Python學習筆記：實驗設計組合(二)

 本文的起源來自於iThelp中網友的提問：

python資料處理請教

程式碼是我自已寫的。

import pandas as pd
import numpy as np

df = pd.DataFrame(
    {
      'GROUP':   ['R','W','W','R','R','W'],
      'GOAT':    ['15/10','10/7/11','9','12','8/13','10'],
      'TYPE':    ['A/B','A/B/B','A','B','B/A','B'],
      'TIME': [2,3,2,2,3,2]
    }
)

df

Out[1]:

	GROUP	GOAT	TYPE	TIME
0	R	15/10	A/B	2
1	W	10/7/11	A/B/B	3
2	W	9	A	2
3	R	12	B	2
4	R	8/13	B/A	3
5	W	10	B	2

In [2]:

df_temp = df.copy()
df_temp['GOAT_sub'] = df_temp['GOAT'].str.split('/').tolist()
df_temp['TYPE_sub'] = df_temp['TYPE'].str.split('/').tolist()

df_temp

Out[2]:

	GROUP	GOAT	TYPE	TIME	GOAT_sub	TYPE_sub
0	R	15/10	A/B	2	[15, 10]	[A, B]
1	W	10/7/11	A/B/B	3	[10, 7, 11]	[A, B, B]
2	W	9	A	2	[9]	[A]
3	R	12	B	2	[12]	[B]
4	R	8/13	B/A	3	[8, 13]	[B, A]
5	W	10	B	2	[10]	[B]

In [5]:

# 設定solution來存最後結果
solution = []
for i in range(len(df_temp)):
    #設定temp_RESULT來存暫時的結果
    temp_RESULT = []
    for j in range(len(df_temp['TYPE_sub'][i])):
        if ((df_temp['GROUP'][i] == 'R') & (df_temp['TYPE_sub'][i][j] == 'A')):
            df_temp['temp'+str(j)] = int(df_temp['GOAT_sub'][i][j])*df_temp['TIME'][i]*0.3
        elif ((df_temp['GROUP'][i] == 'R') & (df_temp['TYPE_sub'][i][j] == 'B')):
            df_temp['temp'+str(j)] = int(df_temp['GOAT_sub'][i][j])*df_temp['TIME'][i]*0.5
        elif ((df_temp['GROUP'][i] == 'W') & (df_temp['TYPE_sub'][i][j] == 'A')):
            df_temp['temp'+str(j)] = int(df_temp['GOAT_sub'][i][j])*df_temp['TIME'][i]*0.4
        elif ((df_temp['GROUP'][i] == 'W') & (df_temp['TYPE_sub'][i][j] == 'B')):
            df_temp['temp'+str(j)] = int(df_temp['GOAT_sub'][i][j])*df_temp['TIME'][i]*0.3
        L = str(df_temp['temp'+str(j)][0])
        temp_RESULT.append(L)
    temp_RESULT = '/'.join(temp_RESULT)
    solution.append(temp_RESULT)

    print(temp_RESULT)

solution

9.0/10.0
12.0/6.3/9.9
7.2
12.0
12.0/11.7
6.0

Out[5]:

['9.0/10.0', '12.0/6.3/9.9', '7.2', '12.0', '12.0/11.7', '6.0']

In [6]:

df['RESULT'] = solution
df

Out[6]:

	GROUP	GOAT	TYPE	TIME	RESULT
0	R	15/10	A/B	2	9.0/10.0
1	W	10/7/11	A/B/B	3	12.0/6.3/9.9
2	W	9	A	2	7.2
3	R	12	B	2	12.0
4	R	8/13	B/A	3	12.0/11.7
5	W	10	B	2	6.0

In [ ]: