Python: right justified and ordered dict

Question

Following is my code:

inv = {'rope': 1, 'torch': 6, 'gold coin': 42, 'dagger': 1}

x=max(inv, key=lambda x: len(x.split()))
y=len(x)

#print(y)

n = "-"

q = n * (y+5)

#print(q)
#print("")

def print_table(inventory, order=None):

    if order=="count,asc":
        for key, value in sorted(inventory.iteritems(), key=lambda (k, v): (v, k)):
            print "%s %s" % (value, key)

print_table(inv,"count,asc")

I want to have something like this:

Inventory:
 count    item name
   ----------------
    45    gold coin
    12        arrow
     6        torch
     2       dagger
     1         rope
     1         ruby
    ----------------
  Total number of items: 67

Order parameter must be a string, which works as following: empty (by default) means the table is unordered "count,desc" table is ordered by count (of items in the inventory) in descending order and "cound,asc" in ascending order

I wrote function that's finding longest string in each of the inner lists to know how wide should be column to fit all strings, but now I'm stuck at this point, what should I do now?

The `pandas` library will help to translate this data into a column/row format where sorting can be performed etc. — tda, Sep 07 '18 at 13:37

Hai Vu · Accepted Answer · 2018-09-09T02:53:57.877

You have the right idea to find the longest key and its length. All you need is to format it:

def print_table(inventory, order=None):
    longest_key = max(inventory, key=len)
    max_width = len(longest_key)

    if order == "count,asc":
        for key, value in sorted(inventory.iteritems(), key=lambda (k, v): (v, k)):
            print '%6d %*s' % (value, max_width, key)

Output:

     1    dagger
     1      rope
     6     torch
    42 gold coin

Update

To address different sort columns and order (ascending, descending), I would like to suggest not passing in a string, but a key function and sort direction (ascending or descending):

def print_table(inventory, order=None, descending=False):
    longest_key = max(inventory, key=len)
    max_width = len(longest_key)

    for key, value in sorted(inventory.iteritems(), key=order, reverse=descending):
        print '%6d %*s' % (value, max_width, key)

# Define a bunch of key functions, used for sorting
by_count = lambda (k, v): (v, k)
by_name = None

inv = {'rope': 1, 'torch': 6, 'gold coin': 42, 'dagger': 1}

print '\nBy count, ascending'
print_table(inv, order=by_count)

print('\nBy name, descending')
print_table(inv, order=by_name, descending=True)

Note: By default, the sorted function will sort a dict.items() by (k, v), so the by_name function can be None, or it can be a more explicit lambda (k, v): (k, v)

What if I want to change order, f.e to descending? – PatrykMandrak Sep 07 '18 at 14:08 — PatrykMandrak, Sep 07 '18 at 14:08

score 0 · Answer 2 · answered Sep 07 '18 at 14:11

If this is what you are trying to do,

reverse=True in your sorted method will do!

def print_table(inventory, order=None):    
    if order=="count,asc":
        for key, value in sorted(inventory.iteritems(), key=lambda (k, v): (v, k)):
            print str(value).rjust(10), key.rjust(10)
    elif order=="count,desc":
        for key, value in sorted(inventory.iteritems(), key=lambda (k, v): (v, k), reverse=True):
            print str(value).rjust(10), key.rjust(10)

print_table(inv,"count,desc")

Python: right justified and ordered dict

2 Answers2

Update