_list):

you need to add a * to unpack the new_list

answered Feb 2, 2021 at 4:38

MusHusKat

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4 Comments

gevizoxu Over a year ago

Wow, I can't believe that was the solution. Why is the wildcard needed there?

MusHusKat Over a year ago

Your code is fine but when you pass new_list as an argument, you really are passing a list. What you want to pass are all the elements of the list. When you add an * you are unpacking the list and passing each element as a param. This is what is expected.

MusHusKat Over a year ago

Look at this function: def test_func(a, b, c): # do something. If I pass a list like test_func(new_list), it will mean test_func(a = new_list). If I unpack the list like test_func(*new_list) it will mean test_func(a=new_list[0], b = new_list[1]....).

MusHusKat Over a year ago

Check this out. Really useful if you know how args and kwargs work: realpython.com/python-kwargs-and-args

Mubashar Javed · Accepted Answer · 2021-02-02 10:17:25Z

1

you can use * to unpack it:

for i in itertools.product(*new_list)

you can also do it by list comprehension

c = ["".join(x) for x in itertools.product(*new_list)]
for i in c:
   print(i)

edited Feb 2, 2021 at 10:17

Mubashar Javed

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answered Feb 2, 2021 at 7:59

Abhishek Kumar Saw

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