Merge 'left', but override 'right' values where possible

Question

Objective

I've reviewed pandas documentation on merge but have a question on overriding values efficiently in a 'left' merge. I can do this simply for one pair of values (as seen here), but it becomes cluttered when trying to do multiple pairs.

Setup

If I take the following dataframes:

a = pd.DataFrame({
   'id': [0,1,2,3,4,5,6,7,8,9],
    'val': [100,100,100,100,100,100,100,100,100,100]
})

b = pd.DataFrame({
    'id':[0,2,7],
    'val': [500, 500, 500]
})

I can merge them:

df = a.merge(b, on=['id'], how='left', suffixes=('','_y'))

to get

   id  val  val_y
0   0  100  500.0
1   1  100    NaN
2   2  100  500.0
3   3  100    NaN
4   4  100    NaN
5   5  100    NaN
6   6  100    NaN
7   7  100  500.0
8   8  100    NaN
9   9  100    NaN

I want to keep left values where no right value exists, but where possible overwrite with the right values.

My desired outcome is:

   id    val
0   0  500.0
1   1  100.0
2   2  500.0
3   3  100.0
4   4  100.0
5   5  100.0
6   6  100.0
7   7  500.0
8   8  100.0
9   9  100.0

My Attempt

I know I can accomplish this with a few lines of code:

df.loc[df.val_y.notnull(), 'val'] = df[df.val_y.notnull()].val_y
df = df.drop(['val_y'], axis = 1)

Or I can use the logic from this question.

But this becomes cluttered when there are multiple column pairings where I want to apply this logic.

For example, using a and b below:

a = pd.DataFrame({
   'id': [0,1,2,3,4,5,6,7,8,9],
    'val': [100,100,100,100,100,100,100,100,100,100],
    'val_2':[200, 200, 200, 200, 200, 200, 200, 200, 200, 200]
})
b = pd.DataFrame({
    'id':[0,2,7],
    'val': [500, 500, 500],
    'val_2': [500,500,500]
})

Is there a quicker, cleaner way to get my desired outcome?

Answer 1

I'd do this using set_index and update:

u = a.set_index('id')
u.update(b.set_index('id'))  # Update a's values with b's values

u.reset_index()

   id    val
0   0  500.0
1   1  100.0
2   2  500.0
3   3  100.0
4   4  100.0
5   5  100.0
6   6  100.0
7   7  500.0
8   8  100.0
9   9  100.0

The update is aligned on the index. For this reason, I set "id" to be the index in both DataFrames before performing the update step.

Note that the "id" column must be unique.

---

Another option is using concat and drop_duplicates:

pd.concat([b, a]).drop_duplicates('id').sort_values('id')

   id  val
0   0  500
1   1  100
1   2  500
3   3  100
4   4  100
5   5  100
6   6  100
2   7  500
8   8  100
9   9  100

Since b overrides a, b must come first in the concat step.

Answer 2

numpy searchsorted and assign

a.iloc[np.searchsorted(a.id,b.id),1]=b.val.values
a
Out[1382]: 
   id  val
0   0  500
1   1  100
2   2  500
3   3  100
4   4  100
5   5  100
6   6  100
7   7  500
8   8  100
9   9  100

Answer 3

Goofing Off with dict

d = dict(a.values)
d.update(dict(b.values))
pd.DataFrame(dict(zip(a, zip(*d.items()))))

   id  val
0   0  500
1   1  100
2   2  500
3   3  100
4   4  100
5   5  100
6   6  100
7   7  500
8   8  100
9   9  100