我有一个数据表,
DT_X = dt.Frame({'variety': ['Caturra',
'Bourbon',
'Typica',
'Catuai',
'Hawaiian Kona',
'Yellow Bourbon',
'Mundo Novo',
'Catimor',
'SL14',
'SL28',
'Pacas',
'Gesha',
'Pacamara',
'SL34',
'Arusha',
'Peaberry',
'Mandheling',
'Sumatra',
'Blue Mountain',
'Ethiopian Yirgacheffe',
'Java',
'Ruiru 11',
'Ethiopian Heirlooms',
'Marigojipe',
'Moka Peaberry',
'Pache Comun',
'Sulawesi',
'Sumatra Lintong'],
'count': [256,
226,
211,
74,
44,
35,
33,
20,
17,
15,
13,
12,
8,
8,
6,
5,
3,
3,
2,
2,
2,
2,
1,
1,
1,
1,
1,
1]})它可以被看作,
Out[8]:
| variety count
-- + --------------------- -----
0 | Caturra 256
1 | Bourbon 226
2 | Typica 211
3 | Catuai 74
4 | Hawaiian Kona 44
5 | Yellow Bourbon 35
6 | Mundo Novo 33
7 | Catimor 20
8 | SL14 17
9 | SL28 15
10 | Pacas 13
11 | Gesha 12
12 | Pacamara 8
13 | SL34 8
14 | Arusha 6
15 | Peaberry 5
16 | Mandheling 3
17 | Sumatra 3
18 | Blue Mountain 2
19 | Ethiopian Yirgacheffe 2
20 | Java 2
21 | Ruiru 11 2
22 | Ethiopian Heirlooms 1
23 | Marigojipe 1
24 | Moka Peaberry 1
25 | Pache Comun 1
26 | Sulawesi 1
27 | Sumatra Lintong 1现在,我想用前4级“Caturra”、“Bourbon”、“Typica”、“Catuai”和“Catuai”来填写品种栏,其余的级别都应该作为其他级别来处理。
预期产出如下:
Out[9]:
| variety count
-- + ------- -----
0 | Caturra 256
1 | Bourbon 226
2 | Typica 211
3 | Catuai 74
4 | Others 236
[5 rows x 2 columns]案例2:
我有一个数据表,
DT_X_1 = dt.Frame({'variety': ['Bourbon',
'Catimor',
'Ethiopian Yirgacheffe',
'Caturra',
'Bourbon',
'SL14',
'Caturra',
'Sumatra',
'Bourbon',
'Caturra',
'SL34',
'Hawaiian Kona',
'Caturra',
'Yellow Bourbon',
'Yellow Bourbon',
'Bourbon',
'SL28',
'Bourbon',
'Caturra',
'SL28',
'Bourbon',
'SL14',
'Caturra',
'Gesha',
'Bourbon',
'Catuai',
'Caturra',
'Bourbon',
'Bourbon',
'Hawaiian Kona']})它可以被看作是
Out[7]:
| variety
-- + ---------------------
0 | Bourbon
1 | Catimor
2 | Ethiopian Yirgacheffe
3 | Caturra
4 | Bourbon
5 | SL14
6 | Caturra
7 | Sumatra
8 | Bourbon
9 | Caturra
10 | SL34
11 | Hawaiian Kona
12 | Caturra
13 | Yellow Bourbon
14 | Yellow Bourbon
15 | Bourbon
16 | SL28
17 | Bourbon
18 | Caturra
19 | SL28
20 | Bourbon
21 | SL14
22 | Caturra
23 | Gesha
24 | Bourbon
25 | Catuai
26 | Caturra
27 | Bourbon
28 | Bourbon
29 | Hawaiian Kona
[30 rows x 1 column]Out[8]:
| variety count
-- + --------------------- -----
0 | Bourbon 9
1 | Catimor 1
2 | Catuai 1
3 | Caturra 7
4 | Ethiopian Yirgacheffe 1
5 | Gesha 1
6 | Hawaiian Kona 2
7 | SL14 2
8 | SL28 2
9 | SL34 1
10 | Sumatra 1
11 | Yellow Bourbon 2
[12 rows x 2 columns]在这里,我想将字段多样性级别从12级降到2级,这是最常见的级别。
预期的产出是,
Out[13]:
| variety
-- + -------
0 | Bourbon
1 | Others
2 | Others
3 | Caturra
4 | Bourbon
5 | Others
6 | Caturra
7 | Others
8 | Bourbon
9 | Caturra
10 | Others
11 | Others
12 | Caturra
13 | Others
14 | Others
15 | Bourbon
16 | Others
17 | Bourbon
18 | Caturra
19 | Others
20 | Bourbon
21 | Others
22 | Caturra
23 | Others
24 | Bourbon
25 | Others
26 | Caturra
27 | Bourbon
28 | Bourbon
29 | Others
[30 rows x 1 column]发布于 2020-07-09 19:49:43
一种方法是首先用字符串"Other“替换从第4开始的所有variety值,然后按variety分组。
>>> DT_X[4:, f.variety] = "Other"
>>> DT_X = DT_X[:, sum(f.count), by(f.variety)]
| variety count
-- + ------- -----
0 | Bourbon 226
1 | Catuai 74
2 | Caturra 256
3 | Other 236
4 | Typica 211
[5 rows x 2 columns]另一种可能是获取原始表,逐行将其拆分为2部分,折叠第二部分并重新绑定回原始表:
>>> dt.rbind(DT_X[:4, :],
dt.Frame(variety=["Other"], count=[DT_X[4:, f.count].sum1()]))
| variety count
-- + ------- -----
0 | Caturra 256
1 | Bourbon 226
2 | Typica 211
3 | Catuai 74
4 | Other 236
[5 rows x 2 columns]案例2
您已经按品种创建了计数表,因此现在只需按计数对其进行排序,并选择2个最常见的品种:
>>> from datatable import by, sort, count, join, update, f, g
>>> counts = DT_X_1[:, count(), by(f.variety)]
>>> frequent = counts[-2:, :, sort(f.count)]
>>> frequent
| variety count
-- + ------- -----
0 | Caturra 7
1 | Bourbon 9
[2 rows x 2 columns](或者,您可以按计数值进行筛选)。
现在,下一步是将此表加入到原始表中,这样我们就有了哪些值是“频繁”的指示符。可以将联接操作与更新结合起来,以便在同一操作中将所有在联接期间不匹配的字段设置为"others"。
>>> frequent.key = "variety"
>>> DT_X_1[g.variety==None, update(variety="others"), join(frequent)]
>>> DT_X_1
| variety
-- + -------
0 | Bourbon
1 | others
2 | others
3 | Caturra
4 | Bourbon
5 | others
6 | Caturra
7 | others
8 | Bourbon
9 | Caturra
10 | others
11 | others
12 | Caturra
13 | others
14 | others
15 | Bourbon
16 | others
17 | Bourbon
18 | Caturra
19 | others
20 | Bourbon
21 | others
22 | Caturra
23 | others
24 | Bourbon
25 | others
26 | Caturra
27 | Bourbon
28 | Bourbon
29 | others
[30 rows x 1 column]https://stackoverflow.com/questions/62818990
复制相似问题