CODEWITHSUNDEEP
pythonpandasdataframe
Alejandro Alcalde
Alejandro Alcalde

Reputation: 6220

Update cells of a pandas DataFrame for a specific row

I want to update some cells in a row for a pandas DataFrame, I am using update to do it, but it always modify the first index only, this is a example:

df = pd.DataFrame(data={'cod':[1000,1001], 'B': ['b1','b2'], 'C':['c1','c2']})
updated_data = pd.DataFrame({'cod':[1001], 'C':['newC1']})
updated_data2 = pd.DataFrame({'cod':[1000], 'B':['newB1']})
df.update(updated_data)
df.update(updated_data2)

After this code, df will have:

      cod      B      C
0  1000.0  newB1  newC1
1  1001.0     b2     c2

When it should be

         cod      B      C
cod                       
1000  1000.0  newB1     c1
1001  1001.0     b2  newC1

In order to achieve, I wrote the following code, but do not know if its the best approach:

df = pd.DataFrame(data={'cod':[1000,1001], 'B': ['b1','b2'], 'C':['c1','c2']})
df = df.set_index(df.cod)
updated_data = pd.DataFrame({'cod':[1001], 'C':['newC1']})
updated_data = updated_data.set_index(updated_data.cod)
df.update(updated_data, overwrite=True)
updated_data = pd.DataFrame({'cod':[1000], 'B':['newB1']})
updated_data = updated_data.set_index(updated_data.cod)
df.update(updated_data, overwrite=True)

It seems to me its very verbose for something simple, is there another approach?

Update

This is the actual code, instead of having two updated_data, in reality is within a loop:

df = pd.DataFrame(data={'cod':[1000,1001], 'B': ['b1','b2'], 'C':['c1','c2']})
df = df.set_index(df.cod)
for i in (1000,1001):
    updated_data = pd.DataFrame({'cod':[i], 'C':['newC1']})
    updated_data = updated_data.set_index(updated_data.cod)
    df.update(updated_data, overwrite=True)

Upvotes: 5

Views: 1939

Answers (1)

DmytroSytro
DmytroSytro

Reputation: 639

In your case you can simply use:

df.loc[df.cod == 1001, 'C'] = 'newC1'
df.loc[df.cod == 1000, 'B'] = 'newB1'

To make it faster, it's better to set index:

df = df.set_index(df.cod)
df.loc[df.index == 1001, 'C'] = 'newC1'
df.loc[df.index == 1000, 'B'] = 'newB1'

You can use list of columns:

df.loc[df.index == 1001, ['C', 'B']] = ['newC', 'newB']

Upvotes: 4

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