Conditionally fill column values based on another columns value in pandas

Taking Tom Kimber's suggestion one step further, you could use a Function Dictionary to set various conditions for your functions. This solution is expanding the scope of the question.

I'm using an example from a personal application.

# write the dictionary

def applyCalculateSpend (df_name, cost_method_col, metric_col, rate_col, total_planned_col):
    calculations = {
            'CPMV'  : df_name[metric_col] / 1000 * df_name[rate_col],
            'Free'  : 0
            }
    df_method = df_name[cost_method_col]
    return calculations.get(df_method, "not in dict")

# call the function inside a lambda

test_df['spend'] = test_df.apply(lambda row: applyCalculateSpend(
row,
cost_method_col='cost method',
metric_col='metric',
rate_col='rate',
total_planned_col='total planned'), axis = 1)

  cost method  metric  rate  total planned  spend
0        CPMV    2000   100           1000  200.0
1        CPMV    4000   100           1000  400.0
4        Free       1     2              3    0.0

You probably want to do

df['Normalized'] = np.where(df['Currency'] == '$', df['Budget'] * 0.78125, df['Budget'])

Similar results via an alternate style might be to write a function that performs the operation you want on a row, using row['fieldname'] syntax to access individual values/columns, and then perform a DataFrame.apply method upon it

This echoes the answer to the question linked here: pandas create new column based on values from other columns

def normalise_row(row):
    if row['Currency'] == '$'
    ...
    ...
    ...
    return result

df['Normalized'] = df.apply(lambda row : normalise_row(row), axis=1) 

An option that doesn't require an additional import for numpy:

df['Normalized'] = df['Budget'].where(df['Currency']=='$', df['Budget'] * 0.78125)