多次元テーブルを持たないパンダの長さ

Question

私はこのような表を持っています。完全なサンプルコードとソースデータは2つの回答を含み、一番下にあります。

  id technology  question response 
0  subj1 technology1  Q1   3 
1  subj1 technology2  Q1   4 
... 
10  subj1 technology3  Q3   6 
11  subj1 technology4  Q3   2 
12  subj1 technology4  Q4   7 
13  subj1 technology3  Q4   5 
14  subj1 technology1  Q4   5 
15  subj1 technology2  Q4   9 
16  subj2 technology2  Q1   1 
17  subj2 technology1  Q1   4 
... 
29  subj2 technology3  Q4   0 

私が欲しいのは「質問」列の異なる値が自分の右の列になるテーブルであり、質問欄セルの値が与えられたために与えられた質問に対する応答値です私は（これまでの提案に基づき、以下の完全なコードで改善したバージョンに注意してください）このようにテーブルを旋回している場合

  id  technology  Q1 Q2 Q3 Q4 
0  subj1 technology1  3 3 2 1 
1  subj1 technology2  4 4 3 1 
... 
10  subj1 technology3  6 3 7 2 
... 
16  subj2 technology2  4 5 7 3 

私は近いものに得ることができます。この（ちょうどイラスト）のように、対象とテクノロジー：

source_data_df_pvt1 = pd.pivot_table(source_data_df, index = ['id'], 
            columns = ['technology', 'question'], 
            values = 'response', aggfunc='first') 

私にこの多次元データフレームを与える

：私は私のデータフレームはしかしMULT次元になりたくない

technology technology1     technology2      technology3    technology4    technology5 
question   Q1 Q2 Q3  Q4   Q1 Q2 Q3 Q4   Q1 Q2 Q3 Q4   Q1 Q2 Q3 Q4   Q1 
id                                  
subj1    3.0 9.0 7.0 5.0   4.0 5.0 3.0 9.0   3.0 8.0 6.0 5.0   5.0 8.0 2.0 7.0   NaN 
subj2    4.0 9.0 8.0 7.0   1.0 5.0 8.0 20.0  20.0 9.0 4.0 0.0   3.0 0.0 8.0 6.0   NaN 
subj3    14.0 NaN 10.0 0.0   NaN NaN NaN NaN   NaN NaN NaN NaN   NaN NaN NaN NaN  15.0 
subj4    13.0 4.0 5.0 11.0  17.0 NaN NaN NaN   0.0 NaN NaN NaN   0.0 NaN NaN NaN   NaN 
subj5    3.0 20.0 4.0 8.0   2.0 20.0 3.0 2.0   3.0 5.0 7.0 5.0   4.0 2.0 7.0 5.0   NaN 
subj6    2.0 8.0 1.0 6.0   0.0 7.0 4.0 1.0  20.0 6.0 1.0 0.0   6.0 8.0 7.0 3.0   NaN 

は、私は、単一のディメンション表が欲しいです。

これはPandasでも可能ですか？

作業溶液を含む完全なサンプルコードと出力、：

import pandas as pd 
import numpy as np 

pd.set_option('display.width', 1000) 

#See https://gist.github.com/NathanDotTo/e506c1946c23234d2c24a2bd27e570a0#file-technology_skills-csv 
sample_data = "../test_data/technology_skills.csv" 

#The sample data has these columns: 
column_names = ["id", "technology", "question", "response"] 

source_data_df = pd.read_csv(sample_data, names = column_names, header = None) 

#Change response to be numeric 
pd.to_numeric(source_data_df.response) 

#Strip white spaces from questions 
source_data_df['question'] = source_data_df['question'].str.strip() 

#Pivot the table to create columns from response values for each question  

source_data_df_pvt1 = pd.pivot_table(source_data_df, index = ['id','technology'], 
           columns = [ 'question'], 
           values = 'response', 
           aggfunc='first', 
           fill_value=np.nan).reset_index() 

print('source_data_df_pvt1 *******************') 
print (source_data_df_pvt1) 

source_data_df_pvt2 = source_data_df.pivot_table(index=['id','technology'], 
           columns='question', 
           values='response', 
           aggfunc='sum', 
           fill_value=np.nan).reset_index() 

print('source_data_df_pvt2 *******************') 
print (source_data_df_pvt2) 

