Question

我想使用列（groupby No来保存我的数据，并将列date1和results的每个结果保存在不同的列中。

以下是具有相应预期输出的输入示例：

我添加了更多数据。并且有很多数据。

Answer 1

这是一种实现方法：

from datetime import datetime

df = pd.DataFrame({'No.' : ['s1', 's2', 's2'], 'date_1' : [datetime.now() for x in range(3)],
                  'results' : [1.2, 9.73, 3.71]})

# Use groupby to get the lists of dates and result
result = df.groupby('No.')[['date_1', 'results']].agg({'date_1' : list, 'results' : list})
# if you are running a pandas version <0.24.2 uncomment the following line and comment the one above
#result = df.groupby('No.')[['date_1', 'results']].agg({'date_1' : lambda x: list(x), 'results' : lambda x: list(x)})

# Look at the number of columns we will have to create
len_max = np.max([len(x) for x in result['results']])

# Create all the required columns  
for i in range(1,len_max):
    result['date__{}'.format(i+1)] = [x[i] if len(x)>i else 0 for x in result['date_1']]
    result['results_{}'.format(i+1)] = [x[i] if len(x)>i else 0 for x in result['results']]

# Modify the first  two columns that still contain the lists of the groupby
result['date_1'] = [x[0] for x in result['date_1']]
result['results'] = [x[0] for x in result['results']]

输出：

                        date_1  results                     date__2  results_2
No.                                                                           
s1  2019-07-29 08:00:45.878494     1.20                           0       0.00
s2  2019-07-29 08:00:45.878499     9.73  2019-07-29 08:00:45.878500       3.71

Answer 2

基于vlemaistre的答案-您可以以更紧凑的方式做到这一点：

import pandas as pd
import numpy as np
from datetime import datetime, timedelta
df = pd.DataFrame({'No.' : ['s1', 's2', 's2'], 'date' : [datetime.now()+timedelta(days=x) for x in range(3)],
                  'results' : [1.2, 9.73, 3.71]})

joint_df = df.groupby('No.')[['date', 'results']].agg(lambda x: list(x))
result = pd.DataFrame(index=joint_df.index)
for column in df.columns.difference({'No.'}):
    result = result.join(pd.DataFrame.from_records(
        list(joint_df[column]), index=joint_df.index).rename(lambda x: column+str(x+1), axis=1), how='outer')

输出为：

    date1                       date2                       results1    results2
No.             
s1  2019-07-29 12:58:28.627950  NaT                         1.20        NaN
s2  2019-07-30 12:58:28.627957  2019-07-31 12:58:28.627960  9.73        3.71

熊猫-如何将行数据转换为列

2 个答案: