Python程序：读取文件

Question

我应该编写一个程序，它不断地向用户提供一个文件名，直到它输入正确的文件名为止。然后，find_min_percent应该接受一个参数，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。这里是我的代码到目前为止

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)) 

我的问题是与readline（）。它说变量文件是未定义的。此代码的大纲不包含“def find_min_percent（line）：”部分中的文件。所以我不知道如何去解决这个问题。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能保持不变

Answer 1

您在一个函数中定义的变量不能从另一个函数访问。为了解决这个问题，你可以举例来说，做到这一点（存储在一个“主要功能”变量返回的值，并把它传递给你的下一个函数）：

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，你用你的line变量的方式很奇怪。它仅在find_min_percent内部使用，但在函数的外部定义，甚至作为参数传递。为什么？你想达到什么目的？

（参见here用于关于功能之外定义变量的访问邮政）

Answer 2

返回file变量超出该函数的范围 固定编码的：

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

的另一种方式这样做：

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，我不是sur e关于您的find_min_percent方法的正确性。另外，如果您手动打开文件（不使用with open），则还需要明确关闭。