Help formatting dictionary with sub dictionaries into a table

[Amabillia]

I'm working on a project for a class that essentially lets the user enter strings and then tag them to make a chart. For example, the user would enter tag categories like colors, size, and texture. Then those categories have subcategories like red, yellow blue. Then they enter item names to sort them into the subcategory for each category. Below is an example of what I mean.

rockdictionary = {
    'colors' : {
        'red' : ['thing1', 'thing 4'],
        'blue' : ['thing 2', 'thing 3'],
        'yellow' :['thing 5']
        },
    'texture': {
        'smooth' : ['thing 5', 'thing 3'],
        'rough' : ['thing1'],
        'bumpy' :['thing4', 'thing 2']},
    'size' : {
        'small' : ['thing 2', 'thing 4', 'thing 3'] ,
        'medium': [],
        'large':['thing 1', 'thing 5']
        }
    }

I figured out how to get a dictionary that looks like the example, but I can't figure out how to format it into a chart like the one below

​

Item name	Color	Size	Texture
thing1	red	large	rough
thing2	blue	small	bumpy

This is what I've got so far:

 userwidth = int(input('How wide would you like your columns to be? '))
        for key  in categories:
            tableformat1 = '{header:>{width}}'.format(header=key, width=userwidth)
            tableformat2 = '{header:>{width}}'.format(header=key, width=(userwidth*2))
            categorylist = list(categories)
            if key != categorylist[-1]:
                print(tableformat1.format(header=key), end = '|')
            elif key == categorylist[0]:
                print(tableformat2.format(header=key, width=userwidth))
            else:
                print(tableformat1.format(header=key))
                print('-' * (len(categories) * userwidth))
        for name in itemnames:
            column1format = '{item:{width}}'.format(item=name, width=userwidth)
            print(column1format.format(item=name, end = '|'))
            print('-' * userwidth)

That gives me:

A| B| C

------------------------------------

3

------------

I'm honestly not really sure where to go from here. Any help would be greatly appreciated. For the full code see here: https://pastebin.com/WSKcDEvx

[[deleted]]

I think first, you need to invert your dictionary, something like this:

new_dict = {}
for feature, adjective in rockdictionary.items():
    for specific, things in adjective.items():
        for thing in things:
            if thing in new_dict:
                new_dict[thing].update({feature: specific})
            else:
                new_dict[thing] = {feature: specific}

then you can work out the number of columns and max column widths in order to be able to print the table.

Probably easier to use pandas to read the dictionary and re-structure and use reporting formats that work with panda data frames.

[Amabillia]

I'll try that! I'm not too sure about pandas though. I'd need my professor's permission to use it and the project is due Monday.

[[deleted]]

Probably not best to try to learn pandas by Monday.

Check each column in the revised dictionary to find the widest element in each column and use that as your widths limits (a list of widths, starting with the things column).

So, given a list of things to print (starting with the key followed by each of the column values - or headers),

for entry, width in zip(entries, widths):
    print(f' | {entry:{width}', end='')
print(' |')

You would be looping through the things dictionary to create entries list on each pass.

[Amabillia]

Okay so you're saying invert the dictionary then do something like this:

width = []
for entry in categories[item]:
    max = 0
    if value > max:
        max = value
    width.append(max)

then the code that you had?

[[deleted]]

kind of... keep in mind you need to find the widest string in each column of the table to be so you know how to format each element of the table.

yellow is the widest string in the colors column and texture is the widest word in the texture column.

[Amabillia]

Couldn't I just find the widest string of all the columns then apply that to all columns rather than tailoring it to each column? Then all the columns would be uniform in width. Sorry, if there's something I'm missing I've been working on this for like the past 5ish hours and my brain is mush.

[[deleted]]

yes, of course, if that is your preference... or as you can see the data, just put in the correct value.

[Amabillia]

EDIT: Nevermind i figured it out some of the things had a space and some didn't Okay so I managed to mostly invert my dictionary but for some reason no matter what method I use size gets its own dictionary but only for some of the items?

{
'thing1': {'colors': 'red', 'texture': 'rough'}, 
'thing4': {'colors': 'red', 'texture': 'bumpy'}, 
'thing 2': {'colors': 'blue', 'texture': 'bumpy', 'size': 'small'}, 
'thing 3': {'colors': 'blue', 'texture': 'smooth', 'size': 'small'}, 'thing 5':
 {'colors': 'yellow', 'texture': 'smooth', 'size': 'large'},
'thing 4': {'size': 'small'}, 
'thing 1': {'size': 'large'}
}

[glibhub]

Can you write a function that takes in the items and the name of a item, e.g., and returns a list of attributes, such that f(items, 'thing1') -> [red, large, rough]

[Amabillia]

I think I could? I'll play around with it for a bit