我有一个数据框,其中一个列包含英文单词。我希望通过NLTKs synsets()函数传递该列中的每个元素。我的问题是synsets()一次只接受一个单词。
例如wordnet.synsets(‘父亲’)
现在,如果我有这样的数据帧:
dc = {'A':[0,9,4,5],'B':['father','mother','kid','sister']}
df = pd.DataFrame(dc)
df
A B
0 0 father
1 9 mother
2 4 kid
3 5 sister我希望通过synsets()函数传递B列,并让另一列包含它的输出。我希望在不遍历数据帧的情况下完成此操作。
我该怎么做?
发布于 2014-06-10 01:54:49
您可以使用apply方法:
In [4]: df['C'] = df['B'].apply(wordnet.synsets)
In [5]: df
Out[5]:
A B C
0 0 father [Synset('father.n.01'), Synset('forefather.n.0...
1 9 mother [Synset('mother.n.01'), Synset('mother.n.02'),...
2 4 kid [Synset('child.n.01'), Synset('kid.n.02'), Syn...
3 5 sister [Synset('sister.n.01'), Synset('sister.n.02'),...然而,拥有一列列表通常不是一个非常有用的数据结构。将每个同义词放在各自的列中可能会更好。您可以通过使回调函数返回一个pd.Series来完成此操作
In [29]: df.join(df['B'].apply(lambda word: pd.Series([w.name for w in wordnet.synsets(word)])))
Out[29]:
A B 0 1 2 3 \
0 0 father father.n.01 forefather.n.01 father.n.03 church_father.n.01
1 9 mother mother.n.01 mother.n.02 mother.n.03 mother.n.04
2 4 kid child.n.01 kid.n.02 kyd.n.01 child.n.02
3 5 sister sister.n.01 sister.n.02 sister.n.03 baby.n.05
4 5 6 7 8
0 father.n.05 father.n.06 founder.n.02 don.n.03 beget.v.01
1 mother.n.05 mother.v.01 beget.v.01 NaN NaN
2 kid.n.05 pull_the_leg_of.v.01 kid.v.02 NaN NaN
3 NaN NaN NaN NaN NaN (我选择只显示每个Synset的name属性;您当然可以使用
df.join(df['B'].apply(lambda word: pd.Series(wordnet.synsets(word))))如果您想要Synset对象本身。)
https://stackoverflow.com/questions/24125862
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