在datafram中将列表元素拆分为子元素

Question

我有一个数据帧：

Filtered_data

['defence possessed russia china','factors driving china modernise']
['force bolster pentagon','strike capabilities pentagon congress detailing china']
[missiles warheads', 'deterrent face continued advances']
......
......

我只想将每个列表元素拆分为子元素（标记化的单词）

Filtered_data

[defence, possessed,russia,factors,driving,china,modernise]
[force,bolster,strike,capabilities,pentagon,congress,detailing,china]
[missiles,warheads, deterrent,face,continued,advances]

这是我的密码我试过了

for text in df['Filtered_data'].iteritems():
for i in text.split():
    print (i)

Answer 1

您可以使用itertools.chain+^{}。与set相比，toolz.unique的好处是它保持了有序性。你知道吗

from itertools import chain
from toolz import unique

df = pd.DataFrame({'strings': [['defence possessed russia china','factors driving china modernise'],
                               ['force bolster pentagon','strike capabilities pentagon congress detailing china'],
                               ['missiles warheads', 'deterrent face continued advances']]})

df['words'] = df['strings'].apply(lambda x: list(unique(chain.from_iterable(i.split() for i in x))))

print(df.iloc[0]['words'])

['defence', 'possessed', 'russia', 'china', 'factors', 'driving', 'modernise']

Answer 2

将列表理解与split和flatening结合使用：

df['Filtered_data'] = df['Filtered_data'].apply(lambda x: [z for y in x for z in y.split()])
print (df)
                                       Filtered_data
0  [defence, possessed, russia, china, factors, d...
1  [force, bolster, pentagon, strike, capabilitie...
2  [missiles, warheads, deterrent, face, continue...

编辑：

对于唯一值，标准方法是使用sets：

df['Filtered_data'] = df['Filtered_data'].apply(lambda x: list(set([z for y in x for z in y.split()])))
print (df)
                                       Filtered_data
0  [russia, factors, defence, driving, china, mod...
1  [capabilities, detailing, china, force, pentag...
2  [deterrent, advances, face, warheads, missiles...

但如果值的排序很重要，请使用^{}：

df['Filtered_data'] = df['Filtered_data'].apply(lambda x: pd.unique([z for y in x for z in y.split()]).tolist())
print (df)
                                       Filtered_data
0  [defence, possessed, russia, china, factors, d...
1  [force, bolster, pentagon, strike, capabilitie...
2  [missiles, warheads, deterrent, face, continue...