Dictionary from 2 lists and a range?

Question

I'm having some issues trying to create a dictionary from 2 lists.

I'm trying to get this as output.

{1: {'name1', 'age1'}, 2: {'name2', 'age2'}, 3: {'name3', 'age3'}}

My current code: (I'm new to coding so this is a mess and very efficient, and yes I do realize I'm running all code below 5 times)

def Update():
    for r in range(1,5):
        v1list.append(StringVar())
        v2list.append(StringVar())
        NameEntry = Entry(root, textvariable=v1list[r-1]).grid(row = r, column = 0, sticky=W)
        AgeEntry = Entry(root, textvariable=v2list[r-1]).grid(row = r, column = 4, sticky=W)
        for var1, var2 in ((a,b) for a in v1list for b in v2list):
            UserInput[r] = {var1.get(), var2.get()}
            print(UserInput)

The output keeps pushing var1 and var2 forward resulting in this as output:

{1: {''}, 2: {''}, 3: {''}, 4: {''}}

I do actually see all inputs in there they just keep moving forward in the dictionary.

Because they are StringVar() and If I don't use .get() it doesn't work. — Eddie The Eagle, Feb 15 '18 at 19:51
`{x:y}` not `{x,y}`, but it looks like there are bugs, since your `varx.get()` always return empty strings... — juanpa.arrivillaga, Feb 15 '18 at 19:56
@juanpa.arrivillaga That's because it's in a loop I think, yeh the entire code is a mess but I'm trying to learn how to do it. When I change it to UserInput = {r:(var1.get(), var2.get())} it does return {1: ('name1', 'age1')} at one point but changes it to "" later on. — Eddie The Eagle, Feb 15 '18 at 20:01
No dude, you're still not getting it: `{var1.get() **:** var2.get()}` — juanpa.arrivillaga, Feb 15 '18 at 20:03

sacuL · Accepted Answer · 2018-02-15T20:32:29.963

Try creating the dictionary through a dictionary comprehension:

names = ['name1', 'name2', 'name3']
ages = ['age1', 'age2', 'age3']

name_age_dict = {i+1:{names[i],ages[i]} for i in range(len(names))}

>>> name_age_dict
{1: {'age1', 'name1'}, 2: {'name2', 'age2'}, 3: {'age3', 'name3'}}

It's cleaner than a for loop, and remains readable. Essentially, it goes through each item in your lists (assuming the lists have the same length), and creates a dictionary of {name, age} for each item in each list, assigning the index number (+1 to skip over the 0) as the key for that dictionary.

Some new python users have a difficult time with the list and dict comprehension syntax, so if you prefer to see it as a loop, this achieves the same thing:

name_age_dict = {}

for i in range(len(names)):
    name_age_dict[i+1] = {names[i], ages[i]}

>>> name_age_dict
{1: {'age1', 'name1'}, 2: {'name2', 'age2'}, 3: {'age3', 'name3'}}

Thank you so much :). Finally got it working after 1 day. – Eddie The Eagle Feb 15 '18 at 20:33 — Eddie The Eagle, Feb 15 '18 at 20:33

score 0 · Answer 2 · answered Feb 15 '18 at 20:18

0

Hope this solves your issue.

    d = {}
    names=['name1','name2','name3','name4','name5']
    age=[1,2,3,4,5]
    for i in range(1,6):
        d[str(i)]={}
        d[str(i)][names[i-1]]=age[i-1]
    print(d)

answered Feb 15 '18 at 20:18

Abhisek Roy

582
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Dictionary from 2 lists and a range?

2 Answers2