Python Pandas：如果字符串值列表== [无]，则从数据框中删除行

Question

我的数据框中有一列包含值列表。

 Tags
 [marvel, comics, comic, books, nerdy]
 [new, snapchat, version, snap, inc]
 [none]
 [new, york, times, ny, times, nyt, times]
 [today, show, today, show, today]
 [none]
 [mark, wahlberg, marky, mark]

我不知道如何从数据框中删除此 [none] 列表。我试过了，

 us_videos = us_videos.drop(us_videos.index[us_videos.tags == 'none'])

但这仅在我将列转换为字符串时才有效。如何做到这一点？

Answer 1

新答案

OP 想从子列表中删除 'none' 并删除仅包含 'none' 的行

us_videos.tags.explode().pipe(lambda s: s[s != 'none']).groupby(level=0).agg(list)

0        [marvel, comics, comic, books, nerdy]
1          [new, snapchat, version, snap, inc]
3    [new, york, times, ny, times, nyt, times]
4            [today, show, today, show, today]
6                [mark, wahlberg, marky, mark]
Name: tags, dtype: object

更pythonic的方式

dat = {}
for k, v in us_videos.tags.iteritems():
    for x in v:
        if x != 'none':
            dat.setdefault(k, []).append(x)

pd.Series(dat, name='tags')

0        [marvel, comics, comic, books, nerdy]
1          [new, snapchat, version, snap, inc]
3    [new, york, times, ny, times, nyt, times]
4            [today, show, today, show, today]
6                [mark, wahlberg, marky, mark]
Name: tags, dtype: object

在理解中使用赋值表达式

pd.Series({
    k: X for k, v in us_videos.tags.iteritems()
    if (X:=[*filter('none'.__ne__, v)])
}, name='tags')

0        [marvel, comics, comic, books, nerdy]
1          [new, snapchat, version, snap, inc]
3    [new, york, times, ny, times, nyt, times]
4            [today, show, today, show, today]
6                [mark, wahlberg, marky, mark]
Name: tags, dtype: object

---

旧答案

explode

us_videos[us_videos.tags.explode().ne('none').any(level=0)]

                                        tags
0      [marvel, comics, comic, books, nerdy]
1        [new, snapchat, version, snap, inc]
3  [new, york, times, ny, times, nyt, times]
4          [today, show, today, show, today]
6              [mark, wahlberg, marky, mark]

---

list.__ne__

us_videos[us_videos.tags.map(['none'].__ne__)]

                                        tags
0      [marvel, comics, comic, books, nerdy]
1        [new, snapchat, version, snap, inc]
3  [new, york, times, ny, times, nyt, times]
4          [today, show, today, show, today]
6              [mark, wahlberg, marky, mark]

Answer 2

首先让我们编写一个函数来去除列表中的'none'：

print(df)

    tags
0   [marvel, comics, comic, books, nerdy]
1   [new, snapchat, version, snap, inc]
2   [none]
3   [new, york, times, ny, times, nyt, times]
4   [today, show, today, show, today, none]


def delete_none(element):
    new = []
    for val in element:
        if val != 'none':
            new.append(val)
    if len(new) == 0:
        return np.nan
    else:
        return new

现在我们在tags 列上应用这个函数：

df.tags.apply(delete_none)

输出：

0         [marvel, comics, comic, books, nerdy]
1           [new, snapchat, version, snap, inc]
2                                           NaN
3    [new, york, times, ny, times, nyt,  times]
4             [today, show, today, show, today]