Sequence item 0: expected str instance, list found (Python)

Sequence item 0: expected str instance, list found (Python) #

The Python "TypeError: sequence item 0: expected str instance, list found" occurs when we call the join() method with an iterable that contains one or more list objects. To solve the error, use a nested join() to join the elements of each list.

Here is an example of how the error occurs.

Copied!
my_list = [['a', 'b'], ['c', 'd']]

# ⛔️ TypeError: sequence item 0: expected str instance, list found
result = ''.join(my_list)

The join method raises a TypeError if there are any non-string values in the iterable.

One way to solve the error is to use a generator expression with a nested join().

Copied!
my_list = [['a', 'b'], ['c', 'd']]

result = ''.join(''.join(l) for l in my_list)

print(result)  # 👉️ "abcd"

We called the join() method with each nested list and then joined the strings.

If your nested lists might contain non-string values, you'll have to convert each value to a string.

Copied!
my_list = [[1, 2], ['c', 'd']]

result = ''.join(''.join(map(str, l)) for l in my_list)

print(result)  # 👉️ "12cd"

The map() function takes a function and an iterable as arguments and calls the function with each item of the iterable.

If you want to join the nested lists into a string, you can pass each list to the str() class.

Copied!
my_list = [['a', 'b'], ['c', 'd']]

result = ''.join(str(l) for l in my_list)

print(result)  # 👉️ "['a', 'b']['c', 'd']"

The str.join method takes an iterable as an argument and returns a string which is the concatenation of the strings in the iterable.

If your list contains numbers, or other types, convert all of the values to string before calling join().

Copied!
my_list = ['a', 'b', 1, 2]


all_strings = list(map(str, my_list))

print(all_strings) # 👉️ ['a', 'b', '1', '2']

result = ''.join(all_strings)

print(result)  # 👉️ "ab12"

The string the method is called on is used as the separator between elements.

Copied!
my_list = ['a', 'b', 'c']

my_str = '-'.join(my_list)

print(my_str)  # 👉️ "a-b-c"

If you don't need a separator and just want to join the iterable's elements into a string, call the join() method on an empty string.

Copied!
my_list = ['a', 'b', 'c']

my_str = ''.join(my_list)

print(my_str)  # 👉️ "abc"

Conclusion #

The Python "TypeError: sequence item 0: expected str instance, list found" occurs when we call the join() method with an iterable that contains one or more list objects. To solve the error, use a nested join() to join the elements of each list.