Python 在dataframe中将列表元素拆分为子元素

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）。因此，我要查找的输出为：-

Filtered_data

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

这是我的代码，我已经试过了

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

---

将列表理解与

split

和flatening一起使用：

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

set

s:

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...

但如果值的顺序很重要，请使用：

---

您可以使用

itertools.chain

+。

toolz.unique

与

set

相比的优点是它保留了顺序

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']

---

为什么投反对票？我是python新手。对不起，如果在这里问一个愚蠢的问题，那么反对票不是因为这个问题愚蠢（事实并非如此），而是因为。我们必须猜测您的数据结构，这使问题变得模棱两可。另一个原因是您需要将代码添加到问题中，您可以尝试…@James-仅添加

set

类似

list（set（[z代表y in x代表z in y.split（）]））

需要您的帮助：-

https://stackoverflow.com/questions/51574485/match-keywords-in-pandas-column-with-another-list-of-elements

。我没有提到解决方案

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']