Python从文件中读取列表

Question

我正在尝试对文件中的列表进行排序，但收到此错误:

Traceback (most recent call last):
  File "/Users/MacbookPro/Documents/Faculta/alg sortare pyth/bubble.py", line 13, in <module>
    f = file.open('lista.txt', 'r')
AttributeError: type object 'file' has no attribute 'open'

这是我的代码:

from timeit import default_timer as timer
import resource
start = timer()
def bubbleSort(alist):
    for passnum in range(len(alist)-1,0,-1):
        for i in range(passnum):
            if alist[i]>alist[i+1]:
                temp = alist[i]
                alist[i] = alist[i+1]
                alist[i+1] = temp


f = file.open('lista.txt', 'r')
long_string = f.readline()
my_list = long_string.split(',')
bubbleSort(alist)
print(alist), resource.getrusage(resource.RUSAGE_SELF).ru_maxrss / 1000
end = timer()
print(end - start)

Answer 1

要打开文件，请使用:

f = open('lista.txt', 'r')

改用上下文管理器:

with open('lista.txt', 'r') as f:
    long_string = f.readline()
    my_list = long_string.split(',')
    ....

上下文管理器方法将自动关闭文件。在写入文件时尤其如此，但这也是最佳实践。