I have following two Data Frames:
df1 = pd.DataFrame({\'ids\':[1,2,3,4,5],\'cost\':[0,0,1,1,0]})
df2 = pd.DataFrame({\'ids\':[1,5],\'cost\':[1,4]})
You can use set_index and combine first to give precedence to values in df2
df_result = df2.set_index('ids').combine_first(df1.set_index('ids'))
df_result.reset_index()
You get
ids cost
0 1 1
1 2 0
2 3 1
3 4 1
4 5 4
Another way to do it, using a temporary merged dataframe which you can discard after use.
import pandas as pd
df1 = pd.DataFrame({'ids':[1,2,3,4,5],'cost':[0,0,1,1,0]})
df2 = pd.DataFrame({'ids':[1,5],'cost':[1,4]})
dftemp = df1.merge(df2,on='ids',how='left', suffixes=('','_r'))
print(dftemp)
df1.loc[~pd.isnull(dftemp.cost_r), 'cost'] = dftemp.loc[~pd.isnull(dftemp.cost_r), 'cost_r']
del dftemp
df1 = df1[['ids','cost']]
print(df1)
OUTPUT-----:
dftemp:
cost ids cost_r
0 0 1 1.0
1 0 2 NaN
2 1 3 NaN
3 1 4 NaN
4 0 5 4.0
df1:
ids cost
0 1 1.0
1 2 0.0
2 3 1.0
3 4 1.0
4 5 4.0
You could do this with a left merge:
merged = pd.merge(df1, df2, on='ids', how='left')
merged['cost'] = merged.cost_x.where(merged.cost_y.isnull(), merged['cost_y'])
result = merged[['ids','cost']]
However you can avoid the need for the merge (and get better performance) if you set the ids as an index column; then pandas can use this to align the results for you:
df1 = df1.set_index('ids')
df2 = df2.set_index('ids')
df1.cost.where(~df1.index.isin(df2.index), df2.cost)
ids
1 1.0
2 0.0
3 1.0
4 1.0
5 4.0
Name: cost, dtype: float64