Question

我有一个数据框为：-

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（标记词）。因此，输出Im寻找为：-

Filtered_data

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

这是我尝试的代码

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

Answer 1

将列表理解与split一起使用并展宽：

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set：

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...

但是如果值的排序很重要，请使用pandas.unique：

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...

Answer 2

您可以使用itertools.chain + toolz.unique。 toolz.unique和set的好处是保留了排序。

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']

在pandas数据框中将列表元素拆分为子元素

2 个答案: