Pandas/Python: Set value of one column based on value in another column

Question

I need to set the value of one column based on the value of another in a Pandas dataframe. This is the logic:

if df[\'c1\'] == \'Value\':
    df[\'c2\'] = 10         


        

Answer 1

try:

df['c2'] = df['c1'].apply(lambda x: 10 if x == 'Value' else x)

Answer 2

one way to do this would be to use indexing with .loc.

Example

In the absence of an example dataframe, I'll make one up here:

import numpy as np
import pandas as pd

df = pd.DataFrame({'c1': list('abcdefg')})
df.loc[5, 'c1'] = 'Value'

>>> df
      c1
0      a
1      b
2      c
3      d
4      e
5  Value
6      g

Assuming you wanted to create a new column c2, equivalent to c1 except where c1 is Value, in which case, you would like to assign it to 10:

First, you could create a new column c2, and set it to equivalent as c1, using one of the following two lines (they essentially do the same thing):

df = df.assign(c2 = df['c1'])
# OR:
df['c2'] = df['c1']

Then, find all the indices where c1 is equal to 'Value' using .loc, and assign your desired value in c2 at those indices:

df.loc[df['c1'] == 'Value', 'c2'] = 10

And you end up with this:

>>> df
      c1  c2
0      a   a
1      b   b
2      c   c
3      d   d
4      e   e
5  Value  10
6      g   g

If, as you suggested in your question, you would perhaps sometimes just want to replace the values in the column you already have, rather than create a new column, then just skip the column creation, and do the following:

df['c1'].loc[df['c1'] == 'Value'] = 10
# or:
df.loc[df['c1'] == 'Value', 'c1'] = 10

Giving you:

>>> df
      c1
0      a
1      b
2      c
3      d
4      e
5     10
6      g

Answer 3

I suggest doing it in two steps:

# set fixed value to 'c2' where the condition is met
df.loc[df['c1'] == 'Value', 'c2'] = 10

# copy value from 'c3' to 'c2' where the condition is NOT met
df.loc[df['c1'] != 'Value', 'c2'] = df[df['c1'] != 'Value', 'c3']

Answer 4

I had a big dataset and .loc[] was taking too long so I found a vectorized way to do it. Recall that you can set a column to a logical operator, so this works:

file['Flag'] = (file['Claim_Amount'] > 0)

This gives a Boolean, which I wanted, but you can multiply it by, say, 1 to make an Integer.

Answer 5

You can use pandas.DataFrame.mask to add virtually as many conditions as you need:

data = {'a': [1,2,3,4,5], 'b': [6,8,9,10,11]}

d = pd.DataFrame.from_dict(data, orient='columns')
c = {'c1': (2, 'Value1'), 'c2': (3, 'Value2'), 'c3': (5, d['b'])}

d['new'] = np.nan
for value in c.values():
    d['new'].mask(d['a'] == value[0], value[1], inplace=True)

d['new'] = d['new'].fillna('Else')
d

Output:

    a   b   new
0   1   6   Else
1   2   8   Value1
2   3   9   Value2
3   4   10  Else
4   5   11  11

Answer 6

Try out df.apply() if you've a small/medium dataframe,

df['c2'] = df.apply(lambda x: 10 if x['c1'] == 'Value' else x['c1'], axis = 1)

Else, follow the slicing techniques mentioned in the above comments if you've got a big dataframe.