02: Filtering and Sorting/Chipotle, Step-4 & 5

[guipsamora]

02: Filtering and Sorting/Chipotle, Step-4 & 5

[bobfang1992]

I agree with you. I think the answer provide is simply wrong.

[maticalderini]

I agree too. However, to get the number of products costing more than $10.00, I believe you could use a simpler command:
chipo.loc[chipo['item_price']/chipo['quantity'] > 10, 'item_name'].drop_duplicates().shape[0]
Hope this helps other newcomers.

[bromero26]

I came up with the same issue. The question needs to be reconsidered, or the proposed solution changed. I believe the number of unique products with a price higher than 10 is 31. I use this code, after changing the column to float:
chipo[chipo['item_price']>10]['item_name'].nunique()

[rahimnathwani]

@bromero26 I get the same answer as you using a slightly more verbose method:

min_max_price_per_item = chipo.groupby('item_name').agg({'item_price': [np.max, np.min]})
min_max_price_per_item[min_max_price_per_item.item_price.amax > 10].shape[0]

But all of those high prices are caused by extras or specific configurations. If you stick to the basics, any item can be had for less than $10:

min_max_price_per_item[min_max_price_per_item.item_price.amin > 10].shape[0]

The question is not well-formed. It could be asking:

1. Which products did at least one person order for more than $10? (A: 31)
2. Which products always cost at least $10, regardless of choice_description? (A: 0)
3. Which product combinations (combination of item_name and choice_description) cost at least $10? (A: 777)

[AndreaAmico]

@rahimnathwani I agree with you, the question is not well-formed. Still, I believe that @matiascalderini suggestion is correct since price vs quantity seems quite linear. Check using water bottles as an example:

(
    chipo.query('item_name == "Bottled Water"')[['quantity', 'item_price']]
    .groupby('quantity')
    .agg(['mean', 'std'])
    .item_price
    .reset_index()
    .plot(x='quantity', y='mean', yerr='std', kind='scatter')
)

Normalizing the cost I get these values:

chipo['price_per_item'] = chipo.item_price/chipo.quantity
A1 = chipo.query('price_per_item > 10').item_name.nunique()
A2 = (chipo.groupby('item_name').price_per_item.min()>10).sum()
chipo['name_with_variants'] = chipo.item_name+chipo.choice_description
A3 = (chipo.groupby('name_with_variants').price_per_item.min()>10).sum()

print(f'A1:{A1}, A2:{A2}, A3:{A3}')

A1:25, A2:0, A3:707

[guipsamora]

Hi everyone, thank you for the comments and feedback. I agree that this question is not so clear too.

Some clarifications:

1. There is a clear distinction of order_id, quantity and product. Example, in a same order_id, you can ask a product in a quantity greater than 1, which will influence the price.

Example:
order _id | quantity | item_name | choice_description | item_price
9 | 2 | Canned Soda | [Sprite] | $2.18
14 | 1 | Canned Soda | [Dr. Pepper] | $1.09

Canned Soda costs $1.09. If I buy 10 sodas, the line will show up $10.90, which is greater than $10, but that doesn't mean that the product Canned Soda costs more than $10.

That is the reason that quantity needs to be considered for this exercise.

2. In order to simplify the exercise take the combination of item_name + choice_description as "one product".

Example:

order _id | quantity | item_name | choice_description | item_price
12 | 1 | Chicken Burrito | [[Tomatillo-Green Chili Salsa (Medium), Tomati... | $10.98
8 | 1 | Chicken Burrito | [Tomatillo-Green Chili Salsa (Medium), [Pinto ... | $8.49

"Chicken Burrito" is the "main" product but depending on the additional items it will cost more or less than $10, so to simplify take the combination item_name + choice_description as "one product".

Considering that what is your suggestion? Send me a PR! 😉

[newera-001]

In the getting and knowing your data part ,the url donnot work.what should I do?