我有一套id有一些category。但是，我希望每個id都具有相同數量的category可以指定為df.id.category.unique().
例如：
Input
df1 = {"id": [1,1,1,2,2,3,3,3,3],
"category": ["a","b","e","a","d","a","b","c","d"]
}
output1 = pd.DataFrame(df1)
output1
Out[57]:
id category
0 1 a
1 1 b
2 1 e
3 2 a
4 2 d
5 3 a
6 3 b
7 3 c
8 3 d
輸出應該是：
Output
df2 = {"id": [1,1,1,1,1,2,2,2,2,2,3,3,3,3,3],
"category": sum([["a","b","c","d","e"] for _ in range(3)], [])}
output2 = pd.DataFrame(df2)
output2
Out[58]:
id category
0 1 a
1 1 b
2 1 c
3 1 d
4 1 e
5 2 a
6 2 b
7 2 c
8 2 d
9 2 e
10 3 a
11 3 b
12 3 c
13 3 d
14 3 e
如果可能，我希望有快速的優化。非常感謝！
uj5u.com熱心網友回復：
您可以按如下方式numpy.tile與numpy.repeat此一起使用
import numpy as np
id_col = np.repeat([1,2,3,4,5],5).reshape(-1,1)
category_col = np.tile(["a","b","c","d","e"],5).reshape(-1,1)
arr = np.hstack([id_col,category_col])
print(arr)
輸出
[['1' 'a']
['1' 'b']
['1' 'c']
['1' 'd']
['1' 'e']
['2' 'a']
['2' 'b']
['2' 'c']
['2' 'd']
['2' 'e']
['3' 'a']
['3' 'b']
['3' 'c']
['3' 'd']
['3' 'e']
['4' 'a']
['4' 'b']
['4' 'c']
['4' 'd']
['4' 'e']
['5' 'a']
['5' 'b']
['5' 'c']
['5' 'd']
['5' 'e']]
uj5u.com熱心網友回復：
一種選擇是pyjanitor的完整功能，以公開丟失的行：
#pip install git https://github.com/pyjanitor-devs/pyjanitor.git
import pandas as pd
import janitor as jn
output1.complete('id', 'category')
Out[1280]:
id category
0 1 a
1 1 b
2 1 e
3 2 a
4 2 d
5 3 a
6 3 b
7 3 c
8 3 d
9 1 c
10 1 d
11 2 b
12 2 c
13 2 e
14 3 e
uj5u.com熱心網友回復：
使用itertools.product：
from itertools import product
df = pd.DataFrame(product(output1['id'].unique(), output1['category'].unique()),
columns=['id','category'])
print (df)
id category
0 1 a
1 1 b
2 1 e
3 1 d
4 1 c
5 2 a
6 2 b
7 2 e
8 2 d
9 2 c
10 3 a
11 3 b
12 3 e
13 3 d
14 3 c
或MultiIndex.from_product與MultiIndex.to_frame：
df = (pd.MultiIndex.from_product([output1['id'].unique(), output1['category'].unique()],
names=['id','category'])
.to_frame(index=False))
print (df)
id category
0 1 a
1 1 b
2 1 e
3 1 d
4 1 c
5 2 a
6 2 b
7 2 e
8 2 d
9 2 c
10 3 a
11 3 b
12 3 e
13 3 d
14 3 c
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