How to assign multiple values to multiple keys with a single command

Question

I have dictionary like:

Info = {
  "City_Name" : {
    "Population" : None,
    "Population_Density" : None
    }
}

I want to assign values to "Population" and "Population_Density" keys. I actually can do that with the use of the following commands:

Info["City_Name"]["Population"] = 20000 
Info["City_Name"]["Population_Density"] = 200

But instead, I want to do that with a single command like:

Info["City_Name"]["Population","Population_Density"] = 20000 , 200

But this doesn't work, the command above generates a new key... (In fact, a function returns me those values, and therefore, I need to do that with a single command)

Edit:

I needed to mention; without using:

Info["City_Name"]["Population"],Info["City_Name"]["Population_Density"] = 20000, 200

The key-names of my dictionary are so long that, it is hard to follow; they take a lot of space. I also need to assig three values to three keys. Therefore, I was wondering if there is any way to do that with just a single modification on the part, that is different than each other (eg; "Population" and "Population_Density").

Try Info["City_Name"]["Population"],Info["City_Name"]["Population_Density"]=20000,200 — santosh-patil
– santosh-patil, Commented Jan 29, 2014 at 20:30

mhlester · Accepted Answer · 2014-01-29 21:52:07Z

2

Only way to do exactly what you're asking is:

Info["City_Name"]["Population"], Info["City_Name"]["Population_Density"] = 20000, 200

Otherwise it takes Info["City_Name"], and creates a new key ("Population", "Population_Density") (a tuple), and assigns another tuple, (20000, 200) to that

Or you can do it in two lines:

d = Info["City_Name"]
d["Population"], d["Population_Density"] = 20000, 200

edited Jan 29, 2014 at 21:52

answered Jan 29, 2014 at 20:30

mhlester

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4 Comments

T-800 Over a year ago

That's true, thanks a lot. But, the key-names of my dictionary are so long that, it is hard to follow; it takes a lot of space. (I am actually assigning three values to three keys). Therefore, I was wondering if there is any way to just to modify the part, which is different than each other. In our case; "Population" and "Population_Density".

T-800 Over a year ago

I tried different combinations of parentheses, curly braces etc. but, it didn't help...

mhlester Over a year ago

Not without storing an intermediate variable for Info["City_Name"] and then using that on a subsequent line

T-800 Over a year ago

If I will not receive any solution, I can do your suggestion. I tried to implement it but I couldn't. How can I do that; how can we store an intermediate variable for such a string?

dusan · Accepted Answer · 2014-01-29 21:47:18Z

1

Try this:

Info["City_Name"].update({"Population": 20000, "Population_Density": 200})

answered Jan 29, 2014 at 21:47

dusan

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5 Comments

T-800 Over a year ago

I get the numerical values from a function, and therefore, I cannot assign the values to key-names like the way you proposed. (For such a use; , I have to assign the values that my function returns to an intermediate variable. And then, I have to assign them the way you proposed. However, it is not the way I want to do it... But, anyway, thanks a lot for your response!)

dusan Over a year ago

So does your function return a list of values e.g. [20000, 200]?

T-800 Over a year ago

No, it actually returns arrays; my function is like: def myfunction(): ... return array1, array2, array3 So, I have an array instead of 20000 and another array of values instead of 200... But leaving the array structure aside, I even do not know how to implement it even if my function would have returned numerical values.

dusan Over a year ago

What about Info["City_Name"].update(dict(zip(["Population", "Population_Density"], myfunction())))?

T-800 Over a year ago

That's it! I have combined your solution with @mhlester 's solution; so; d= Info["City_Name"] and then: d.update(dict(zip(["Population", "Population_Density"], myfunction()))) . Now, it is exactly as I like to have. Thanks a lot!

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