Reputation: 147
I am trying to figure out how to do a vlookup to pick out the latest price to fill up a second table. An example below. For item #1, the latest price is at Month 6 (=$6)
while item #2 is at Month 5 (=$4)
. What's the best way to fill up Table B? Note: There might be occasion that item_id
cannot be found in Table A if the item is new.
Any guidance? Many Thanks.
Table A (Reference)
| Item_ID | Month | Price |
|---------|-------|-------|
| 1 | 4 | 10 |
| 1 | 5 | 8 |
| 1 | 6 | 6 |
| 2 | 5 | 4 |
Table B (To Fill)
| Shop_ID | Item_ID | Price |
|---------|---------|-------|
| 1 | 1 | 6 |
| 1 | 2 | 4 |
Upvotes: 1
Views: 1510
Reputation: 862751
To fill column Price
in df2
we can create a Pandas series with Item_ID and Price. Use drop_duplicates
for last row for each Item_ID
and create Series
by set_index
and selecting column. Lastly create new column with map
.
Full example:
import pandas as pd
# Sample data
data1 = dict(Item_ID=[1,1,1,2], Month=[4,5,6,5], Price = [10,8,6,4])
data2 = dict(Shop_ID=[1,1],Item_ID=[1,2])
# Create dfs
df1 = pd.DataFrame(data1)
df2 = pd.DataFrame(data2)
# Crete a series with Item_ID as index and Price as value
s = df1.drop_duplicates('Item_ID', keep='last').set_index('Item_ID')['Price']
# Create new column in df2
df2['Price'] = df2['Item_ID'].map(s)
print (df2)
Returns:
Shop_ID Item_ID Price
0 1 1 6
1 1 2 4
Further Details
If necessary use sort_values
first
s = (df1.sort_values(['Item_ID','Month'])
.drop_duplicates('Item_ID', keep='last')
.set_index('Item_ID')['Price'])
the serie s
looks like this:
Item_ID
1 6
2 4
Name: Price, dtype: int64
Upvotes: 2
Reputation: 19770
You can first find the latest information, then merge it to create the table:
import pandas
tableA = pandas.DataFrame({'Item_ID': {0: 1, 1: 1, 2: 1, 3: 2},
'Month': {0: 4, 1: 5, 2: 6, 3: 5},
'Price': {0: 10, 1: 8, 2: 6, 3: 4}})
tableB = pandas.DataFrame({'Item_ID': {0: 1, 1: 2},
'Price': {0: 6, 1: 4},
'Shop_ID': {0: 1, 1: 1}})
latest = tableA.loc[tableA.groupby('Item_ID')['Month'].idxmax()]
result = tableB[['Shop_ID', 'Item_ID']].merge(latest[['Item_ID', 'Price']],
on='Item_ID')
This yields
Shop_ID Item_ID Price
0 1 1 6
1 1 2 4
Upvotes: 1