python - pandas 应用返回 NaN

Question

我有一个 json，我将其转换为字典，然后使用字典中存在的某些键值对创建一个数据框

# json
a = """{
    "cluster_id": 3,
    "cluster_observation_data": [[1, 2, 3, 4, 5, 6, 7, 8], [2, 3, 4, 5, 6, 7, 8, 1]],
    "cluster_observation_label": [0, 1],
    "cluster_centroid": [1, 2, 3, 4, 5, 6, 7, 10],
    "observation_id":["id_xyz_999","id_abc_000"]
}"""

# convert to dictionary
data = json.loads(a)
sub_dict = dict((k, data[k]) for k in ('cluster_observation_data', 'cluster_observation_label'))
train = pd.DataFrame.from_dict(sub_dict, orient='columns')

将其转换为 ddataframe 后，我尝试计算其与 cluster_centroid 的欧几里得距离存在于 data字典。该函数工作正常，但在最后 train我得到 NaN 的数据框

def distance_from_center(row):
    centre = data['cluster_centroid']
    obs_data = row[0]
    print('obs_data', obs_data)
    print('\n\n\n\n')
    print('center', centre)
    # print(type(obs_data))
    # print(type(centre))
    dist = sum([(a - b)**2 for a, b in zip(centre, obs_data)])
    print(dist)
    return dist

train.loc[:, 'center_dist'] = train.loc[:, ['cluster_observation_data']].apply(distance_from_center)

我不知道我哪里出了问题。即使是一个小提示也可以。

Answer 1

您需要传递轴，例如:

train.loc[:, 'center_dist'] = train.loc[:, ['cluster_observation_data']].apply(distance_from_center, 1)

原因是您想将函数单独应用于每个列表。 Documentation说:

1 or ‘columns’: apply function to each row