更新

使用循环并改变列表：

scores = df.set_index('Name')['Score']

for l in df['marks']:
    for i, t in enumerate(l):
        l[i] = t+(scores.get(t[0]),)

输出：

Name  Age  Score                             marks
0   Nik   33     90  [(Nik, 100, 90), (Naik, 85, 92)]
1  Kate   34     95                 [(Kate, 100, 95)]
2  Evan   43     93                 [(Evan, 100, 93)]
3  Naik   44     92  [(Nik, 77, 90), (Naik, 100, 92)]

以前的回答

我将使用zip的列表解析和next的生成器：

df['first'] = [next((x for x in m if x[0]==n), None)
               for n, m in zip(df['Name'], df['marks'])]

输出：

Name  Age  Score                     marks        first
0   Nik   33     90  [(Nik, 100), (Naik, 85)]   (Nik, 100)
1  Kate   34     95             [(Kate, 100)]  (Kate, 100)
2  Evan   43     93             [(Evan, 100)]  (Evan, 100)
3  Naik   44     92  [(Nik, 77), (Naik, 100)]  (Naik, 100)

如果你想要号码：

df['first'] = [next((x[1] for x in m if x[0]==n), None)
               for n, m in zip(df['Name'], df['marks'])]

输出：

Name  Age  Score                     marks  first
0   Nik   33     90  [(Nik, 100), (Naik, 85)]    100
1  Kate   34     95             [(Kate, 100)]    100
2  Evan   43     93             [(Evan, 100)]    100
3  Naik   44     92  [(Nik, 77), (Naik, 100)]    100

或者使用纯pandas（可能效率较低）：

df['first'] = (df
   .explode('marks')
   .loc[lambda d: d['marks'].str[0].eq(d['Name'])]
   ['marks'].str[1]
   .groupby(level=0).first()
)