Update Python dictionary with appended list value

Question

I have a dataframe with price quotes for a variety of parts and makers. ~10k parts and 10 makers, so my dataset contains up to 100k rows, looking roughly like this:

Part	Maker	Price
1	Alpha	1.00
2	Alpha	1.30
3	Alpha	1.25
1	Bravo	1.10
2	Bravo	1.02
3	Bravo	1.15
4	Bravo	1.19
1	Charlie	.99
2	Charlie	1.10
3	Charlie	1.12
4	Charlie	1.19

I am wanting to return two dictionaries based on the best price, Part/Maker and Part/Price. My main issue is when two makers have the same best price.

I want my result to end up like this:

1:.99

2:1.1

3: 1.02

4:1.19

and the second one to be:

1:Charlie

2: Charlie

3: Bravo

4: [Bravo, Charlie]

The first dictionary is easy. Second one is what I’m stuck on. Here’s what I have so far:

winning_price_dict={}
winning_mfg_dict={}
for index, row in quote_df.iterrows():
   if row['Part'] not in winning_price_dict:
       winning_price_dict[row['Part']] = row['Proposed Quote']
       winning_mfg_dict[row['Part']] = list(row['Maker'])
   if winning_price_dict[row['Part']]>row['Proposed Quote']:
       winning_price_dict[row['Part']] = row['Proposed Quote']
       winning_mfg_dict[row['Part']] = row['Maker']
   if winning_price_dict[row['Part']]==row['Proposed Quote']:
       winning_price_dict[row['Part']] = row['Proposed Quote']
       winning_mfg_dict[row['Part']] = winning_mfg_dict[row['Part']].append(row['Maker']) #this is the only line that I don't believe works

When I run it as is, it says ‘str’ object has no attribute ‘append’. However, I thought that it should be a list because of the list(row[‘Maker’]) command.

When I change the relevant lines to this:

for index, row in quote_df.iterrows():
if row['Part'] not in winning_price_dict:
    winning_mfg_dict[row['Part']] = list(row['Mfg'])
if winning_price_dict[row['Part']]>row['Proposed Quote']:
    winning_mfg_dict[row['Part']] = list(row[['Mfg']])
if winning_price_dict[row['Part']]==row['Proposed Quote']:
    winning_mfg_dict[row['Part']] = list(winning_mfg_dict[row['Part']]).append(row['Mfg'])

The winning_mfg_dict is all the part numbers and NoneType values, not the maker names.

What do I need to change to get it to return the list of suitable makers?

Thanks!

Asked By: Stephen Juza

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Source

Answer 1

In your original code, the actual problem was on line 9 of the first fragment: you set vale to a string, not to a list. Also, calling list(some_string) dos not what you expect: it creates a list of single chars, not a [some_string].

I took the liberty to improve the overall readability by extracting common keys to variables, and joined two branches with same bodies. Something like this should work:

winning_price_dict = {}
winning_mfg_dict = {}

for index, row in quote_df.iterrows():
    # Extract variables, saving a few accesses and reducing line lengths
    part = row['Part']
    quote = row['Proposed Quote']
    maker = row['Maker']
    
    if part not in winning_price_dict or winning_price_dict[part] > quote:
        # First time here or higher value found - reset to initial
        winning_price_dict[part] = quote
        winning_mfg_dict[part] = [maker]
    elif winning_price_dict[part] == quote:
        # Add one more item with same value
        # Not updating winning_price_dict - we already know it's proper
        winning_mfg_dict[part].append(maker)

Answered By: SUTerliakov

Answer 2

You can use groupby to get all quotes for one part

best_quotes = quote_df.groupby("part").apply(lambda df: df[df.price == df.price.min()])

Then you get a dataframe with the part number and the previous index as Multiindex. The lambda function selects only the quotes with the minimum price.

You can get the first dictionary with

winning_price_dict = {part : price for (part, _), price in best_quotes.price.iteritems()}

and the second one with

winning_mfg_dict = {part:list(best.loc[part]["maker"]) for part in  best_quotes.index.get_level_values("part")}

Answered By: steflbert

Update Python dictionary with appended list value

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