Question

我想从字典中删除NaN。

my_dict = {'House': ['has keys',
  'check lights',
  nan,
  nan,
  nan],
 'The Office': ['reading',
  nan,
  nan,
  nan,
  'coffee breaks']}

我相信NaN是浮点数，而不是字符串。我尝试过：

import math

my_dict['House'] = [
    x
    for x in dict2['House']
    if not (isinstance(x, float) and math.isnan(x))
]

然后我得到：

my_dict = {'House': ['has keys',
  'check lights',],
 'The Office': ['reading',
  nan,
  nan,
  nan,
  'coffee breaks']}

我希望它看起来像下面的样子，但是我不知道如何让我的for循环遍历所有键，而不仅仅是House：

my_dict = {'House': ['has keys',
  'check lights'],
 'The Office': ['reading',
  'coffee breaks']}

Answer 1

这应该起作用，它将过滤字典中的所有值，并删除NaN数字：

{ k: [x for x in v if not isinstance(x, float) or not math.isnan(x)] for k, v in my_dict.items() }

这是结果：

{'House': ['has keys', 'check lights'],
 'The Office': ['reading', 'coffee breaks']}

Answer 2

自从标记了const params = new HttpParams().set("observe", "events");以来，您可以执行以下操作：

pandas

print(df)

          House     The Office
0      has keys        reading
1  check lights            NaN
2           NaN            NaN
3           NaN            NaN
4           NaN  coffee breaks

df_new=df.apply(lambda x: pd.Series(x.dropna().values))
print(df_new)

Answer 3

您处在正确的轨道上，但是只选中“ house”，您需要将逻辑应用于所有按键：

import math
for tv_show in my_dict:
    my_dict[tv_show] = [
        x
        for x in dict2[tv_show]
        if not (isinstance(x, float) and math.isnan(x))
    ]

Answer 4

您也可以采用其他方法，只保留字符串值（当然，只要您只需要字符串值）：

>>> for k, v in my_dict.items():
>>>     my_dict[k] = [val for val in v if isinstance(val, str)]

>>> my_dict
# {'House': ['has keys', 'check lights'], 'The Office': ['reading', 'coffee breaks']}

即使这也可以在dictionary comprehension中实现，但我认为在这种情况下，单线逻辑有点密集，但是，它看起来像这样：

>>> {k: [val for val in v if isinstance(val, str)] for k, v in my_dict.items()}

从字典中删除NaN-Python

4 个答案: