快速简单的代码修复方法是使用numpy.select：

df['Label'] = np.select([df['Description'].str.contains('|'.join(list1)),
                         df['Description'].str.contains('|'.join(list2))],
                        ['weather', 'equipment'], 'Other')

但是，你可以做得更好。
您可以使用自动构建的正则表达式和Map字典：

import re

d = {'weather': ['wind','air'], 'equipment': ['crane','machine']}

# reverse d
d2 = {k: v for v,l in d.items() for k in l}
# {'wind': 'weather', 'air': 'weather',
#  'crane': 'equipment', 'machine': 'equipment'}

pattern = f"({'|'.join(map(re.escape, sorted(d2, key=len, reverse=True)))})"
# '(machine|crane|wind|air)'

df['Label'] = (df['Description'].str.extract(pattern, expand=False)
               .map(d2).fillna('Other')
               )

如果每个句子可以有多个匹配项：

df['Label'] = (df['Description'].str.extractall(pattern)[0]
               .groupby(level=0).agg(lambda x: ','.join(x.drop_duplicates()))
              )

输出：

Description      Label
0           There was a heavy wind due to cyclone.    weather
1                Pollution hamper the air quality.    weather
2    The machine failure was due to short circuit.  equipment
3             The game was called off due to wind.    weather
4               Players played the game very well.      Other
5  the crane operator took the crane to wrong side  equipment

它似乎是这样工作的：

df["Label"] = "Other"
df["Label"][df["Description"].str.contains('|'.join(list1))] = "weather"
df["Label"][df["Description"].str.contains('|'.join(list2))] = "equipment"

我将使用.apply()来实现这一点，您可以对填充label列所使用的逻辑进行更多控制：

df['label'] = df['Description'].apply(lambda x: 'weather' if any(word in x for word in list1) else ('equipment' if any(word in x for word in list2) else 'other'))