Python 2D Array - Get Value by Key

Question

Suppose it is the following structure given:

from django.utils.translation import ugettext_lazy as _

# Constants for all available difficulty types.
SIMPLE = 1
MEDIUM = 2
DIFFICULT = 3

# Names for all available difficulty types.
DIFFICULTIES = (
    (SIMPLE, _("simple")),
    (MEDIUM, _("medium")),
    (DIFFICULT, _("difficult")),
)

How do you get the string value, if a constant is given?

A loop is easy to program, but is there a shorter python-like way with a single expression?

The expression

DIFFICULTIES[SIMPLE][1]

returns the string "medium". What is obviously wrong.

What are the underscores doing in there? _("simple") doesn't look like syntactically valid Python to me. — Junuxx
– Junuxx, Commented Nov 11, 2012 at 14:43
I'd guess that the underscores were there as references to gettext's _ function for internationalization (see here). — DSM
– DSM, Commented Nov 11, 2012 at 14:53

DSM · Accepted Answer · 2012-11-11 15:33:35Z

2

Of course you can use a dict, but the array is given.

So exchange it. (I'm assuming you kept the receipt..)

>>> dict(DIFFICULTIES)
{1: 'simple', 2: 'medium', 3: 'difficult'}
>>> d = dict(DIFFICULTIES)
>>> d[MEDIUM]
'medium'

Searching through an unsorted tuple for something simply isn't the right way to go about things. I suppose you could do

>>> next(v for k,v in DIFFICULTIES if k == MEDIUM)
'medium'

if you wanted to avoid a for loop with a colon, but that's a little silly.

answered Nov 11, 2012 at 15:33

DSM

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1 Comment

tom Over a year ago

Great, thanks! I've searched this expression dict(DIFFICULTIES)[SIMPLE]

Artsiom Rudzenka · Accepted Answer · 2012-11-11 15:25:47Z

1

it just because you specified a tuple, an indexing starting from 0, you either need to switch to dictionary or modify your constants with correct values:

DIFFICULTIES = {SIMPLE: "simple", MEDIUM: "medium", DIFFICULT: "difficult"}

OR:

SIMPLE, MEDIUM, DIFFICULT = range(3)

edited Nov 11, 2012 at 15:25

answered Nov 11, 2012 at 14:43

Artsiom Rudzenka

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2 Comments

Artsiom Rudzenka Over a year ago

Then what about second suggestion about const chg

tom Over a year ago

Would be a possibility, but I do not know how it impacts on the whole project, because other dependencies exist. I will solve it with a loop. Thought there would be a shorter solution.

Chris · Accepted Answer · 2012-11-11 14:43:54Z

0

Use a dictionary:

SIMPLE, MEDIUM, DIFFICULT = range(3)
DIFFICULTIES = {
    SIMPLE: _("simple"),
    MEDIUM: _("medium"),
    DIFFICULT: _("difficult")
}
DIFFICULTIES[SIMPLE] # will return "simple"

answered Nov 11, 2012 at 14:43

Chris

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3 Comments

Junuxx Over a year ago

NameError: name '_' is not defined

Chris Over a year ago

I just copied the code from the question. Guess those _ and the brackets gotta go

Chris Over a year ago

Then the only thing you can do is to go with Artsiom Rudzenka's answer (let the index start at 0, not at 1)

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