Question

以下是我必须执行的任务的代码段。我想为这项任务提供最好的时间复杂性。是否有图书馆或更好的实施？

import pandas as pd
import numpy as np

df1 = pd.DataFrame({"v_id": [1,1,1,2,2,3,3,3,3,5,5], \
                    "date": ['01-01-2017', '02-01-2017', '03-01-2017',\
                    '02-01-2017', '03-01-2017',
                    '01-01-2017', '02-01-2017', '03-01-2017', '04-01-2017',\
                    '05-01-2017','04-01-2017'],\
                    "value":[0.9,1.5,2.4,7.1,0.4,1.8,5.1, 6.4, 7.7, 3.9, 0.3]})

dim1, dim2 = df1.v_id.unique(), df1.date.unique()
dim1_dict, dim2_dict = dict(zip(dim1, list(range(0, len(dim1))))), \
                        dict(zip(dim2, list(range(0, len(dim2)))))

value_result, date_result = np.empty((len(dim1), len(dim2)), dtype=object),\
                            np.empty((len(dim1), len(dim2)), dtype=object)


for i in range(0, len(df1)):
    date_result[dim1_dict.get(df1.loc[i].v_id), \
    dim2_dict.get(df1.loc[i].date)] = df1.loc[i].date
    value_result[dim1_dict.get(df1.loc[i].v_id), \
    dim2_dict.get(df1.loc[i].date)] = df1.loc[i].value

数据框 df1 如下所示：

我的目标是获得 date_result （v_id x 日期）

array([['01-01-2017', '02-01-2017', '03-01-2017', None, None],
       [None, '02-01-2017', '03-01-2017', None, None],
       ['01-01-2017', '02-01-2017', '03-01-2017', '04-01-2017', None],
       [None, None, None, '04-01-2017', '05-01-2017']], dtype=object)

＆安培; value_result 矩阵（v_id x 值）。

array([[0.90000000000000002, 1.5, 2.3999999999999999, None, None],
       [None, 7.0999999999999996, 0.40000000000000002, None, None],
       [1.8, 5.0999999999999996, 6.4000000000000004, 7.7000000000000002,
        None],
       [None, None, None, 0.29999999999999999, 3.8999999999999999]], dtype=object)

Answer 1

您可以使用pivot_table执行此计算：

import numpy as np
import pandas as pd
df1 = pd.DataFrame({"v_id": [1,1,1,2,2,3,3,3,3,5,5], 
                    "date": ['01-01-2017', '02-01-2017', '03-01-2017',
                    '02-01-2017', '03-01-2017',
                    '01-01-2017', '02-01-2017', '03-01-2017', '04-01-2017',
                    '05-01-2017','04-01-2017'],
                    "value":[0.9,1.5,2.4,7.1,0.4,1.8,5.1, 6.4, 7.7, 3.9, 0.3]})

date_result = (df1.assign(date2=df1['date'])
                  .pivot_table(columns='date', index='v_id', 
                               values='date2', aggfunc='first').values)
value_result = df1.pivot_table(columns='date', index='v_id', 
                               values='value', aggfunc='first').values

print(date_result)
print(value_result)

产生date_result：

array([['01-01-2017', '02-01-2017', '03-01-2017', None, None],
       [None, '02-01-2017', '03-01-2017', None, None],
       ['01-01-2017', '02-01-2017', '03-01-2017', '04-01-2017', None],
       [None, None, None, '04-01-2017', '05-01-2017']], dtype=object)

和<{p>的value_result

array([[ 0.9,  1.5,  2.4,  nan,  nan],
       [ nan,  7.1,  0.4,  nan,  nan],
       [ 1.8,  5.1,  6.4,  7.7,  nan],
       [ nan,  nan,  nan,  0.3,  3.9]])

请注意，value_result是具有浮点dtype的NumPy数组，缺少的值由nan表示，而不是None。您可以使用

将其转换为object dtype且缺少值None的NumPy数组

value_result = np.where(pd.isnull(value_result), None, value_result)

产生

array([[0.9, 1.5, 2.4, None, None],
       [None, 7.1, 0.4, None, None],
       [1.8, 5.1, 6.4, 7.7, None],
       [None, None, None, 0.3, 3.9]], dtype=object)

Python中的数据转换。还有更好的方法吗？

1 个答案: