使用函数读取和写入文件

Question

我正在尝试创建 2 个函数。

1. readfiles(file_path)，读取由 file_path 指定的文件，并返回包含文件中每一行的字符串列表。

2. writefiles(lines, file_path) 将列表行的内容逐行写入 file_path 指定的文件。

当一个接一个使用时，输出文件应该是输入文件的精确副本（包括格式）

这是我目前所拥有的。

file_path = ("/myfolder/text.txt", "r")
def readfiles(file_path):
    with open file_path as f:
        for line in f:
            return line
            lst = list[]
            lst = line
            lst.append(line)
            return lst
read_file(file_path)

lines = lst []
def writefiles(lines, file_path):
    with open ("file_path", "w") as f: 
    for line in lst:
        f.write(line)
        f.write("\n")

当我用它来阅读时，我可以让它发挥作用

with open("/myfolder/text.txt", "r") as f:
    for line in f:
        print(line, end='')

这个是写

with open ("/myfolder/text.txt", "w") as f: 
    for line in f:
        f.write(line)
        f.write("\n")

但是当我尝试将它们放入函数时，一切都搞砸了。 我不知道为什么，我知道这是一个简单的问题，但它只是不适合我。我已经阅读了有关它的文档，但我没有完全遵循它并且我束手无策。我的功能有什么问题？

我收到不同的错误

 lst = list[]
               ^
SyntaxError: invalid syntax

到

lst or list is not callable 

我也知道有类似的问题，但我发现的问题似乎没有定义函数。

Answer 1

你的代码的问题解释为 cmets

file_path = ("/myfolder/text.txt", "r") # this is a tupple of 2 elements should be file_path = "/myfolder/text.txt"
def readfiles(file_path):
    with open file_path as f: # "open" is a function and will probably throw an error if you use it without parenthesis
    # use open this way: open(file_path, "r")
        for line in f:
            return line # it will return the first line and exit the function
            lst = list[] # "lst = []" is how you define a list in python. also you want to define it outside the loop
            lst = line # you are replacing the list lst with the string in line
            lst.append(line) # will throw an error because lst is a string now and doesn't have the append method
            return lst
read_file(file_path) # should be lines = read_file(file_path)

lines = lst [] # lines is an empty list 
def writefiles(lines, file_path):
    with open ("file_path", "w") as f: 
    for line in lst: # this line should have 1 more tabulation
        f.write(line) # this line should have 1 more tabulation
        f.write("\n") # this line should have 1 more tabulation

代码如下所示

def readfiles(file_path):
    lst = []
    with open(file_path) as f:
        for line in f:
            lst.append(line.strip("\n"))
    return lst


def writefiles(lines, file_path):
    with open(file_path, "w") as f:
        for line in lines:
            f.write(line + "\n")


file_path = "/myfolder/text.txt"
filepathout = "myfolder/text2.txt"
lines = readfiles(file_path)
writefiles(lines, filepathout)

一种更 Pythonic 的方式

# readlines is a built-in function in python
with open(file_path) as f:
    lines = f.readlines()

# stripping line returns
lines = [line.strip("\n") for line in lines]

# join will convert the list to a string by adding a \n between the list elements
with open(filepathout, "w") as f:
    f.write("\n".join(lines))

关键点：

- the function stops after reaching the return statement

- be careful where you define your variable. 
  i.e "lst" in a for loop will get redefined after each iteration

定义变量：

- for a list: list_var = []

- for a tuple: tup_var = (1, 2)

- for an int: int_var = 3

- for a dictionary: dict_var = {}

- for a string: string_var = "test"

Answer 2

这里有几个学习点会有所帮助。

在你的阅读功能中，你有点接近。但是，您不能将return 语句放入循环中。一旦函数第一次到达任何地方，它就会结束。此外，如果您要创建一个容器来保存读取的内容列表，则需要在开始循环之前创建它。最后，不要命名任何东西list。它是一个关键字。如果您想创建一个新列表项，只需执行以下操作：results = list() 或 results = []

所以在伪代码中，你应该：

Make a list to hold results
Open the file as you are now
Make a loop to loop through lines
    append to the results list
return the results (outside the loop)

您的writefiles 非常接近。您应该循环访问作为函数参数的 lines 变量。现在你引用lst 这不是你的函数的参数。

祝你好运！