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

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

You can use np.where() to set values based on a specified condition:

#df
   c1  c2  c3
0   4   2   1
1   8   7   9
2   1   5   8
3   3   3   5
4   3   6   8

Now change values (or set) in column ['c2'] based on your condition.

df['c2'] = np.where(df.c1 == 8,'X', df.c3)

   c1  c2  c3
0   4   1   1
1   8   X   9
2   1   8   8
3   3   5   5
4   3   8   8