用户自定义格式的输入.format（）

Question

我得到了一个包含4个变量的元组列表，这些元组将被打印在一行中，我希望用户指定这些打印格式。

这是我到目前为止的代码：

mylist = [ 
    ("Monday", "13", "November", "2017"), 
    ("Tuesday", "14", "November", "2017"), 
    ("Wednesday", "15", "November", "2017")] 
# Note: The real list have a thousand of items, not just 3. 

custom_format = self.GUI_field_where_user_types.get_text() 
for item in mylist: 
    weekday = item[0] 
    daynumber = item[1] 
    month = item[2] 
    year = item[3] 
    print(eval(custom_format)) 

# Example custom_format: '"{0}, {1} of {2} from year {3}".format(weekday, daynumber, month, year)' 

做检查，我只是想与上面的输入，它显然作品，但我不想使用eval，因为它显然太危险了。所以我的问题是：如何让用户以一种安全的方式定义格式？

注：用户应该能够在.format（）方法"{0}, {1} of {2} from year {3}"之前只能编辑字符串，阻止访问任何蟒蛇命令，但让他只能使用变量，他想要的，并不总是4其中。一些实例中可能是：

• {0}，{1}的{2}
• {3}，{1}在一个月{2}
• {0}，{1}的{2 }从今年{3}
• {1}任何{2} - {3}

Answer 1

这很简单。只需拨打.format输入字符串，并将项目传递与使用参数参数拆包语法：

mylist = [ 
    ("Monday", "13", "November", "2017"), 
    ("Tuesday", "14", "November", "2017"), 
    ("Wednesday", "15", "November", "2017")] 

fmt = input('Format? ') 
for item in mylist: 
    print(fmt.format(*item)) 

执行：

Format? {0}, {1} of {2} 
Monday, 13 of November 
Tuesday, 14 of November 
Wednesday, 15 of November 

处理用户输入无效的格式，你可以捕获该异常再试一次：

while True: 
    try: 
     fmt = input('Format? ') 
     for item in mylist: 
      print(fmt.format(*item)) 
     break 
    except (IndexError,ValueError,KeyError): 
     print('Invalid format.') 

执行：

Format? {A} 
Invalid format. 
Format? {4} 
Invalid format. 
Format? { 
Invalid format. 
Format? The year is {3} 
The year is 2017 
The year is 2017 
The year is 2017 

Answer 2

使用的方法来安排格式：

def fill_format(custom_format, date): 
    used_vars = [] 
    used_indexes = [] 

    #Look for what is used 
    for i in range(len(date)): 
     if "{"+str(i)+"}" in custom_format: 
      used_vars.append(date[i]) 
      used_indexes.append(i) 

    #Replace indexes starting from 0 
    for new_index, used_index in enumerate(used_indexes): 
     custom_format = custom_format.replace("{"+str(used_index)+"}", "{"+str(new_index)+"}") 

    return custom_format.format(*used_vars) 

mylist = [ 
    ("Monday", "13", "November", "2017"), 
    ("Tuesday", "14", "November", "2017"), 
    ("Wednesday", "15", "November", "2017")] 

custom_format = "{3}, {1} at month {2}" 
for item in mylist: 
    print(fill_format(custom_format, item)) 

输出：

2017, 13 at month November 
2017, 14 at month November 
2017, 15 at month November 

编辑：

如果日期一向4个时间单位，您可以使用：

def fill_format(custom_format, date): 
    return custom_format.replace("{0}", date[0]).replace("{1}", date[1]).replace("{2}", date[2]).replace("{3}", date[3]) 

Answer 3

我的建议是：坚持以标准化datetime格式。使用日期时间，我们可以使用强大的strftime根据一组预定义的模式打印日期时间。考虑下面这个例子：

import datetime 

mylist = [ 
    ("Monday", "13", "November", "2017"), 
    ("Tuesday", "14", "November", "2017"), 
    ("Wednesday", "15", "November", "2017")] 

custom_format = "%A, %d of %B from year %Y" 

for item in mylist: 
    # Create datetime object 
    dt = datetime.datetime.strptime('-'.join([*item]),"%A-%d-%B-%Y") 
    # Print with format 
    print(dt.strftime(custom_format)) 

打印：

Monday, 13 of November from year 2017 
Tuesday, 14 of November from year 2017 
Wednesday, 15 of November from year 2017 

---

如果你想创建“自己的语言”，使用字典映射到通用的标准。考虑下面这个例子：

import datetime 
import re 

mylist = [ 
    ("Monday", "13", "November", "2017"), 
    ("Tuesday", "14", "November", "2017"), 
    ("Wednesday", "15", "November", "2017")] 

d = {"WEEKDAY": "%A", 
    "DAYNUMBER": "%d", 
    "MONTH": "%B", 
    "YEAR": "%Y"} 

input_ = "YEAR-%m-DAYNUMBER is equal to WEEKDAY, DAYNUMBER of MONTH from year YEAR" 

# Use re library to change WEEKDAY --> %A and so on... 
pattern = re.compile(r'\b(' + '|'.join(d.keys()) + r')\b') 
custom_format = pattern.sub(lambda x: d[x.group()], input_) 

for item in mylist: 
    # Create datetime object 
    dt = datetime.datetime.strptime('-'.join([*item]),"%A-%d-%B-%Y") 
    # Print with format 
    print(dt.strftime(custom_format)) 

输出：

2017-11-13 is equal to Monday, 13 of November from year 2017 
2017-11-14 is equal to Tuesday, 14 of November from year 2017 
2017-11-15 is equal to Wednesday, 15 of November from year 2017 

这给了所有你需要用最少的代码的灵活性。