Python CSV 将单列读取为列表

Question

我有这个代码:

类型(i[0]) == 'str'。我怎样才能将其作为列表获取？然后得到 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]))

Answer 1

它返回 str 而不是 list 的原因是因为 csv 编写器将始终执行与 str(['a','b'] 等效的操作)，即 "['a', 'b']",c,d，因此您可以使用 ast.literal_eval 来获取 list 来自字符串列表，就像第一个元素一样 -

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])