How to pass the name of a dictionary

Question

In the following code block i want to transmit just the name of a dictionary (result_dict) but not the content.

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

when i call the function like:

define_var(result_dict, 'start', 3)

it prints:

reason1 = {'start_2': 'test2', 'start_3': 'test3', 'start_1': 'test1'}['start_1']

but i want to print it like that:

reason1 = result_dict[start_1]

reason2 = result_dict[start_2]

and so on

The idea of the "name" of the dictionary makes no sense for Python. — Ffisegydd, Feb 09 '15 at 10:01
From your code there is no way for the function to magically know that you think your dictionary is called `result_dict`, unless you pass in the string `"result_dict"`. — khelwood, Feb 09 '15 at 10:03
when i call the function i write result_dict as first parameter. but it prints me the content of result_dict instead of 'result_dict' — Alex Held, Feb 09 '15 at 10:04
The "name" of a variable (e.g. dictionary) is not an intricate part of the variable. Instead, it is a tag, floating in the namescope of the code that points to the variable. When you write ``define_var(result_dict)`` the function receives a pointer to the variable. In this sense, the "name" of the dictionary has no meaning to Python. — MrGumble, Feb 09 '15 at 12:32

Padraic Cunningham · Accepted Answer · 2015-02-09T12:17:52.707

0

You can find the object in globals:

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']

If you print(id(result_dict)) and print(id(dictionary)) in the function when you pass in result_dict you will see they both are the same object with the same id's so we just use is to get the name from the global dict.

If you want to assign the result to a reason variable you can return it, something like:

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)

Or use a dict:

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

edited Feb 09 '15 at 12:17

answered Feb 09 '15 at 10:22

Padraic Cunningham

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thanks man! i'm still new to python. don't know how it works, but it works and i'll find out! is there any way i can give you a honor or something similar? – Alex Held Feb 09 '15 at 10:27
@AlexHeld, it just checks identity using `is` which checks if two objects are in fact the same object to find the name of the `dictionary` passed from the `globals()` dict. – Padraic Cunningham Feb 09 '15 at 10:29
damn. using above but i got problems with assigning the variable using `eval("reason%d = %s['%s_%d']" % (i + 1, d_name, entry, counter"`. i know eval is a security risk but i won't publish the final program so it doesn't matter. – Alex Held Feb 09 '15 at 10:55
@AlexHeld "honor or something similar" - accept the answer (and upvote it) – Holloway Feb 09 '15 at 11:10
@AlexHeld, why do you need eval? – Padraic Cunningham Feb 09 '15 at 11:31
@Padraic Cunningham, i need it for assign the variable. or is there a better and safer way to do it? – Alex Held Feb 09 '15 at 12:01
@AlexHeld, use a dict – Padraic Cunningham Feb 09 '15 at 12:18

How to pass the name of a dictionary

1 Answers1