如何传递字典的名称

Question

在下面的代码块中，我想传输字典的名称（result_dict），但不传输内容。

def define_var(dictionary, entry, counter): 
    for i in range(counter): 
    print "reason%d = %s['%s_%d']" % (i + 1, dictionary, entry, counter) 

当我调用函数等：

define_var(result_dict, 'start', 3) 

它打印：

reason1 = { 'START_2'： 'TEST2'， 'start_3'： 'TEST3'，' START_1 '： 'TEST1'}' START_1' ]

，但我想打印这样的：

reason1 = result_dict [START_1]

reason2 = result_dict [START_2]

等

Answer 1

您可以在全局对象：

def define_var(dictionary, entry, counter): 
    d_name = next(k for k, v in globals().items() if v is dictionary) 
    for i in range(counter): 
     print("reason%d = %s['%s_%d']" % (i + 1, d_name, entry, counter)) 

print(define_var(result_dict, 'start', 3)) 


reason1 = result_dict['start_3'] 
reason2 = result_dict['start_3'] 
reason3 = result_dict['start_3'] 

如果你在print(id(result_dict))和print(id(dictionary))中的功能当你通过result_dict 你会看到他们两个都是相同的对象，所以我们只用is从global字典中获得名称。

如果你想结果分配给一个理由的变量可以退货吗，是这样的：

def define_var(dictionary, entry, counter): 
    d_name = next(k for k, v in globals().items() if v is dictionary) 
    return "%s['%s_%d']" % (d_name, entry, counter) 

reason = define_var(result_dict,2,3) 
print(reason) 

或者使用的字典：

def define_var(dictionary, entry, counter): 
    d_name = next(k for k, v in globals().items() if v is dictionary) 
    reasons = {} 
    for i in range(counter): 
     reasons["reason{}".format(i)] = "{}[{}]".format(d_name, entry) 
    return reasons