Question

请任何人帮我阅读python中的.text文件

这是我的代码

flink open(2of12inf.txt, "rU")

但我收到错误

Answer 1

您忘记了=转让声明和引号：

flink = open('2of12inf.txt', "rU")

最佳做法是将文件作为上下文管理器（with statement）打开，以便自动关闭：

with open('2of12inf.txt', "rU") as flink:
    # do something with the open file object

# flink will be closed automatically.

flink是file object，因此您可以使用.read()，.readline()等方法来阅读它。或者你可以遍历对象（迭代）每次获得单行：

with open('2of12inf.txt', "rU") as flink:
    for line in flink:
        # do something with each line.

我使用文件的绝对路径而不是相对路径来避免意外：

with open('/path/to/directory/with/2of12inf.txt', "rU") as flink:

或者您可以使用os.path library构建绝对路径：

import os.path

filename = os.path.expanduser('~/2of12inf.txt')

with open(filename, "rU") as flink:

例如，

在当前用户主目录中打开名为2of12inf.text的文件。

Answer 2

使用以下示例：

#!/usr/bin/python
# open file
f = open ("/etc/passwd","r")

#Read whole file into data
data = f.read()

# Print it
print data

# Close the file
f.close()

Answer 3

您应该将文件名括在引号中（并在flink和open之间添加赋值运算符）：

flink = open("2of12inf.txt", "rU")

正如IT Ninja所说，我还强烈建议使用with构造来打开文件：

with open("2of12inf.txt", "rU") as flink:
    # do stuff...

这将像处理try-finally块一样关闭文件。

试图在python中读取文件

3 个答案: