how to print items within a nested dictionary where values are of list | Sololearn: Learn to code for FREE!

0

how to print items within a nested dictionary where values are of list

hi i tried to describe as best i could - so within a dictionary with lists as values how can i print side by side the correct indexes within each list while calling out each key- this is more of a how it works question i suppose https://code.sololearn.com/cdNfA32UeR9x/?ref=app

11/28/2021 12:48:53 PM

ren[paused]

11 Answers

New Answer

+8

ren[paused] , to get the output as mentioned from you, we can use the zip() function to combine it like: 2021-06-10, 42.8 2021-06-11, 102.03 2021-06-12, 240.38 2021-06-13, 80.9 do a try with zip(), if you get stuck somewhere just post your attempt. happy coding!

+2

Not sure what you mean, but look here https://code.sololearn.com/cyfM8V4Dz7Nx/?ref=app

+1

Oh okay i gotcha. Pandas is good for this just because it's written in a way where everything is uniform. Using different data containers without the use of pandas just means you have to write specific code for whatever data set you're working with.

+1

gah i forgot bout that one 🤯

+1

i guess part of the idea is what do you do / best practice when you dont know entirely what oncoming data could be, but you need to build a container for it. that is what i think it feels like id be in for

+1

here is my attempt - https://code.sololearn.com/cuWfeYJY3Pew/?ref=app

+1

Nope! Lothar remembered a really useful one! As long as each list contains the same amount of values it doesnt matter how many keys you have. The zip method will stick each key[index] value into a tuple with the others of the same index. I forgot about that too haha

0

yea like that! i am learning pandas and i wondered how else could it be done - i didnt know how to phrase my question better

0

i kind of understand the basic idea for zip here? but when you dont know how many lists i guess you have to loop through them instead of statically input them like i just did :D

0

from previous (sounds like this could be a job for unpack operator?)

0

but do you need to resort to unpacking when it gets longer and you have 'a' : [0,1,2] , 'b : [1,3,10] , c : [0,8,9] and like many of these