結果は以下の通りです：ボーナスとして

source_data_df_pvt1 ******************* 
question  id technology Q1 Q2 Q3 Q4 
0   subj1 technology1 3 9.0 7.0 5.0 
1   subj1 technology2 4 5.0 3.0 9.0 
2   subj1 technology3 3 8.0 6.0 5.0 
3   subj1 technology4 5 8.0 2.0 7.0 
4   subj2 technology1 4 9.0 8.0 7.0 
5   subj2 technology2 1 5.0 8.0 20.0 
6   subj2 technology3 20 9.0 4.0 0.0 
7   subj2 technology4 3 0.0 8.0 6.0 
8   subj3 technology1 14 NaN 10.0 0.0 
9   subj3 technology5 15 NaN NaN NaN 
10  subj4 technology1 13 4.0 5.0 11.0 
11  subj4 technology2 17 NaN NaN NaN 
12  subj4 technology3 0 NaN NaN NaN 
13  subj4 technology4 0 NaN NaN NaN 
14  subj5 technology1 3 20.0 4.0 8.0 
15  subj5 technology2 2 20.0 3.0 2.0 
16  subj5 technology3 3 5.0 7.0 5.0 
17  subj5 technology4 4 2.0 7.0 5.0 
18  subj6 technology1 2 8.0 1.0 6.0 
19  subj6 technology2 0 7.0 4.0 1.0 
20  subj6 technology3 20 6.0 1.0 0.0 
21  subj6 technology4 6 8.0 7.0 3.0 

source_data_df_pvt2 ******************* 
question  id technology Q1 Q2 Q3 Q4 
0   subj1 technology1 3 9.0 7.0 5.0 
1   subj1 technology2 4 5.0 3.0 9.0 
2   subj1 technology3 3 8.0 6.0 5.0 
3   subj1 technology4 5 8.0 2.0 7.0 
4   subj2 technology1 4 9.0 8.0 7.0 
5   subj2 technology2 1 5.0 8.0 20.0 
6   subj2 technology3 20 9.0 4.0 0.0 
7   subj2 technology4 3 0.0 8.0 6.0 
8   subj3 technology1 14 NaN 10.0 0.0 
9   subj3 technology5 15 NaN NaN NaN 
10  subj4 technology1 13 4.0 5.0 11.0 
11  subj4 technology2 17 NaN NaN NaN 
12  subj4 technology3 0 NaN NaN NaN 
13  subj4 technology4 0 NaN NaN NaN 
14  subj5 technology1 3 20.0 4.0 8.0 
15  subj5 technology2 2 20.0 3.0 2.0 
16  subj5 technology3 3 5.0 7.0 5.0 
17  subj5 technology4 4 2.0 7.0 5.0 
18  subj6 technology1 2 8.0 1.0 6.0 
19  subj6 technology2 0 7.0 4.0 1.0 
20  subj6 technology3 20 6.0 1.0 0.0 
21  subj6 technology4 6 8.0 7.0 3.0 

、これは私が取得できるようにしたかったです、中使用法の簡潔さという用語。両方のピボットスタイルで同じように機能します。 unstackと

for row in source_data_df_pvt1.itertuples(): 
    print(row) 
    print(row.id) 
    print(row.technology) 
    print(row.Q1) 
    print(row.Q2) 
    print(row.Q3) 
    print(row.Q4) 

Answer 1

使用set_index：

df = (source_data_df.set_index(['id','technology','question'])['response'] 
        .unstack(fill_value=0) 
        .reset_index()) 

が、エラーを取得した場合

：

ValueError: Index contains duplicate entries, cannot reshape

それはトリプルid、technology、questionで重複があることを意味しますので、必要に応じて削除の重複やで集合体first：

同じである何

source_data_df = source_data_df.drop_duplicates(['id','technology','question']) 
df = (source_data_df.set_index(['id','technology','question'])['response'] 
        .unstack(fill_value=0) 
        .reset_index()) 

：

df = pd.pivot_table(source_data_df, index = ['id','technology'], 
            columns = [ 'question'], 
            values = 'response', 
            aggfunc='first', 
            fill_value=0).reset_index() 
print (df) 
question  id technology Q1 Q3 Q4 
0   subj1 technology1 3 0 5 
1   subj1 technology2 4 0 9 
2   subj1 technology3 0 6 5 
3   subj1 technology4 0 2 7 
4   subj2 technology1 4 0 0 
5   subj2 technology2 1 0 0 
6   subj2 technology3 0 0 0