How to update values in a specific row in a Python Pandas DataFrame?

Question

With the nice indexing methods in Pandas I have no problems extracting data in various ways. On the other hand I am still confused about how to change data in an existing DataFrame.

In the following code I have two DataFrames and my goal is to update values in a specific row in the first df from values of the second df. How can I achieve this?

import pandas as pd
df = pd.DataFrame({'filename' :  ['test0.dat', 'test2.dat'], 
                                  'm': [12, 13], 'n' : [None, None]})
df2 = pd.DataFrame({'filename' :  'test2.dat', 'n':16}, index=[0])

# this overwrites the first row but we want to update the second
# df.update(df2)

# this does not update anything
df.loc[df.filename == 'test2.dat'].update(df2)

print(df)

gives

   filename   m     n
0  test0.dat  12  None
1  test2.dat  13  None

[2 rows x 3 columns]

but how can I achieve this:

    filename   m     n
0  test0.dat  12  None
1  test2.dat  13  16

[2 rows x 3 columns]

Answer 1

So first of all, pandas updates using the index. When an update command does not update anything, check both left-hand side and right-hand side. If you don't update the indices to follow your identification logic, you can do something along the lines of

>>> df.loc[df.filename == 'test2.dat', 'n'] = df2[df2.filename == 'test2.dat'].loc[0]['n']
>>> df
Out[331]: 
    filename   m     n
0  test0.dat  12  None
1  test2.dat  13    16

If you want to do this for the whole table, I suggest a method I believe is superior to the previously mentioned ones: since your identifier is filename, set filename as your index, and then use update() as you wanted to. Both merge and the apply() approach contain unnecessary overhead:

>>> df.set_index('filename', inplace=True)
>>> df2.set_index('filename', inplace=True)
>>> df.update(df2)
>>> df
Out[292]: 
            m     n
filename           
test0.dat  12  None
test2.dat  13    16

Answer 2

In SQL, I would have do it in one shot as

update table1 set col1 = new_value where col1 = old_value

but in Python Pandas, we could just do this:

data = [['ram', 10], ['sam', 15], ['tam', 15]] 
kids = pd.DataFrame(data, columns = ['Name', 'Age']) 
kids

which will generate the following output :

    Name    Age
0   ram     10
1   sam     15
2   tam     15

now we can run:

kids.loc[kids.Age == 15,'Age'] = 17
kids

which will show the following output

Name    Age
0   ram     10
1   sam     17
2   tam     17

which should be equivalent to the following SQL

update kids set age = 17 where age = 15

Answer 3

If you have one large dataframe and only a few update values I would use apply like this:

import pandas as pd

df = pd.DataFrame({'filename' :  ['test0.dat', 'test2.dat'], 
                                  'm': [12, 13], 'n' : [None, None]})

data = {'filename' :  'test2.dat', 'n':16}

def update_vals(row, data=data):
    if row.filename == data['filename']:
        row.n = data['n']
    return row

df.apply(update_vals, axis=1)

Answer 4

Update null elements with value in the same location in other. Combines a DataFrame with other DataFrame using func to element-wise combine columns. The row and column indexes of the resulting DataFrame will be the union of the two.

df1 = pd.DataFrame({'A': [None, 0], 'B': [None, 4]})
df2 = pd.DataFrame({'A': [1, 1], 'B': [3, 3]})
df1.combine_first(df2)
     A    B
0  1.0  3.0
1  0.0  4.0

more information in this link

Answer 5

There are probably a few ways to do this, but one approach would be to merge the two dataframes together on the filename/m column, then populate the column 'n' from the right dataframe if a match was found. The n_x, n_y in the code refer to the left/right dataframes in the merge.

In[100] : df = pd.merge(df1, df2, how='left', on=['filename','m'])

In[101] : df
Out[101]: 
    filename   m   n_x  n_y
0  test0.dat  12  None  NaN
1  test2.dat  13  None   16

In[102] : df['n'] = df['n_y'].fillna(df['n_x'])

In[103] : df = df.drop(['n_x','n_y'], axis=1)

In[104] : df
Out[104]: 
    filename   m     n
0  test0.dat  12  None
1  test2.dat  13    16

Answer 6

If you want to put anything in the iith row, add square brackets:

df.loc[df.iloc[ii].name, 'filename'] = [{'anything': 0}]

Answer 7

I needed to update and add suffix to few rows of the dataframe on conditional basis based on the another column's value of the same dataframe -

df with column Feature and Entity and need to update Entity based on specific feature type

df.loc[df.Feature == 'dnb', 'Entity'] = 'duns_' + df.loc[df.Feature == 'dnb','Entity']