Sort by price and remove prices

Question

I have a shopping list:

items = [
    ['Aspirin', 'Walgreens', 6.00],
    ['book lamp', 'Amazon', 2.87],
    ['popsicles', 'Walmart', 5.64],
    ['hair brush', 'Amazon', 6.58],
    ['Listerine', 'Walmart', 3.95],
    ['gift bag', 'Target', 1.50]
]

I want to sort the items from cheapest to highest price, and remove the prices. (I don't need them anymore then, I'll just buy from top down until I run out of money). Goal is:

items = [
    ['gift bag', 'Target'],
    ['book lamp', 'Amazon'],
    ['Listerine', 'Walmart'],
    ['popsicles', 'Walmart'],
    ['Aspirin', 'Walgreens'],
    ['hair brush', 'Amazon']
]

A way that works but looks clumsy (demo/template):

import operator
items = sorted(items, key=operator.itemgetter(2))
for i in range(len(items)):
    items[i] = items[i][:2]

Is there a shorter way?

@closer Bah, I wish you had more faith in me :-). There's a neat and educational solution that I'd like people to find and see. It's not on SO, I checked the database. Even Google finds only one case. — Kelly Bundy, Feb 10 '23 at 11:59
Can you clarify the "buy until I run out of money" part, please? The code you've provided makes no comparison to a total price. — gvee, Feb 10 '23 at 12:02
@gvee That's just the reason for why I don't need the prices anymore. Not part of the task. — Kelly Bundy, Feb 10 '23 at 12:03
By creating a new object a naively `map(operator.itemgetter(0, 1), sorted(...))` — cards, Feb 10 '23 at 12:10
If anyone wonders about that one previous case [Google](https://www.google.com/search?q=python+%22key%3Dlist.pop%22) found: it was in [2007-April.txt - Python mailing list](https://mail.python.org/pipermail/python-list/2007-April.txt). — Kelly Bundy, Feb 10 '23 at 19:15
Oops, actually I overlooked one there, and it *is* [on Codegolf](https://codegolf.stackexchange.com/a/222973/110278), haha @Juraj — Kelly Bundy, Feb 10 '23 at 19:18

score 5 · Accepted Answer · edited Feb 26 '23 at 05:23

5

I don't know how I really feel about using key for side-effects, but:

items.sort(key=list.pop)
print(items)

Result:

[['gift bag', 'Target'], ['book lamp', 'Amazon'], ['Listerine', 'Walmart'], ['popsicles', 'Walmart'], ['Aspirin', 'Walgreens'], ['hair brush', 'Amazon']]

edited Feb 26 '23 at 05:23

Sunderam Dubey

1
11
20
40

answered Feb 10 '23 at 17:32

slothrop

3,218
1
18
11

score 2 · Answer 2 · answered Feb 10 '23 at 11:56

2

Use the beauty of list comprehension in python.

items = sorted(items, key=lambda x: x[2])
items = [x[:2] for x in items]

answered Feb 10 '23 at 11:56

Kaustubh Dwivedi

418
4
16

2

Or in one line `items = [x[:2] for x in sorted(items, key=lambda x: x[2])]` – Stef Feb 10 '23 at 12:47

Kulasangar · Answer 3 · 2023-02-11T06:19:58.967

What if you use the in built sorted method:

items = [
    ['Aspirin', 'Walgreens', 6.00],
    ['book lamp', 'Amazon', 2.87],
    ['popsicles', 'Walmart', 5.64],
    ['hair brush', 'Amazon', 6.58],
    ['Listerine', 'Walmart', 3.95],
    ['gift bag', 'Target', 1.50]
]
# this sorts by the third element from your each list
new_list = sorted(items, key=lambda x: x[2]) 
    
for i in new_list:
    del i[2]

print(new_list)

Output:

score 2 · Answer 4 · answered Feb 10 '23 at 15:31

2

Someone suggested I could use more 'for' and fewer builtins:

_,items = zip(*sorted((z, x) for *x,z in items))

answered Feb 10 '23 at 15:31

Stef

13,242
2
17
28

Nah, no `for` intended. I was just joking regarding your first code here and [this](https://stackoverflow.com/questions/75385452/how-would-i-get-a-list-of-pairs-where-every-element-from-list1-is-paired-with-so/75395480#comment133025713_75385676). – Kelly Bundy Feb 10 '23 at 15:42
This is nice, but the intended solution is half as long. – Kelly Bundy Feb 10 '23 at 15:44

Jamiu S. · Answer 5 · 2023-02-10T13:02:29.103

1

Here is another approach:

items = list(map(lambda x: x[:2], sorted(items, key=lambda x: x[2])))
print(items)

You can also try operator

import operator
items = list(map(operator.itemgetter(0, 1), sorted(items, key=operator.itemgetter(2))))

[['gift bag', 'Target'], ['book lamp', 'Amazon'], ['Listerine', 'Walmart'], ['popsicles', 'Walmart'], ['Aspirin', 'Walgreens'], ['hair brush', 'Amazon']]

edited Feb 10 '23 at 13:02

answered Feb 10 '23 at 12:19

Jamiu S.

5,257
5
12
34

What about this? – Jamiu S. Feb 10 '23 at 12:24
Well, I dislike list(map(lambda, but I'm upvoting everything better than what I showed, and this still qualifies :-) – Kelly Bundy Feb 10 '23 at 12:37

score 1 · Answer 6 · answered Feb 10 '23 at 12:53

1

zipzipzip

from operator import itemgetter
from itertools import islice

print(
    list(zip(*islice(zip(*sorted(items, key=itemgetter(-1))),2)))
)
# [('gift bag', 'Target'), ('book lamp', 'Amazon'), ('Listerine', 'Walmart'), ('popsicles', 'Walmart'), ('Aspirin', 'Walgreens'), ('hair brush', 'Amazon')]

answered Feb 10 '23 at 12:53

Stef

13,242
2
17
28

1

Just plugging builtins together without looping yourself? Who does that? :-) – Kelly Bundy Feb 10 '23 at 12:55
But if that's your guess at what I meant: No, there's something **much** better. – Kelly Bundy Feb 10 '23 at 12:57
@KellyBundy Not really a guess, because `zip(*islice(zip(*...), 2))` doesn't seem much better than `map(itemgetter(0,1), ...)` – Stef Feb 10 '23 at 12:59

cards · Answer 7 · 2023-02-10T17:52:22.230

1

The inevitable numpy version. Select the first two columns by infering a reordering wrt to the values of the last columns.

import numpy as np

a = np.array(items)

a = a[:,:2][a[:,-1].argsort()]

edited Feb 10 '23 at 17:52

answered Feb 10 '23 at 16:55

cards

3,936
1
7
25

1

You can shorten that. – Kelly Bundy Feb 10 '23 at 17:02
Advice: slice thrice ≠ nice. Slice twice suffice. – Kelly Bundy Feb 10 '23 at 17:55
1

(Not sure it's called slicing, might be indexing, but I only realized afterwards.) – Kelly Bundy Feb 10 '23 at 17:57

Sort by price and remove prices

7 Answers7