Python程序：读取文件

Question

我应该为一个程序编写代码，该程序会不断地向用户询问文件名，直到输入正确的文件名。然后 find_min_percent 应该采用一个参数，即 GDP.txt 文件中的一行 (str)，然后遍历该行以找到最小值并返回该值。到目前为止，这是我的代码

line = " " 

def open_file(): 
    ''' Repeatedly prompt until a valid file name allows the file to be opened.'''
    while True: 
        user_input = input('Enter a file name: ')
        try: 
            file = open(user_input, 'r')
            return file 
            break
        except FileNotFoundError: 
            print('Error. Please try again') 
            open_file()

def find_min_percent(line): 
    '''Find the min percent change in the line; return the value and the index.'''
    percent_lst = []
    line = file.readline(9)
    percent_lst += [line]
    percent_int = [float(i) for i in percent_lst]
    min_value = 10000
    for percent in percent_int:
        if percent < min_value:
            min_value = percent
            return min_value 


print(open_file())
print (find_min_percent(line))

我的问题在于 readline()。它说变量文件未定义。此代码的大纲不包括“def find_min_percent(line):”部分中的文件。所以我不知道我将如何解决这个问题。我也不能在函数之外设置行，因为我必须在程序后面的其他函数中使用相同的行变量来读取其他行。所以我不知道该怎么做才不会保留

Answer 1

您在一个函数中定义的变量不能从另一个函数访问。例如，要解决这个问题，您可以这样做（将返回的值存储在“主函数”变量中并将其传递给您的下一个函数）：

def find_min_percent(line): 
    '''Find the min percent change in the line; return the value and the index.'''
    percent_lst = []
    # You can use f from this function, as long as you don't modify it
    line = f.readline(9)
    percent_lst += [line]
    percent_int = [float(i) for i in percent_lst]
    min_value = 10000
    for percent in percent_int:
        if percent < min_value:
            min_value = percent
            return min_value 


f = open_file()
print(f)
print (find_min_percent(line))

顺便说一句，您使用line 变量的方式很奇怪。它仅在find_min_percent 内部使用，但在函数外部定义，甚至作为参数传递。为什么？你想达到什么目的？

（请参阅here 了解有关访问函数外部定义的变量的帖子）

Answer 2

返回的file变量超出函数作用域 固定代码：

line = " " 

def open_file(): 
    ''' Repeatedly prompt until a valid file name allows the file to be opened.'''
    while True: 
        user_input = input('Enter a file name: ')
        try: 
            file = open(user_input, 'r')
            return file 
            break
        except FileNotFoundError: 
            print('Error. Please try again') 
            open_file()

def find_min_percent(line,file): 
    '''Find the min percent change in the line; return the value and the index.'''
    percent_lst = []
    line = file.readline(9)
    percent_lst += [line]
    percent_int = [float(i) for i in percent_lst]
    min_value = 10000
    for percent in percent_int:
        if percent < min_value:
            min_value = percent
            return min_value 

temp=open_file()
print(temp)
print (find_min_percent(line,temp))

Answer 3

另一种方法：

def open_file(): 
    ''' Repeatedly prompt until a valid file name allows the file to be opened.'''
    while True: 
        user_input = input('Enter a file name: ')
        try: 
            file = open(user_input, 'r')
            print('user_input: ', user_input)
            line = file.readline(9)
            file.close()
            return find_min_percent(line)
        except FileNotFoundError: 
            print('Error. Please try again') 
            open_file()

def find_min_percent(line): 
    '''Find the min percent change in the line; return the value and the index.'''
    percent_lst = []
#    line = file.readline(9)
    percent_lst += [line]
    percent_int = [float(i) for i in percent_lst]
    min_value = 10000
    for percent in percent_int:
        if percent < min_value:
            min_value = percent
            return min_value 


print(open_file())

请注意，我不确定您的 find_min_percent 方法的正确性。此外，如果您手动打开文件（不使用with open），您也需要显式关闭。