What I'd like to do is the following:
test_dict = dict(value = 1,
property = properties_list[value-1]
)
Naturally, I couldn't do it this way, since it throws me an error saying that name 'value' is not defined.
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Gabriel Dante
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3What are you trying to accomplish by doing this? – Scott Hunter Jan 11 '21 at 14:25
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2The question is a little hard to follow. What exactly are you trying to achieve? – mandulaj Jan 11 '21 at 14:25
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2Declare `value = 1` before and then do `test_dict = dict(value=value, property =properties_list[value-1])`. Is there any problem with that? – Countour-Integral Jan 11 '21 at 14:25
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I have a previously defined list (e.g. properties_list) from which I extract the value of the 'property' key of the dict. But the selected item of the properties_list depend on the 'value' key of the dict. Is it clearer? – Gabriel Dante Jan 11 '21 at 14:27
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@Countour-Integral I don't see a problem with that, I'd just like to know if there's a way to do it without having to create another variable outside the dict – Gabriel Dante Jan 11 '21 at 14:28
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you need to be sure all your keys are int and not string , or am I wrong – pippo1980 Jan 11 '21 at 14:29
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test_dict[value] - 1 ?? no its the key value not the key – pippo1980 Jan 11 '21 at 14:30
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@pippo1980 that doesn't work, since the test_dict is not yet defined – Gabriel Dante Jan 11 '21 at 14:33
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But since `value = 1`, did you mean: `property = properties_list[0]`? – quamrana Jan 11 '21 at 14:35
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@quamrana yes, that's why I subtract 1 from ```value```. – Gabriel Dante Jan 11 '21 at 14:37
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So you meant: `dict(value = 1, property = properties_list[0])`? – quamrana Jan 11 '21 at 14:38
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I am at loss , do you want your property value to be the 'value key' [value -1] or 'value key' -1 ?? , just change name to your first dict_test key to be clearer, (I know dictionary dont have first items, I meant first in your code) – pippo1980 Jan 11 '21 at 14:40
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I am bookmarking this – pippo1980 Jan 11 '21 at 14:42
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Python-3.8: `dict(value=(x := calc()), property=properties_list[x - 1])`. – ekhumoro Jan 11 '21 at 14:44
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The ```property``` will be chosen on the ```properties_list``` accordingly to the ```value```. So if ```value = 1``` then ```property = properties_list[0]```, if ```value = 2``` then ```property = properties_list[1]``` and so on – Gabriel Dante Jan 11 '21 at 14:45
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1@GabrielDante no, there is no other way if you want to define it all in one go. – juanpa.arrivillaga Jan 11 '21 at 14:47
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@ekhumoro well, that creates a variable too. – juanpa.arrivillaga Jan 11 '21 at 14:48
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@juanpa.arrivillaga So would creating an entry in the dict. The value has to go *somewhere*, right? – ekhumoro Jan 11 '21 at 14:52
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[What do I do when I need a self referential dictionary?](https://stackoverflow.com/q/3738381/984421). – ekhumoro Jan 11 '21 at 14:58
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@ekhumoro um, no, creating an entry in the dict does not create a variable? I'm not sure how you figure that. – juanpa.arrivillaga Jan 11 '21 at 14:59
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@juanpa.arrivillaga Because the dict is a variable (i.e. like `test_dict` in the answer below). – ekhumoro Jan 11 '21 at 15:00
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@ekhumoro right, but it doesn't create *an additional variable* which the OP says they are trying to avoid. So there is on variable, `test_dict`, with your solution, there are two, `test_dict` and `x`, which is how I've interpreted the OP's comments as trying to avoid – juanpa.arrivillaga Jan 11 '21 at 15:02
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@juanpa.arrivillaga This is getting silly, but if you must: `x = dict(value=(x := calc()), property=properties_list[x - 1])`. Now there's only one variable ;-) – ekhumoro Jan 11 '21 at 15:07
2 Answers
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Since you can add keys after the dictionary is defined you could do
test_dict = dict(value = 1)
test_dict['property'] = properties_list[test_dict['value']-1]
jodag
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You could do it by using an assignment expression (aka “the walrus operator”) which were introduced in Python 3.8.
properties_list = [42, 13]
test_dict = dict(value=(_:=1), property=properties_list[_-1])
print(test_dict) # -> {'value': 1, 'property': 42}
martineau
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@martineau the OP seems to want to avoid that, although it isn't completely clear – juanpa.arrivillaga Jan 11 '21 at 15:06
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