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
```