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Match three 'isin' statements and return index where they match, then replace value if all conditions are satisfied

I want to find where three different columns in two dataframes match and then replace a value in another column if they match. How do I combine the three conditions below, perhaps in a loop form, to check if each row matches all three conditions?

First two conditions - matching the values in two different columns:

DATA1[DATA1[‘f_code’].isin(DATA2[‘F’])] # Column 1
DATA1[DATA1[‘b_id’].isin(DATA2[‘B’])] # Column 2


Last condition which I am writing wrong. I want the value in DATA1 'species' to match the header of DATA2 if the above two conditions are true
DATA1[DATA1[‘species’] == DATA2[header = ‘SO2’]

Then, if the conditions are met in that row, I want to replace the value the column named 'value' in DATA1 with the value that was under the 'SO2' header for that row.

How can I do this?

I tried this (not written correctly):

for a, elem in enumerate(DATA1['f_code']):
if DATA1['f_code'].loc[a] in DATA2['F'] and DATA1['b_id'].loc[a] in DATA2['B']
replace(DATA1['val'] with DATA2['value']


Example:

DATA1

Species Value f_code B
50000 0.00 55933 TPP1
**NOX 0.38 55933 TPP1
NOX 0.48 55933 TPP1**
SO2 0.02 55933 TPP1
50000 0.00 55933 TPP2
NOX 0.31 55933 TPP2
SO2 0.01 55933 TPP2
50000 0.07 6002 1
NOX 2377.90 6002 1
SO2 2354.10 6002 1
50000 0.07 6002 2
NOX 2998.60 6002 2
SO2 531.00 6002 2


DATA2

F B Nox SO2
6002 1 1151.97 152.98
6002 2 1550.17 213.3
6002 3 2428.71 226.26
6002 4 2135.05 219.42
55510 GT-1 0.85 0.05
55542 1 3.05 0.10
55542 2 3.17 0.11
55933 TPP1 11.33 0.64
55933 TPP2 12.67 0.71


RESULT

Species Value f_code B
50000 0.00 55933 TPP1
**NOX 5.01 55933 TPP1
NOX 6.33 55933 TPP1**
SO2 0.64 55933 TPP1
50000 0.00 55933 TPP2
NOX 12.67 55933 TPP2
SO2 0.71 55933 TPP2
50000 0.07 6002 1
NOX 1151.97 6002 1
SO2 152.98 6002 1
50000 0.07 6002 2
NOX 1550.17 6002 2
SO2 213.31 6002 2


Note: For the bolded part, note that there were two values with the same f_code and b_id in DATA1, but only one value in DATA2. In this situation, I want to replace the values in DATA1 with the weighted value in DATA2

11.33*(0.38/(0.38+0.48)) = 5.01
11.33*(0.48/(0.38+0.48)) = 6.33


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