使用groupby.rank和map：

labels = ['First', 'Second', 'Third', 'Fourth', 'Fifth']
df['RESULT'] = (df.groupby('DURATION')['COEFF']
                  .rank('dense', ascending=False).sub(1)
                  .map(dict(enumerate(labels)))
               )

输出量：

DURATION  CLUSTER  COEFF  RESULT
0          3        0  0.340  Second
1          3        1 -0.005  Fourth
2          3        2  1.000   First
3          3        3  0.330   Third
4          4        0 -0.020   Third
5          4        1 -0.280  Fourth
6          4        2  0.220  Second
7          4        3  0.480   First
8          5        0  0.650   First
9          5        1 -0.260  Fourth
10         5        2  0.100   Third
11         5        3  0.150  Second

基于https://stackoverflow.com/a/74547858/7237062出色的答案（* 我自己也不会这么快找到这个答案 *），我建议使用这个Ordinal numbers replacement来完全自动化这个过程。

import pandas as pd
# see answer https://stackoverflow.com/a/20007730/7237062, others exist
# code golfed version of an "ordinal" function (int -> ordinal string in english)
ordinal = lambda n: "%d%s" % (n,"tsnrhtdd"[(n//10%10!=1)*(n%10<4)*n%10::4])
# copy pasta of OP input data
df = pd.read_clipboard()  # let pandas read the clipboard
df["RESULT"] = (df.groupby('DURATION')['COEFF']
                  .rank('dense', ascending=False)
                  .sub(1) # mozway's answer so far !
                  .astype(int)
                  + 1 # +1 so ordinals start at 1 (instead of 0)
                  ).apply(ordinal)

结果：

DURATION  CLUSTER  COEFF RESULT
0          3        0  0.340    2nd
1          3        1 -0.005    4th
2          3        2  1.000    1st
3          3        3  0.330    3rd
4          4        0 -0.020    3rd
5          4        1 -0.280    4th
6          4        2  0.220    2nd
7          4        3  0.480    1st
8          5        0  0.650    1st
9          5        1 -0.260    4th
10         5        2  0.100    3rd
11         5        3  0.150    2nd