一个作业中列表的小问题？

Python如果列表的第六列大于等于90，则输出第一列的代码怎么写？代码输出的结果为列表
import codecs
f = codecs.open('movie.txt', mode='r', encoding='utf-8')

line = f.readline()

list1 = []
while line:
a = line.split()
b = a[0:6]

list1.append(b)
line = f.readline()
f.close()
for i in list1:
print(i)

lst = ... 你的列表
result = [x[6] for x in lst if x[1] >= 90]
for x in result:
  print(x)

import codecs
f = codecs.open('movie.txt', mode='r', encoding='utf-8')

line = f.readline()

list1 = []
while line:
    a = line.split()
    b = a[0:6]

    list1.append(b) 
    line = f.readline()
f.close()
#不知道你的第6列第1列是从0开始算还是从1开始算，如果不是0开始的，修改成5和0
result = [x[6] for x in list1 if int(x[1]) >= 90]
print(result)

如果还有问题，贴出你的txt和错误信息看看

import codecs
f = codecs.open('movie.txt', mode='r', encoding='utf-8')

line = f.readline()

list1 = []
while line:
    a = line.split()
    b = a[0:6]

    list1.append(b) 
    line = f.readline()
f.close()
for i in list1:
    if int(i[5])>=90:
        print(i[0])