输出：

Question

执行此python代码时，我看到以下错误。这是什么问题？我用过＆＃34; sys.stdout.close（）＆＃34;我仍然看到这些错误。

#! /usr/bin/python
import sys
a = [ 10, 12, 13, 14]
sys.stdout=open("file.txt","w")
print("++++++++")
print("***xyz***")
print("++++++++")
sys.stdout.close()
for i in a:
    print i

输出：

    Traceback (most recent call last):
    File "./test3.py", line 10, in <module>
    print i
    ValueError: I/O operation on closed file`

Answer 1

您正在尝试在关闭它之后写入stdout（您的文件）。在第8行关闭文件，在第10行，您调用print。

如果要将列表a写入文件，则应在for循环后将其关闭。

Answer 2

考虑使用with open，因为您不必担心关闭它。如果您的列表需要是一个列表，那么考虑将其腌制而不是将其写入文件。 Pickling将您的数据序列化。

#!python3

# import module
from os import system
import pickle
# clear the screan
system('cls')

a = [ 10, 12, 13, 14]

# write a list to file, but it has to be written as a string
with open('file.txt', 'w') as wf:
    wf.write(str(a))

# when you open your file up, the data is a string
with open('file.txt', 'r') as fp:
    for item in fp:
        print(item)
        print(type(item))

# if you want to retain your data as a list, then pickle it
output = open('file.pkl', 'wb')
pickle.dump(a, output)
output.close()

# open up a pickled file
pkl_file = open('file.pkl', 'rb')
data = pickle.load(pkl_file)
print(data)
print(type(data))
pkl_file.close()

ValueError：I / O操作

输出：

2 个答案: