How to create new row out of calculation across other rows in Pandas Data Frame?

Assuming you are happy to modify the input DataFrame in place, this will produce the exact output you've requested:

df.loc['Perc.'] = (
    df.loc['>48h'] / df.loc['<48h']
).apply(lambda x: f"{x:.0%}")

This assumes that the first column in your printed output is the index of the DataFrame.

To unpack this a bit:

1. The assignment df.loc['Perc.'] = will create a new row populated with the value on the right-hand side of the =.
2. df.loc['>48h'] / df.loc['<48h'] computes the desired ratio, as a float
3. The .apply call calls a function on each element of the new row, to convert the raw percentage number to a string formatted in the manner you requested. To unpack this a little further:
   • f"{x}" converts x to a string (it is essentially equivalent to str(x))
   • f"{x:.0%}" enhances the above by adding the .0% format specifier, which will display the value as a percentage with 0 decimal places (i.e. multiply by 100, round to 0 decimal places, and then add a % sign). There are various other format specifiers available for floating point data.

Note that if you want to do any further processing on the resulting value, I would recommend not doing step 3 - once you've converted your data into a string it obviously can't be e.g. multiplied by other data.

If you want the best of both worlds, keeping the internal value as a float ratio, but displaying them as a percentage, you can do this:

df.loc['Perc.'] = (
    df.loc['>48h'] / df.loc['<48h']
)

display_str = df.T.to_string(formatters={"Perc.": lambda x: f"{x:.0%}"})
print(display_str)

This will modify df to add the Perc. row, but its values will stay as the actual float ratio. Then display_str will be an appropriately-formatted string representation of the whole dataframe, including the percentage.

Note that formatters accepts a dict keyed by column names, not rows, so you must first transpose your data (df.T), the resulting output will be transposed:

   Para  <48h  >48h Perc.
0  1.21  22.0   0.0    0%
1  2.21  25.0   1.0    4%
2  3.21  38.0   3.0    8%

If you're working inside a Jupyter notebook, you can also use the df.style.format method to achieve something similar; calling

df.T.style.format({"Perc.": "{:.0%}"})

will return a pandas.io.formats.style.Styler object, but if you're in a Jupyter notebook, this will be rendered like a DataFrame with the relevant column formatted as above. Again though, this only works on a column-by-column basis, so you'll need to transpose.

Use DataFrame.loc for select rows:

#If first column is not index create it
#df = df.set_index('Para')


df.loc['Perc'] = df.loc['>48h'].div(df.loc['<48h']).mul(100).round()
print (df)
      01.21  02.21  03.21
Para                     
<48h   22.0   25.0   38.0
>48h    0.0    1.0    3.0
Perc    0.0    4.0    8.0

Better is transpose first, so is possible select by columns:

#If first column is not index create it
#df = df.set_index('Para')

df = df.T

df['Perc'] = df['>48h'].div(df['<48h']).mul(100).round()
print (df)
Para   <48h  >48h  Perc
01.21    22     0   0.0
02.21    25     1   4.0
03.21    38     3   8.0

para = ['01.21', '02.21', '03.21']
a = np.array(([22, 25, 38]))
b = np.array(([0, 1, 3]))
df = pd.DataFrame([para, a, b], index=['Para', '<48h', '>48h'], columns=['col0', 'col1', 'col2'])

perc = np.array(((b/a)*100))

df2 = pd.DataFrame([perc], index=['Perc.'], columns=['col0', 'col1', 'col2'])
df = df.append(df2)
print(df)

Output:

        col0   col1     col2
Para   01.21  02.21    03.21
<48h      22     25       38
>48h       0      1        3
Perc.      0      4  7.89474

It's not best practice to store data like this, read Hadley's tidy data paper.

But in a line you could do it thusly.

df.loc['perc'] = df.iloc[2]/df.iloc[1]

df.loc[key] = row

this is how we create new row in pandas data frame