Python CSV read a single column as a list

November 6, 2022

I have this code:

type(i[0]) == 'str'. How can I get it as a list? and later get print(i[0][0]) == 'a'

import csv
i = [['a', 'b'], 'c', 'd']
with open('a.csv', 'a', newline='', encoding='utf-8') as file:
    writer = csv.writer(file, delimiter=';')
    writer.writerow(i)

file.close()
with open('a.csv', 'r', newline='', encoding='utf-8') as file:
    for i in csv.reader(file, delimiter=';'):
        print(type(i[0]))

>Solution :

The reason it is returning str instead of list because the csv writer will always do the equivalent of str(['a','b']), which is "['a', 'b']",c,d, so you can use ast.literal_eval to get the list from the string list like this for the first element-

import csv
import ast
i = [['a', 'b'], 'c', 'd']
with open('a.csv', 'a', newline='', encoding='utf-8') as file:
    writer = csv.writer(file, delimiter=';')
    writer.writerow(i)

file.close()
with open('a.csv', 'r', newline='', encoding='utf-8') as file:
    for i in csv.reader(file, delimiter=';'):
        i[0] = ast.literal_eval(i[0])
        print(type(i),type(i[0]), i[0][0])