Question

我有一个json，我正在转换为字典，然后我使用字典中存在的某些键值对创建数据框

# 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，我试图计算其与cluster_centroid字典中存在的data的欧氏距离。该函数工作正常，但在最终的train数据框中，我得到了NaNs

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)

我无法确定我出错的地方。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)

原因是您希望将功能应用于每个列表。 Documentation说：

1或'columns'：将函数应用于每一行

Answer 2

只需将 distance_from_center（）中 obs_data 的值从行[0] 更改为行在调用该方法时已经占用了该特定列。然后它应该工作得很好。我尝试了它，它在我的系统中工作。

import json
import pandas as pd
# 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')

def distance_from_center(row):
    centre = data['cluster_centroid']
    obs_data = row
    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)

输出：

obs_data [1, 2, 3, 4, 5, 6, 7, 8]
center [1, 2, 3, 4, 5, 6, 7, 10]
4

obs_data [2, 3, 4, 5, 6, 7, 8, 1]
center [1, 2, 3, 4, 5, 6, 7, 10]
88

大熊猫适用于返回NaN

2 个答案: