Pandas 为每个单元格添加行，其中 function 应用于该特定单元格，而新行的其他元素等于起始行

Question

I want to loop through each element of a pandas dataframe row such that only that element is stressed (ie: it's multiplied by 10%) while the other elements of the row are kept equal.

I'm planning to use this for sensitivity analysis.

Example: df = pd.DataFrame({'AGE':[5,10],'POP':[100,200]})

AGE	POP
5	100
10	200

Final desired output:

AGE	POP
5	100
10	200
5*1.1	100
5	100*1.1
10*1.1	200
10	200*1.1

Answer 1

You can use a cross merge andconcat :

pd.concat([df, 
           (df.merge(pd.Series([1.1, 1], name='factor'), how='cross')
              .pipe(lambda d: d.mul(d.pop('factor'), axis=0))
            )], ignore_index=True)

Output:

    AGE    POP
0   5.0  100.0
1  10.0  200.0
2   5.5  110.0
3   5.0  100.0
4  11.0  220.0
5  10.0  200.0

Answer 2

If you have 2 columns, you can multiply with the [1, stress] and its reverse those columns, concatanate them while sorting to preserve multiplied column order. Lastly, prepend the original frame as well:

stress = 1.1
factor = [stress, 1]
pd.concat([df,
           pd.concat([df.mul(factor),
                      df.mul(factor[::-1])]).sort_index()
          ], ignore_index=True)

    AGE    POP
0   5.0  100.0
1  10.0  200.0
2   5.5  100.0
3   5.0  110.0
4  11.0  200.0
5  10.0  220.0

---

Generalizing to N columns could be via a comprehension:

def gen_factors(stress, N):
    for j in range(N):
        # make all-1s list, except j'th is `stress`
        f = [1] * N
        f[j] = stress
        yield f

stress = 1.1
N = len(df.columns)
pd.concat([df,
           pd.concat(df.mul(factor)
                     for factor in gen_factors(stress, N)).sort_index()
          ], ignore_index=True)

Example run for a 3-column frame:

>>> df

   AGE  POP  OTHER
0    5  100      7
1   10  200      8

>>> # output of above:

    AGE    POP  OTHER
0   5.0  100.0    7.0
1  10.0  200.0    8.0
2   5.5  100.0    7.0
3   5.0  110.0    7.0
4   5.0  100.0    7.7
5  11.0  200.0    8.0
6  10.0  220.0    8.0
7  10.0  200.0    8.8